Fundamentals of
thermodynamics
Sixth Edition
FUNDAMENTALS OF
THERMODYNAMICS
Sixth edition
Richard E. Sonntag
Claus borgnakke
University of Michigan
GORDON J. VAN WYLEN
Hope College (emeritus)
WILEY
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Richard E. Sonntag, Claus Borgnakke, and Gordon J. Van Wylea
Fundamentals of Thermodynamics, sixth edition
ISBN 0-471-15232-3
Printed in the United States of America.
10 9876543
preface
In this sixth edition we have retained the basic objective of the earlier editions:
• to present a comprehensive and rigorous treatment of classical thermodynamics
while retaining an engineering perspective, and in doing so
• to lay the groundwork for subsequent studies in such fields as fluid mechanics, heat
transfer, and statistical thermodynamics, and also
• to prepare the student to effectively use thermodynamics in the practice of engineering.
We have deliberately directed our presentation to students. New concepts and defi-
nitions are presented in the context where they are first relevant in a natural progression.
The first thermodynamic properties to be denned (Chapter 2) are those that can be readily
measured: pressure, specific volume, and temperature. In Chapter 3, tables of thermody-
namic properties are introduced, but only in regard to these measurable properties. Inter-
nal energy and enthalpy are introduced in connection with the first law, entropy with the
second law, and the Helmholtz and Gibbs functions in the chapter on thermodynamic re-
lations. Many real world realistic examples have been included in the book to assist the
student in gaining an understanding of thermodynamics, and the problems at the end of
each chapter have been carefully sequenced to correlate with the subject matter and are
grouped and identified as such. The early chapters in particular contain a much larger
number of examples, illustrations, and problems than in previous editions, and through-
out the book, chapter-end summaries are included, followed by a set of concept/study
problems that should be of benefit to the students.
New Features in this edition
End-of-Chapter Summaries
The new end-of-chapter summaries provide a short review of the main concepts covered
in the chapter, with highlighted key words. To further enhance the summary we have
listed the set of skills that the student should have mastered after studying the chapter.
These skills are among the outcomes that can be tested with the accompanying set of
study-guide problems in addition to the main set of homework problems.
Main Concepts and Formulas
Main concepts and formulas are included at the end of each chapter for reference.
Vi M PREFACE
Study Guide Problems
We have made a set of study guide problems for each chapter as a quick check of the
chapter material. These are selected to be short and directed toward a very specific con-
cept. A student can answer all of these questions to assess their level of understanding,
and determine if any of the subjects need to be studied further. These problems are also
suitable to use together with the rest of the homework problems in assignments and in-
cluded in the solution manual.
Homework Problems
The number of homework problems has been greatly expanded and now exceeds 2,400. A
large number of introductory problems have been added to cover all aspects of the chapter
material. We have furthermore separated the problems into sections according to subject
for easy selection according to the particular coverage given. A number of more compre-
hensive problems have been retained and grouped in the end as review problems.
Tables
The tables of the substances have been expanded to include the specific internal energy in
the superheated vapor region. The ideal gas tables have been printed on a mass basis as
well as a mole basis, to reflect their use on mass basis early in the text, and mole basis for
the combustion and chemical equilibrium chapters.
Revisions
In this edition we have incorporated a number of developments and approaches included
in our recent textbook, Introduction to Engineering Thermodynamics, Richard E. Sonntag
and Claus Borgnakke, John Wiley & Sons, Inc. (2001). In Chapter 3, we first introduce
thermodynamic tables, and then note the behavior of superheated vapor at progressively
lower densities, which leads to the definition of the ideal gas model, then the compres-
sibility factor and equations of state. In Chapter 5, the result of ideal gas energy depend-
ing only on temperature follows the examination of steam table values at different
temperatures and pressures. Second law presentation in Chapter 7 is streamlined, with
better integration of the concepts of thermodynamic temperature and ideal gas tempera-
ture. Coverage of ideal gas and ideal gas mixtures focuses on unit mass basis, instead of
mole basis, and is simpler. Development of availability and reversible work in Chapter 10
focuses on the steady-state process, and leads to the general expression for exergy. We
have therefore included a new section on the general exergy balance to amplify the con-
cept of transport and destruction of exergy. The chapter with property relations is slightly
reorganized and streamlined to also focus on properties on a mass basis. Due to current
technology developments we have expanded our discussion of the fuel cells and also up-
dated the chapter with combustion.
Expanded Software Included
In this edition we have included the expanded software CATT2 that includes a number of
additional substances besides those included in the printed tables in Appendix B. A
number of hydrocarbon fuels, refrigerants, and cryogenic fluids are included and a
preface E vii
printed version is available in the booklet Thermodynamic and Transport Properties,
Claus Borgnakke and Richard E. Sonritag, John Wiley & Sons, Inc. (1997).
Flexibility m Coverage and scope
We have attempted to cover fairly comprehensively the basic subject matter of classical
thermodynamics, and believe that the book provides adequate preparation for study of the
application of thermodynamics to the various professional fields as well as for study of
more advanced topics in thermodynamics, such as those related to materials, surface phe-
nomena, plasmas, and cryogenics. We also recognize that a number of colleges offer a
single introductory course in thermodynamics for all departments, and we have tried to
cover those topics that the various departments might wish to have included in such a
course. However, since specific courses vary considerably in prerequisites, specific objec-
tives, duration, and background of the students, we have arranged the material, particu-
larly in the later chapters, so that there is considerable flexibility in the amount of material
that may be covered.
Units
Our philosophy regarding units in this edition has been to organize the book so that the
course or sequence can be taught entirely in SI units (Le Systeme International d'UnitSs).
Thus, all the text examples are in SI units, as are the complete problem sets and the ther-
modynamic tables. In recognition, however, of the continuing need for engineering gradu-
ates to be familiar with English Engineering units, we have included an introduction to
this system in Chapter 2. We have also repeated a sufficient number of examples, prob-
lems, and tables in these units, which should allow for suitable practice for those who
wish to use these units. For dealing with English units, the force-mass conversion ques-
tion between pound mass and pound force is treated simply as a units conversion, without
using an explicit conversion constant. Throughout, symbols, units and sign conventions
are treated as in previous editions.
Supplements and Additional Support
Additional support is made available through a companion website at www.wiley.com/
college/sonntag. Tutorials and reviews of all the basic material as indicated in the main
text by the ThermoNet icon are accessible through the companion website. The website
allows students to go through a self-paced study developing the basic skill set associated
with the various subjects usually covered in a first course in thermodynamics.
The chapter on compressible flow is also available at www.wiley.com/colIege/
sonntag and revised with summary, study guide problems, and new homework problems.
We recognize that in many cases this chapter is not included in the thermodynamics
courses, but may instead be covered elsewhere in the curriculum.
We have tried to include material appropriate and sufficient for a two-semester
course sequence, and to provide flexibility for choice of topic coverage. Instructors may
want to visit the companion website at www.wiley.com/college/sonntag for information
and suggestions on possible course structure and schedules, additional study problem ma-
terial, and current errata for the book.
viii ffl Preface
Acknowledgments
We acknowledge with appreciation the suggestions, counsel, and encouragement of many
colleagues, both at the University of Michigan and elsewhere. This assistance has been
very helpful to us during the writing of this edition, as it was with the earlier editions of
the book. Both undergraduate and graduate students have been of particular assistance,
for their perceptive questions have often caused us to rewrite or rethink a given portion of
the text, or to try to develop a better way of presenting the material in order to anticipate
such questions or difficulties. Finally, for each of us, the encouragement and patience of
our wives and families have been indispensable and have made this time of writing pleas-
ant and enjoyable, in spite of the pressures of the project. A special thanks to a number of
colleagues at other institutions, who have reviewed the book and provided input to the re-
visions. Some of the reviewers are
Edward E. Anderson, Texas Tech University
Haim H. Bau, University of Pennsylvania
Pei-Feng Hsu, Florida Institute of Technology
Gerald J. Micklow, University of Alabama
Jayathi Y. Murthy, Carnegie Mellon University
Anthony J. Wheeler, San Francisco State University
We also wish to thank our editor, Joseph Hayton, for his effort in the planning and the
support and encouragement during the production of this edition.
Our hope is that this book will contribute to the effective teaching of thermodynam-
ics to students who face very significant challenges and opportunities during their profes-
sional careers. Your comments, criticism, and suggestions will also be appreciated and
you may channel that through Claus Borgnakke, claus@umich.edu.
richard e. sonntag
Claus borgnakke
Gordon J. Van wylen
Ann Arbor, Michigan
April 2002
Contents
1 Some Introductory Comments 1
1.1 The Simple Steam Power Plant, 1
1.2 Fuel Cells, 2
1 .3 The Vapor-Compression-Refrigeration Cycle, 5
1.4 The Thermoelectric Refrigerator, 7
1.5 The Air Separation Plant, 8
1.6 The Gas Turbine, 9
1.7 The Chemical Rocket Engine, 11
1.8 Other Applications and Environmental Issues, 72
2 SOME CONCEPTS AND DEFINITIONS 14
2. 1 A Thermodynamic System and the Control Volume, 14
2.2 Macroscopic Versus Microscopic Point of View, 15
2.3 Properties and State of a Substance, 16
2.4 Processes and Cycles, 17
2.5 Units for Mass, Length, Time, and Force, 18
2.6 Energy, 21
2.7 Specific Volume and Density, 23
2.8 Pressure, 25
2.9 Equality of Temperature, 31
2.10 The Zeroth Law of Thermodynamics, 31
2. 11 Temperature Scales, 32
Problems, 34
3 PROPERTIES OF A PURE SUBSTANCE 43
3 . 1 The Pure Substance, 44
3.2 Vapor-Liquid-Solid-Phase Equilibrium in a Pure Substance, 44
3.3 Independent Properties of a Pure Substance, 51
3.4 Tables of Thermodynamic Properties, 51
3.5 Thermodynamic Surfaces, 59
3.6 The P- V-T Behavior of Low- and Moderate-Density Gases, 61
3.7 Computerized Tables, 69
Problems, 72
4 WORK AND HEAT 84
4.1 Definition of Work, 84
4.2 Units for Work, 86
4.3 Work Done at the Moving Boundary of a Simple Compressible System, 87
4.4 Other Systems that Involve Work, 96
4.5 Concluding Remarks Regarding Work, 98
ix
X m CONTENTS
4.6 Definition of Heat, 100
4.7 Heat Transfer Modes, 101
4.8 Comparison of Heat and Work, JOS
Problems, 105
5 The
First Law of thermodynamics
116
5.1 The First Law of Thermodynamics for a Control Mass Undergoing
a Cycle, 116
5.2 The First Law of Thermodynamics for a Change in State of a Control Mass, 117
5.3 Internal Energy— A Thermodynamic Property, 124
5.4 Problem Analysis and Solution Technique, 126
5.5 The Thermodynamic Property Enthalpy, 130
5.6 The Constant-Volume and Constant-Pressure Specific Heats, 133
5.7 The Internal Energy, Enthalpy, and Specific Heat of Ideal Gases, 135
5.8 The First Law as a Rate Equation, 141
5.9 Conservation of Mass, 143
Problems, 145
6 First Law analysis for a Control volume 162
6.1 Conservation of Mass and the Control Volume, 162
6.2 The First Law of Thermodynamics for a Control Volume, 165
6.3 The Steady-State Process, 1 67
6.4 Examples of Steady-State Processes, 169
6.5 The Transient Process, 183
Problems, 195
7 The Second Law of thermodynamics 214
7.1 Heat Engines and Refrigerators, 214
7.2 The Second Law of Thermodynamics, 220
7.3 The Reversible Process, 223
7.4 Factors that Render Processes Irreversible, 224
7.5 The Camot Cycle, 227
7.6 Two Propositions Regarding the Efficiency of a Carnot Cycle, 229
7.7 The Thermodynamic Temperature Scale, 230
7.8 The Ideal-Gas Temperature Scale, 233
7.9 Ideal versus Real Machines, 236
Problems, 240
8. 1 The Inequality of Clausius, 251
8.2 Entropy — A Property of a System, 255
8.3 The Entropy of a Pure Substance, 257
8.4 Entropy Change in Reversible Processes, 259
8.5 The Thermodynamic Property Relation, 263
8 ENTROPY
251
CONTENTS B XI
8.6 Entropy Change of a Control Mass During an Irreversible Process, 264
8.7 Entropy Generation, 266
8.8 Principle of the Increase of Entropy, 268
8.9 Entropy Change of a Solid or Liquid, 272
8.10 Entropy Change of an Ideal Gas, 273
8.11 The Reversible Polytropic Process for an Ideal Gas, 2 78
8.12 Entropy as a Rate Equation, 282
Problems, 285
SECOND LAW ANALYSIS FOR A CONTROL VOLUME 302
9. 1 The Second Law of Thermodynamics for a Control Volume, 302
9.2 The Steady-State Process and the Transient Process, 304
9.3 The Reversible Steady-State Process, 313
9.4 Principle of the Increase of Entropy, 316
9.5 Efficiency, 317
9.6 Some General Comments Regarding Entropy, 323
Problems, 325
IRREVERSIBILITY AND AVAILABILITY 343
1 0. 1 Available Energy, Reversible Work, and Irreversibility, 343
10.2 Availability and Second-Law Efficiency, 355
10.3 Exergy Balance Equation, 363
Problems, 370
POWER AND REFRIGERATION SYSTEMS 382
11.1 Introduction to Power Systems, 382
11 .2 The Rankine Cycle, 384
1 1 .3 Effect of Pressure and Temperature on the Rankine Cycle, 388
11.4 The Reheat Cycle, 393
11.5 The Regernative Cycle, 396
1 1 .6 Deviation of Actual Cycles from Ideal Cycles, 403
1 1 .7 Cogeneration, 409
11.8 Air-Standard Power Cycles, 41
11.9 The Brayton Cycle, 411
11.10 The Simple Gas-Turbine Cycle with a Regenerator, 418
11.11 Gas-Turbine Power Cycle Configurations, 421
1 1.12 The Air-Standard Cycle for Jet Propulsion, 424
11.13 Reciprocating Engine Power Cycles, 426
11.14 The Otto Cycle, 427
1 1.15 The Diesel Cycle, 431
11.16 The Stirling Cycle, 433
11.17 Introduction to Refrigeration Systems, 434
1 1.18 The Vapor-Compression Refrigeration Cycle, 435
11.19 Working Fluids for Vapor-Compression Refrigeration Systems, 438
H Contents
1 1.20 Deviation of the Actual Vapor-Compression Refrigeration Cycle
from the Ideal Cycle, 439
1 1 .2 1 The Ammonia Absorption Refrigeration Cycle, 441
1 1.22 The Air-Standard Refrigeration Cycle, 442
1 1.23 Combined-Cycle Power and Refrigeration Systems, 446
Problems, 450
GAS MIXTURES 473
12.1 General Considerations and Mixtures of Ideal Gases, 473
12.2 A Simplified Model of a Mixture Involving Gases and a Vapor, 480
12.3 The First Law Applied to Gas- Vapor Mixtures, 485
12.4 The Adiabatic Saturation Process, 488
12.5 Wet-Bulb and Dry-Bulb Temperatures, 490
12.6 The Psychrometric Chart, 491
Problems, 494
THERMODYNAMIC RELATIONS 511
13.1 The Clapeyron Equation, 51 1
13.2 Mathematical Relations for a Homogeneous Phase, 515
13.3 The Maxwell Relations, 516
13.4 Thermodynamic Relations Involving Enthalpy, Internal Energy,
and Entropy, 519
13.5 Volume Expansivity and Isothermal and Adiabatic Compressibility, 524
13.6 Real Gas Behavior and Equations of State, 527
13.7 The Generalized Chart for Changes of Enthalpy at Constant
Temperature, 532
13.8 The Generalized Chart for Changes of Entropy at Constant
Temperature, 535
13.9 Developing Tables of Thermodynamic Properties from Experimental
Data, 538
13.10 The Property Relation for Mixtures, 540
13.11 Pseudopure Substance Models for Real-Gas Mixtures, 543
Problems, 550
CHEMICAL REACTIONS 561
14.1 Fuels, 561
14.2 The Combustion Process, 564
14.3 Enthalpy of Formation, 572
14.4 First-Law Analysis of Reacting Systems, 574
14.5 Enthalpy and Internal Energy of Combustion; Heat of Reaction, 581
14.6 Adiabatic Flame Temperature, 585
14.7 The Third Law of Thermodynamics and Absolute Entropy, 557
14.8 Second-Law Analysis of Reacting Systems, 589
14.9 Fuel Cells, 596
14.10 Evaluation of Actual Combustion Processes, 599
Problems, 604
Contents H xiii
1 5 introduction to Phase and Chemical equilibrium 61 7
15.1 Requirements for Equilibrium, 617
15.2 Equilibrium Between Two Phases of a Pure Substance, 619
15.3 Metastable Equilibrium, 623
15.4 Chemical Equilibrium, 625
15.5 Simultaneous Reactions, 634
15.6 Ionization, 638
Problems, 643
1 6 COMPRESSIBLE FLOW W16-1
{available on the website: www.wiley.com/college/sonntag)
16.1 Stagnation Properties, W16-1
16.2 The Momentum Equation for a Control Volume, W16-3
16.3 Forces Acting on a Control Surface, W16-6
16.4 Adiabatic, One-Dimensional, Steady-State Flow
of an Incompressible Fluid Through a Nozzle, W16-8
16.5 Velocity of Sound in an Ideal Gas, W16-10
1 6.6 Reversible, Adiabatic, One-Dimensional Flow of an Ideal Gas through
a Nozzle, W16-12
16.7 Mass Rate of Flow of an Ideal Gas through an Isentropic Nozzle, W16-16
16.8 Normal Shock in an Ideal Gas Flowing through a Nozzle, W16-20
16.9 Nozzle and Diffuser Coefficients, W16-26
16.10 Nozzle and Orifices as Flow-Measuring Devices, W 16-28
Problems, W16-37
CONTENTS OF APPENDIX
Appendix A SI units: Single State properties 653
Appendix B SI Units: Thermodynamic Tables 673
Appendix C Ideal- Gas Specific Heat 723
Appendix D equations of State 725
appendix E Figures 731
Appendix F English Unit Tables 737
ANSWERS TO SELECTED PROBLEMS 779
INDEX 789
Symbols
a
acceleration
A
area
a, a
specmc rivininoitz luiiouuii diiu iuidi ncniuiuiii iuli^iiuii
Ar
air- fuel ratio
D
B s
adiabatic bulk modulus
B T
isothermal bulk modulus
c
c
c
c
C P
»"l £' t ' t~l f nr»">P^l ITP t" t~l A 1 ^ 1 T T 1~\ f Clt
COnaldm-prCabUIc speOllH* HCat
n
^rtnptant irrtlitTYIP cnpflfif^ n**?lt"
constant- vuiunit spcoiiio iicai
r
zero-pressure constant-pressure specific heat
zero-pressure constant-volume specific heat
e, E
specific energy and total energy
F
force
FA
fuel-air ratio
g
acceleration due to gravity
g>G
specific Gibbs function and total Gibbs function
specific enthalpy and total enthalpy
i
electrical current
I
irreversibility
J
proportionality factor to relate units of work to units of heat
k
specific heat ratio: C p /C v
K
equilibrium constant
KE
kinetic energy
L
length
m
mass
m
mass flow rate
M
molecular weight
M
Mach number
n
number of moles
n
polytropic exponent
P
pressure
Pi
partial pressure of component i in a mixture
PE
potential energy
Pr
relative pressure as used in gas tables
q,Q
heat transfer per unit mass and total heat transfer
Q
rate of heat transfer
XV
XVi M SYMBOLS
Q H , Q L heat transfer with high-temperature body and heat transfer with
low-temperature body; sign determined from context
R gas constant
R universal gas constant
s, S specific entropy and total entropy
5 gen entropy generation
S £ea rate of entropy generation
/ time
T temperature
it, U specific internal energy and total internal energy
v, V specific volume and total volume
v r relative specific volume as used in gas tables
V velocity
w, W work per unit mass and total work
W rate of work, or power
>t^ ev reversible work between two states
x quality
y gas-phase mole fraction
Z elevation
Z compressibility factor
Z electrical charge
Script Letters
% electrical potential
if surface tension
3" tension
Greek Letters
a residual volume
<x p volume expansivity
j3 coefficient of performance for a refrigerator
(3' coefficient of performance for a heat pump
j3 s adiabatic compressibility
j3 r isothermal compressibility
17 efficiency
fx chemical potential
v stoichiometric coefficient
p density
<J> equivalence ratio
4> relative humidity
4>, 3> exergy or availability for a-control mass
tp flow availability
co humidity ratio or specific humidity
co a centric factor
SUBSCRIPTS c property at the critical point
c.v. control volume
e state of a substance leaving a control volume
/ formation
Symbols H XVii
/ property of saturated liquid
fg difference in property for saturated vapor and saturated liquid
g property of saturated vapor
i state of a substance entering a control volume
i property of saturated solid
if difference in property for saturated liquid and saturated solid
ig difference in property for saturated vapor and saturated solid
r reduced property
s isentropic process
property of the surroundings
stagnation property
Superscripts _
bar over symbol denotes property on a molal basis (over V, H, S, U, A, G, the
bar denotes partial molal property)
° property at standard-state condition
* ideal gas
* property at the throat of a nozzle
rev reversible
FUNDAMENTALS OF
THERMODYNAMICS
Sixth Edition
Some Introductory Comments
In the course of our study of thermodynamics! a number of the examples and problems
presented refer to processes that occur in equipment such as a steam power plant, a fuel
cell, a vapor-compression refrigerator, a thermoelectric cooler, a turbine or rocket en-
gine, and an air separation plant. In this introductory chapter, a brief description of this
equipment is given. There are at least two reasons for including such a chapter. First,
many students have had limited contact with such equipment, and the solution of
problems will be more meaningful when they have some familiarity with the actual
processes and the equipment. Second, this chapter will provide an introduction to ther-
modynamics, including the use of certain terms (which will be more formally defined
in later chapters), some of the problems to which thermodynamics can be applied, and
some of the things that have been accomplished, at least in part, from the application of
thermodynamics.
Thermodynamics is relevant to many other processes than those cited in this chap-
ter. It is basic to the study of materials, chemical reactions, and plasmas. The student
should bear in mind that this chapter is only a brief and necessarily very incomplete intro-
duction to the subject of thermodynamics.
1,1 The Simple Steam Power Plant
A schematic diagram of a recently installed steam power plant is shown in Fig. 1.1. High-
pressure superheated steam leaves the steam drum at the top of the boiler, also referred to
as a steam generator, and enters the turbine. The steam expands in the turbine and in
doing so does work, which enables the turbine to drive the electric generator. The steam,
now at low pressure, exits the turbine and enters the heat exchanger, where heat is trans-
ferred from the steam (causing it to condense) to the cooling water. Since large quantities
of cooling water are required, power plants have traditionally been located near rivers or
lakes, leading to thermal pollution of those water supplies. More recently, condenser
cooling water is recycled by evaporating a fraction of the water in large cooling towers,
thereby cooling the remainder of the water that remains as a liquid. In the power plant
shown in Fig. 1.1, the plant is designed to recycle the condenser cooling water by using
the heated water for district space heating.
The pressure of the condensate leaving the condenser is increased in the pump, en-
abling it to return to the steam generator for reuse. In many cases, an economizer or water
preheater is used in the steam cycle, and in many power plants, the air that is used for
combustion of the fuel may be preheated by the exhaust combustion-product gases.
These exhaust gases must also be purified before being discharged to the atmosphere,
such that there are many complications to the simple cycle.
1
2 H Chapter One some introductory Comments
FIGURE 1.1 Schematic diagram of a steam power plant.
Figure 1.2 is a photograph of the power plant depicted in Fig. 1.1. The tall building
shown at the left is the boiler house, next to which are buildings housing the turbine and
other components. Also noted are the tall chimney or stack, and the coal supply ship at
dock. This particular power plant is located in Denmark, and at the time of its installation,
set a world record of efficiency, converting 45% of the 850 MW of coal combustion en-
ergy into electricity. Another 47% is reusable for district space heating, an amount that in
older plants was simply thrown away to the environment with no benefit.
The steam power plant described utilizes coal as the combustion fuel. Other plants
use natural gas, fuel oil, or biomass as the fuel. A number of power plants around the
world operate on the heat released from nuclear reactions instead of fuel combustion. Fig-
ure 1.3 is a schematic diagram of a nuclear marine propulsion power plant. A secondary
fluid circulates through the reactor, picking up heat generated by the nuclear reaction in-
side. This heat is then transferred to the water in the steam generator. The steam cycle
processes are the same as in the previous example, but in this application the condenser
cooling water is seawater, that is then returned at higher temperature to the sea.
1.2 Fuel Cells
When a conventional power plant is viewed as a whole, as shown in Fig. 1 .4, fuel and air
enter the power plant and products of combustion leave the unit. There is also a transfer of
heat to the cooling water, and work is done in the form of the electrical energy leaving the
power plant. The overall objective of a power plant is to convert the availability (to do
work) of the fuel into work (in the form of electrical energy) in the most efficient manner,
taking into consideration cost, space, safety, and environmental concerns.
fuel Cells g 3
FIGURE 1.2 The Esbjerg, Denmark power station. (Courtesy Vestkraft 1996.)
4 M CHAPTER ONE SOME INTRODUCTORY COMMENTS
FIGURE 1.3 Schematic diagram of a shipboard nuclear propusion system. (Courtesy Babcock & Wilcox Co.)
We might well ask whether all the equipment in the power plant, such as the steam
generator, the turbine, the condenser, and the pump, is necessary. Is it possible to produce
electrical energy from the fuel in a more direct manner?
The fuel cell accomplishes this objective. Figure 1.5 shows a schematic arrange-
ment of a fuel cell of the ion-exchange-membrane type. In this fuel cell, hydrogen and
oxygen react to form water. Hydrogen gas enters at the anode side and is ionized at the
surface of the iort-exchange membrane, as indicated in Fig. 1.5. The electrons flow
through the external circuit to the cathode while the positive hydrogen ions migrate
through the membrane to the cathode, where both react with oxygen to form water.
There is a potential difference between the anode and cathode, and thus there is a
flow of electricity through a potential difference; this, in thermodynamic terms, is called
work. There may also be a transfer of heat between the fuel cell and the surroundings.
At the present time the fuel used in fuel cells is usually either hydrogen or a mixture
of gaseous hydrocarbons and hydrogen. The oxidizer is usually oxygen. However, current
development is directed toward the production of fuel cells that use hydrocarbon fuels and
Products of
combustion
f
~ Electrical energy
~ (work)
FIGURE 1.4
Schematic diagram of a Heat transfer to
power plant. cooling water
The Vapor-Compression Refrigeration Cycle M 5
air. Although the conventional (or nuclear) steam power plant is still used in large-scale
power-generating systems and conventional piston engines and gas turbines are still used
in most transportation power systems, the fuel cell may eventually become a serious com-
petitor. The fuel cell is already being used to produce power for space and other special
applications.
Thermodynamics plays a vital role in the analysis, development, and design of
all power-producing systems, including reciprocating internal-combustion engines and
gas turbines. Considerations such as the increase of efficiency, improved design, opti-
mum operating conditions, environmental pollution, and alternate methods of power
generation involve, among other factors, the careful application of the fundamentals of
thermodynamics.
1.3 The Vapor-Compression
Refrigeration Cycle
A simple vapor-compression refrigeration cycle is shown schematically in Fig. 1.6. The
refrigerant enters the compressor as a slightly superheated vapor at a low pressure. It then
leaves the compressor and enters the condenser as a vapor at some elevated pressure,
where the refrigerant is condensed as heat is transferred to cooling water or to the sur-
roundings. The refrigerant then leaves the condenser as a high-pressure liquid. The pres-
sure of the liquid is decreased as it flows through the expansion valve, and as a result,
some of the liquid flashes into cold vapor. The remaining liquid, now at a low pressure
and temperature, is vaporized in the evaporator as heat is transferred from the refrigerated
space. This vapor then reenters the compressor.
In a typical home refrigerator the compressor is located in the rear near the bot-
tom of the unit. The compressors are usually hermetically sealed; that is, the motor and
6 K Chapter One Some Introductory Comments
FIGURE 1.6
Schematic diagram of a
simple refrigeration cycle.
High-pressure
liquid
Expansion
valve
Low-pressure
mixture of
liquid and vapor
Heat transfer to ambient
air or to cooling water
High-pressure vapor
Low-pressure
vapor
Heat transfer from
refrigerated space
Work in
The Thermoelectric Refrigerator H 7
compressor are mounted in a sealed housing, and the electric leads for the motor pass
through this housing. This seal prevents leakage of the refrigerant. The condenser is
also located at the back of the refrigerator and is arranged so that the air in the room
flows past the condenser by natural convection. The expansion valve takes the form of
a long capillary tube, and the evaporator is located around the outside of the free2ing
compartment inside the refrigerator.
Figure 1 .7 shows a large centrifugal unit that is used to provide refrigeration for an
air-conditioning unit. In this unit, water is cooled and the circulated to provide cooling
where needed.
1.4 The Thermoelectric Refrigerator
We may well ask the same question about the vapor-compression refrigerator that we
asked about the steam power plant — is it possible to accomplish our objective in a more
direct manner? Is it possible, in the case of a refrigerator, to use the electrical energy
(which goes to the electric motor that drives the compressor) to produce cooling in a more
direct manner and thereby to avoid the cost of the compressor, condenser, evaporator, and
all the related piping?
The thermoelectric refrigerator is such a device. This is shown schematically in Fig.
1.8fl. The thermoelectric device, like the conventional thermocouple, uses two dissimilar
materials. There are two junctions between these two materials in a thermoelectric refrig-
erator. One is located in the refrigerated space and the other in ambient surroundings.
When a potential difference is applied, as indicated, the temperature of the junction lo-
cated in the refrigerated space will decrease and the temperature of the other junction will
Heat transfer from
refrigerated space
Mil
Cold junction
Material A —
H
\
Material B — .
H
V Metai
/electrodes
/
Hot Junction r
Hot junction
1 ' Heat transfer to ambient ^
i
-o o-
Heat transfer from
high-temperature body
Material A
Materials
Metal
electrodes
— YvWA-
Load
ib)
FIGURE 1,8 (a) A thermoelectric refrigerator, (b) A thermoelectric power generation device.
8 M chapter One some introductory Comments
increase. Under steady-state operating conditions, heat will be transferred from the refrig-
erated space to the cold junction. The other junction will be at a temperature above the
ambient, and heat will be transferred from the junction to the surroundings.
A thermoelectric device can also be used to generate power by replacing the refrig-
erated space with a body that is at a temperature above the ambient. Such a system is
shown in Fig. 1 .86.
The thermoelectric refrigerator cannot yet compete economically with conventional
vapor-compression units. However, in certain special applications, the thermoelectric re-
frigerator is already is use and, in view of research and development efforts under way in
this field, it is quite possible that thermoelectric refrigerators will be much more exten-
sively used in the future.
lo5 The Air Separation Plant
One process of great industrial significance is the air separation. In an air separation plant,
air is separated into its various components. The oxygen, nitrogen, argon, and rare gases
so produced are used extensively in various industrial, research, space, and consumer-
goods applications. The air separation plant can be considered an example from two
major fields, the chemical process industry and cryogenics. Cryogenics is a term applied
to technology, processes, and research at very low temperatures (in general, below 150
K). In both chemical processing and cryogenics, thermodynamics is basic to an under-
standing of many phenomena that occur and to the design and development of processes
and equipment.
A number of different designs of air separation plants have been developed. Con-
sider Fig. 1.9, which shows a somewhat simplified sketch of a type of plant that is fre-
quently used. Air from the atmosphere is compressed to a pressure of 2 to 3 MPa. It is
then purified, particularly to remove carbon dioxide (which would plug the flow passages
as it solidifies when the air is cooled to its liquefaction temperature). The air is then com-
pressed to a pressure of 15 to 20 MPa, cooled to the ambient temperature in the after-
cooler, and dried to remove the water vapor (which would also plug the flow passages as
it freezes).
The basic refrigeration in the liquefaction process is provided by two different
processes. In one process the air in the expansion engine expands. During this process the
air does work and as a result the temperature of the air is reduced. In the other refrigera-
tion process air passes through a throttle valve that is so designed and so located that there
is a substantial drop in the pressure of the air and, associated with this, a substantial drop
in the temperature of the air.
As shown in Fig. 1.9, the dry, high-pressure air enters a heat exchanger. The air
temperature drops as it flows through the heat exchanger. At some intermediate point in
the heat exchanger, part of the air is bled off and flows through the expansion engine. The
remaining air flows through the rest of the heat exchanger and through the throttle valve.
The two streams join (both are at the pressure of 0.5 to 1 MPa) and enter the bottom of the
distillation column, which is referred to as the high-pressure column. The function of the
distillation column is to separate the air into its various components, principally oxygen
and nitrogen. Two streams of different composition flow from the high-pressure column
through throttle valves to the upper column (also called the low-pressure column). One of
these streams is an oxygen-rich liquid that flows from the bottom of the lower column,
The Gas Turbine M 9
FIGURE 1.9 A
simplified diagram of a
liquid oxygen plant.
Htgh-pressure
compressor
Air
purifier
Low-pressure
compressor
Q
Fresh air
intake
Hydrocarbon
absorber
and the other is a nitrogen-rich stream that flows through the subcooler. The separation is
completed in the upper column. Liquid oxygen leaves from the bottom of the upper, col-
umn, and gaseous nitrogen leaves from the top of the column. The nitrogen gas flows
through the subcooler and the main heat exchanger. It is the transfer of heat to this cold
nitrogen gas that causes the high-pressure air entering the heat exchanger to become
cooler.
Not only is a thermodynamic analysis essential to the design of the system as a
whole, but essentially every component of such a system, including the compressors, the
expansion engine, the purifiers and driers, and the distillation column, operates according
to the principles of thermodynamics. In this separation process we are also concerned
with the thermodynamic properties of mixtures and the principles and procedures by
which these mixtures can be separated. This is the type of problem encountered in petro-
leum refining and many other chemical processes. It should also be noted that cryogenics
is particularly relevant to many aspects of the space program, and a thorough knowledge
of thermodynamics is essential for creative and effective work in cryogenics.
1.6 The Gas Turbine
The basic operation of a gas turbine is similar to that of the steam power plant, except that
air is used instead of water. Fresh atmospheric air flows through a compressor that brings
it to a high pressure. Energy is then added by spraying fuel into the air and igniting it so
the combustion generates a high-temperature flow. This high-temperature, high-pressure
gas enters a turbine, where it expands down to the exhaust pressure, producing a shaft
10 U CHAPTER ONE SOME INTRODUCTORY COMMENTS
FIGURE 1.10 A 150-
MW gas turbine.
(Courtesy Westinghouse
Electric Corporation.)
work output in the process. The turbine shaft work is used to drive the compressor and
other devices, such as an electric generator that may be coupled to the shaft. The energy
that is not used for shaft work comes out in the exhaust gases, so these have either a high
temperature or a high velocity. The purpose of the gas turbine detennines the design so
that the most desirable energy form is maximized. An example of a large gas turbine for
stationary power generation is shown in Fig. 1.10. The unit has sixteen stages of compres-
sion and four stages in the turbine and is rated at 150 MW. Notice that since the combus-
tion of fuel uses the oxygen in the air, the exhaust gases cannot be recirculated as the
water is in the steam power plant.
FIGURE 1.11 A
turbofanjet engine.
(Courtesy General
Electric Aircraft Engines.)
The Chemical Rocket Engine □ 11
A gas turbine is often the preferred power-generating device where a large amount
of power is needed, but only a small, physical size is possible. Examples are jet engines,
turbofan jet engines, offshore oilrig power plants, ship engines, helicopter engines,
smaller local power plants, or peak-load power generators in larger power plants. Since
the gas turbine has relatively high exhaust temperatures, it can also be arranged so the ex-
haust gases are used to heat water that runs in a steam power plant before it exhausts to
the atmosphere.
In the examples mentioned previously, the jet engine and turboprop applications
utilize part of the power to discharge the gases at high velocity. This is what generates the
thrust of the engine that moves the airplane forward. The gas turbines in these applica-
tions are therefore designed differently than for the stationary power plant, where the en-
ergy is taken out as shaft work to an electric generator. An example of a turbofan jet
engine used in a commercial airplane is shown in Fig. 1.11. The large front-end fan also
blows air past the engine, providing cooling and gives additional thrust.
1.7 The Chemical rocket Engine
The advent of missiles and satellites brought to prominence the use of the rocket engine as
a propulsion power plant. Chemical rocket engines may be classified as either liquid pro-
pellant or solid propellant, according to the fuel used.
Figure 1.12 shows a simplified schematic diagram of a liquid-propellant rocket. The
oxidizer and fuel are pumped through the injector plate into the combustion chamber
where combustion takes place at high pressure. The high-pressure, high-temperature
products of combustion expand as they flow through the nozzle, and as a result they leave
the nozzle with a high velocity. The momentum change associated with this increase in
velocity gives rise to the forward thrust on the vehicle.
FIGURE 1.12
Simplified schematic
diagram of a liquid- High-velocity
propellant rocket engine. exhaust gases
12 M Chapter one some introductory Comments
The oxidizer and fuel must be pumped into the combustion chamber, and some aux-
iliary power plant is necessary to drive the pumps. In a large rocket this auxiliary power
plant must be very reliable and have a relatively high power output, yet it must be light in
weight. The oxidizer and fuel tanks occupy the largest part of the volume of an actual
rocket, and the range and payload of a rocket are determined largely by the amount of ox-
idizer and fuel that can be carried. Many different fuels and oxidizers have been consid-
ered and tested, and much effort has gone into the development of fuels and oxidizers that
will give a higher thrust per unit mass rate of flow of reactants. Liquid oxygen is fre-
quently used as the oxidizer in liquid-propel lant rockets, and liquid hydrogen is frequently
used as the fuel.
Much work has also been done on solid-propellant rockets. They have been very
successfully used for jet-assisted takeoffs of airplanes, military missiles, and space vehi-
cles. They are much simpler in both the basic equipment required for operation and the lo-
gistic problems involved in their use, but they are more difficult to control.
1.8 Other Applications and
environmental issues
There are many other applications in which thermodynamics is relevant. Many municipal
landfill operations are now utilizing the heat produced by the decomposition of biomass
waste to produce power, and they also capture the methane gas produced by these chemi-
cal reactions for use as a fuel. Geofhermal sources of heat are also being utilized, as are
solar- and windmill-produced electricity. Sources of fuel are being converted from one
form to another, more usable or convenient form, such as in the gasification of coal or the
conversion of biomass to liquid fuels. Hydroelectric plants have been in use for many
years, as have other applications involving water power. Thermodynamics is also relevant
to such processes as the curing of a poured concrete slab, which produces heat, the cool-
ing of electronic equipment, in various applications in cryogenics (cryo-surgery, food
fast-freezing), and many other diverse applications.
We must also be concerned with environmental issues related to these many devices
and applications of thermodynamics. For example, the construction and operation of the
steam power plant creates electricity, which is so deeply entrenched in our society that we
take its ready availability for granted. In recent years, however, it has become increas-
ingly apparent that we need to consider seriously the effects of such an operation on our
environment. Combustion of hydrocarbon fuels releases carbon dioxide into the atmos-
phere, where its concentration is increasing. Carbon dioxide, as well as other gases, ab-
sorbs infrared radiation from the surface of the earth, holding it close to the planet and
creating the "greenhouse effect," which in turn is believed to cause global warming and
critical climatic changes around the earth. Power plant combustion, particularly of coal,
releases sulfur dioxide, which is absorbed in clouds and later falls as acid rain in many
areas. Combustion processes in power plants and gasoline and diesel engines also gener-
ate pollutants other than these two. Species such as carbon monoxide, nitric oxides, and
partly burned fuels together with particulates all contribute to atmospheric pollution and
are regulated by law for many applications. Catalytic converters on automobiles help to
minimize the air pollution problem. In power plants, Fig. 1 . 1 indicates the fly ash cleanup
and also the flue gas clean up processes that are now incorporated to address these prob-
other applications and environmental issues 13 13
lems. Thermal pollution associated with power plant cooling water requirements was dis-
cussed in Section 1.1,
Refrigeration and air-conditioning systems, as well as other industrial processes,
have used certain chlorofluorocarbon fluids that eventually find their way to the upper at-
mosphere and destroy the protective ozone layer. Many countries have already banned the
production of some of these compounds, and the search for improved replacement fluid
continues.
These are only some of the many environmental problems caused by our efforts to
produce goods and effects intended to improve our way of life. During our study of ther-
modynamics", which is the science of the conversion of energy from one form to another,
we must continue to reflect on these issues. We must consider how we can eliminate or
at least minimize damaging effects, as well as use our natural resources, efficiently and
responsibly.
Some Concepts
AND DEFINITIONS
One excellent definition of thermodynamics is that it is the science of energy and entropy.
Since we have not yet defined these terms, an alternate definition in already familiar
terms is: Thermodynamics is the science that deals with heat and work and those proper-
ties of substances that bear a relation to heat and work. Like all sciences, the basis of ther-
modynamics is experimental observation. In thermodynamics these findings have been
formalized into certain basic laws, which are known as the first, second, and third laws of
thermodynamics. In addition to these laws, the zeroth law of thermodynamics, which in
the logical development of thermodynamics precedes the first law, has been set forth.
In the chapters that follow, we will present these laws and the thermodynamic
properties related to these laws and apply them to a number of representative examples.
The objective of the student should be to gain both a thorough understanding of the fun-
damentals and an ability to apply these fundamentals to thermodynamic problems. The
examples and problems further this twofold objective. It is not necessary for the student
to memorize numerous equations, for problems are best solved by the application of the
definitions and laws of thermodynamics. In this chapter some concepts and definitions
basic to thermodynamics are presented.
2.1 A Thermodynamic System
AND THE CONTROL VOLUME
A thermodynamic system comprises a device or combination of devices containing a
quantity of matter that is being studied. To define this more precisely, a control volume is
chosen so that it contains the matter and devices inside a control surface. Everything ex-
ternal to the control volume is the surroundings, with the separation given by the control
surface. The surface may be open or closed to mass flows, and it may have flows of en-
ergy in terms of heat transfer and work across it. The boundaries may be movable or sta-
tionary. In the case of a control surface that is closed to mass flow, so that no mass can
escape or enter the control volume, it is called a control mass containing the same amount
of matter at all times.
Selecting the gas in the cylinder of Fig. 2.1 as a control volume by placing a control
surface around it, we recognize this as a control mass. If a Bunsen burner is placed under
the cylinder, the temperature of the gas will increase and the piston will rise. As the pis-
ton rises, the boundary of the control mass moves. As we will see later, heat and work
cross the boundary of the control mass during this process, but the matter that composes
the control mass can always be identified and remains the same.
14
Macroscopic versus microscopic point of View B 15
Weights
Piston
FIGURE 2.1 Example
of a control mass.
System
boundary
Gas
L.
FIGURE 2.2 Example
of a control volume
O
Low-pressure
air in
High-pressure
air out
Motor
An isolated system is one that is not influenced in any way by the surroundings.
This means that no mass, heat, or work cross the boundary of the system. In many cases a
thermodynamic analysis must be made of a device, such as an air compressor, which has a
flow of mass into it, out of it, or both, as shown schematically in Fig. 2.2. The procedure
followed in such an analysis is to specify a control volume that surrounds the device
under consideration. The surface of this control volume is the control surface, which may
have mass momentum, and also heat and work, cross it.
Thus the more general control surface defines a control volume, where mass may
flow tn or out, with a control mass as the special case of no mass flow in or out. Hence the
control mass contains a fixed mass at all times, which explains its name. The difference in
the formulation of the analysis is considered in detail in Chapter 6. The terms closed sys-
tem (fixed mass) and open system (involving a flow of mass) are sometimes used to make
this distinction. Here, we use the term system as a more general and loose description for
a mass, device, or combination of devices that then is more precisely defined, when a con-
trol volume is selected. The procedure that will be followed in the presentation of the first
and the second laws of thermodynamics is first to present these laws for a control mass
and then to extend the analysis to the more general control volume.
2,2 macroscopic versus microscopic
Point of view
An investigation into the behavior of a system may be undertaken from either a micro-
scopic or a macroscopic point of view. Let us briefly describe a system from a microscopic
point of view. Consider a system consisting of a cube 25 mm on a side and containing a
16 H ChapterTwo Some Concepts and Definitions
monatomic gas at atmospheric pressure and temperature. This volume contains approxi-
mately 10 20 atoms. To describe the position of each atom, we need to specify three coordi-
nates; to describe the velocity of each atom, we specify three velocity components.
Thus, to describe completely the behavior of this system from a microscopic point
of view we must deal with at least 6 X 10 20 equations. Even with a large digital computer,
this is a quite hopeless computational task. However, there are two approaches to this
problem that reduce the number of equations and variables to a few that can be computed
relatively easily. One approach is the statistical approach, in which, on the basis of statis-
tical considerations and probability theory, we deal with "average" values for all particles
under consideration. This is usually done in connection with a model of the atom under
consideration. This is the approach used in the disciplines known as kinetic theory and
statistical mechanics.
The other approach to reducing the number of variables to a few that can be handled
is the macroscopic point of view of classical thermodynamics. As the word macroscopic
implies, we are concerned with the gross or average effects of many molecules. These ef-
fects can be perceived by our senses and measured by instruments. However, what we re-
alty perceive and measure is the time-averaged influence of many molecules. For
example, consider the pressure a gas exerts on the walls of its container. This pressure re-
sults from the change in momentum of the molecules as they collide with the wall. From a
macroscopic point of view, however, we are not concerned with the action of the individ-
ual molecules but with the time-averaged force on a given area, which can be measured
by a pressure gauge. In fact, these macroscopic observations are completely independent
of our assumptions regarding the nature of matter.
Although the theory and development in this book is presented from a macroscopic
point of view, a few supplementary remarks regarding the significance of the microscopic
perspective are included as an aid to the understanding of the physical processes involved.
Another book in this series, Infroduction to Thermodynamics: Classical and Statistical,
by R. E. Sormtag and G. J, Van Wylen, includes thermodynamics from the microscopic
and statistical point of view.
A few remarks should be made regarding the continuum. From the macroscopic
view, we are always concerned with volumes that are very large compared to molecular
dimensions and, therefore, with systems that contain many molecules. Because we are not
concerned with the behavior of individual molecules, we can treat the substance as being
continuous, disregarding the action of individual molecules. This continuum concept, of
course, is only a convenient assumption that loses validity when the mean free path of the
molecules approaches the order of magnitude of the dimensions of the vessel, as, for ex-
ample, tn high-vacuum technology. In much engineering work the assumption of a con-
tinuum is valid and convenient, going hand in hand with the macroscopic view.
2.3 PROPERTIES AND STATE OF A SUBSTANCE
If we consider a given mass of water, we recognize that this water can exist in various
forms. If it is a liquid initially, it may become a vapor when it is heated or a solid when it
is cooled. Thus, we speak of the different phases of a substance. A phase is defined as a
quantity of matter that is homogeneous throughout. When more than one phase is present,
the phases are separated from each other by the phase boundaries. In each phase the sub-
stance may exist at various pressures and temperatures or, to use the thermodynamic term,
Processes and Cycles H 17
in various states. The state may be identified or described by certain observable, macro-
scopic properties; some familiar ones are temperature, pressure, and density. Iniaier
chapters other properties will be introduced. Each of the properties of a substance (^)
given state has only one definite value, and these properties always have the same value
for a given state, regardless of how the substance arrived at the state. In fact, a property
can be defined as any quantity that depends on the state of the system and is independent
of the path (that is, the prior history) by which the system arrived at the given state. Con-
versely, the state is specified or described by the properties. Later we will consider the
number of independent properties a substance can have, that is, the minimum number of
properties that must be specified to fix the state of the substance. r— — >
(^3 Thermodynamic properties can be divided into two general classes, intensive ! 0)|
extensive properties. An intensive property is independent of th e mass; the value of an ex-
tensive property varies directly with the mass. Thus, if a quantity of matter in a given state
is divided into two equal parts, each part will have the same value of intensive properties
as the original and half the value of the extensive properties. Pressure, temperature, and
density are examples of intensive properties. Mass and total volume are examples of ex-
tensive properties. Extensive properties per unit mass, such as specific volume, are inten-
sive properties.
Frequently we will refer not only to the properties of a substance but to the proper-
ties of a system. When we do so we necessarily imply that the value of the property has
significance for the entire system, and this implies equilibrium. For example, if the gas
that composes the system (control mass) in Fig. 2.1 is in thermal equilibrium, the temper-
ature will be the same throughout the entire system, and we may speak of the temperatv^
as a property of the system. We may also consider mechanical equilibrium, which is
lated to pressure. If a sys tem is_jn mechanical equilibrium, there is no tendency for tie
pressure at any pdlhTto changeTwim~tffie^^ the systemTBTsoIated^
roundings. There will be a variation in pressure with elevation because of the influence of
gravitational forces, although under equilibrium conditions there will be no tendency for
the pressure at any location to change. However, in many thermodynamic problems, this
variation in pressure with elevation is so small that is can be neglected. Chemical equilib-
rium is also important and will be considered in Chapter 15. When a system is in equilib-
rium regarding all possible changes of state, we say that the system is in thermodynamic
equilibrium.
2.4 PROCESSES ANB CYCLES
Whenever one or more of the properties of a system change, we say that a change in state
has occurred. For example, when one of the weights on the piston in Fig. 2,3 is removed,
the piston ris[
volume incre;
T^Tjd a change in state occurs, for the pressure decreases and the specific
jrriJ The path of the successi on of s tates throug h which the system passes is
ca lled the process.
Ext us consider the equilibrium of a system as it undergoes a change in state. The
moment the weight is removed from the piston in Fig. 2.3, mechanical equilibrium does
not exist, and as a result the piston is moved upward until mechanical equilibrium is again
restored. The question is this: Since the properties describe the state of a system only
when it is in equilibrium, how can we describe the states of a system during a process if
the actual process occurs only when equilibrium does not exist? One step in the answer to
18 B chapter two Some concepts and definitions
Weights
Piston
FIGURE 2.3 Example
of a system that may
undergo a quasi-
equilibrium process.
System
boundary
Gas
L:
this question concerns the definition of an ideal process, which we call a guasi-equilibrium
process. A quasi-eqmHbrh^pracess is one in which the deviation from thermodynamic
■quiHbjiirm isjjrfmitesimal, and all the states the system passes througj^uring a quasi-
j^mljbrium process^may be^considered equihbjium states. Many actual processes closely
approach a quasi-equilibrium process and may be so treated with essentially no error. If
the weights on the piston in Fig. 2.3 are small and are taken off one by one, the process
could be considered quasi-equilibrium. However, if all the weights were removed at once,
the piston would rise rapidly until it hit the stops. This would be a nonequilibrium process,
and the system would not be in equilibrium at any time during this change of state.
For nonequilibrium processes, we are limited to a description of the system before
the process occurs and after the process is completed and equilibrium is restored. We are
unable to specify each state through which the system passes or the rate at which the
process occurs. However, as we will see later, we are able to describe certain overall ef-
fects that occur during the process.
Several processes are described by the fact that one property remains constant.
The prefix iso- is used to describe such a process. An isothermal process is a constant-
temperature process, an isobaric (sometimes called isopiestic) process is a constant-pressure
process, and an isochoric process is a constant-volume process.
Wherijtjsjfstorunjy^ of different changes of
state or prwej>sj^sjmrlj^ system has undergone a cycle.
Therefore, at the conclusion of a cycle, all the propertiesliave The same" value they had at
the beginning. Steam (water) that circulates through a steam power plant undergoes a cycle.
A distinction should be made between a thermodynamic cycle, which has just been
described, and a mechanical cycle. A four-stroke-cycle internal-combustion engine goes
through a mechanical cycle once every two revolutions. However, the working fluid does
not go through a thermodynamic cycle in the engine, since air and fuel are burned and
changed to products of combustion that are exhausted to the atmosphere. In this text the
term cycle will refer to a thermodynamic "cycle" unless otherwise designated.
2.5 Units for Mass, Length,
time, and force
Since we are considering thermodynamic properties from a macroscopic perspective, we
are dealing with quantities that can, either directly or indirectly, be measured and counted.
Therefore, the matter of units becomes an important consideration. In the remaining sec-
tions of this chapter we will define certain thermodynamic properties and the basic units.
UNITS FOR MASS, LENGTH, TIME, AND FORCE H 19
Because the relation between force and mass is often a difficult matter for students, it is
considered in this section in some detail.
Force, mass, length, and time are related by Newton's second law of motion, which
states that the force acting on a body is proportional to the product of the mass and the ac-
celeration in the direction of the force:
F cc ma
The concept of time is well established. The basic unit of time is the second (s),
which in the past was defined in terms of the solar day, the time interval for pne complete
revolution of the earth relative to the sun. Since this period varies with the season of the
year, an average value over a one-year period is called the mean solar day, and the mean
solar second is 1/86 400 of the mean solar day. (The measurement of the earth's rotation
is sometimes made relative to a fixed star, in which case the period is called a sidereal
day.) In 1967, the General Conference of Weights and Measures (CGPM) adopted a defi-
nition of the second as the time required for a beam of cesium-133 atoms to resonate
9 192 631 770 cycles in a cesium resonator.
For periods of time less than a second, the prefixes milli, micro, nano, or pico, as
listed in Table 2.1, are commonly used. For longer periods of time, the units minute
(min), hour (h) f or day (day) are frequently used. It should be pointed out that the prefixes
in Table 2.1 are used with many other units as well.
The concept of length is also well established. The basic unit of length is the meter
(m). For many years the accepted standard was the International Prototype Meter, the dis-
tance between two marks on a platinum-indium bar under certain prescribed conditions.
This bar in maintained at the International Bureau of Weights and Measures, in Sevres,
France. In I960, the CGPM adopted a definition of the meter as a length equal to
1 650 763.73 wavelengths in a vacuum of the orange-red line of krypton-86. Then in
1983, the CGPM adopted a more precise definition of the meter in terms of the speed of
light (which is now a fixed constant): The meter is the length of the path traveled by light
in a vacuum during a time interval of 1/299 792 458 of a second.
The fundamental unit of mass is the kilogram (kg). As adopted by the first CGPM in
1889 and restated in 1901, it is the mass of a certain platinum-iridium cylinder main-
tained under prescribed conditions at the International Bureau of Weights and Measures.
A related unit that is used frequently in thermodynamics is the mole (mol), defined as an
amount of substance containing as many elementary entities as there are atoms in 0.012
kg of carbon- 12. These elementary entities must be specified; they may be atoms, mole-
cules, electrons, ions, or other particles or specific groups. For example, one mole of di-
atomic oxygen, having a molecular weight of 32 (compared to 12 for carbon), has a mass
of 0.032 kg. The mole is often termed a gram mole, since it is an amount of substance in
TABLE 2.1
Unit Prefixes
Factor
Prefix
Symbol
Factor
Prefix
Symbol
10 12
tera
T
10" 3
milli
m
10*
giga
G
10~ 6
micro
10 6
mega
M
I0" 9
nano
n
10 J
kilo
k
1(T 12
pico
P
20 M CHAPTER TWO SOME CONCEPTS AND DEFINITIONS
grams numerically equal to the molecular weight. In this text, when using the metric SI
system we will find it preferable to use the kilomole (kmol), the amount of substance in
kilograms numerically equal to the molecular weight, rather than the mole.
The system of units in use presently throughout most of the world is the metric
International System, commonly referred to as SI units (from Le Systeme International
d'Unites). In this system, the second, meter, and kilogram are the basic units for time,
length, and mass, respectively, as just defined, and the unit of force is defined directly
from Newton's second law.
Therefore, a proportionality constant is unnecessary, and we may write that law as
an equality:
F=ma (2.1)
The unit of force is the newton (N), which by definition is the force required to accelerate
a mass of one kilogram at the rate of one meter per second per second:
1N = lkgm/s 2
It is worth noting that SI units derived from proper nouns use capital letters for symbols;
others use the lowercase letters. The liter, with the symbol L, is an exception.
The traditional system of units used in the Untied States is the English Engineering
System. In this system the unit of time is the second, which has been discussed earlier.
The basic unit of length is the foot (ft), which at present is defined in terms of the meter as
1 ft = 0.3048 m
The inch (in.) is defined in terms of the foot
12 in. = 1 ft
The unit of mass in this system is the pound mass (Ibm). It was originally the mass of a
certain platinum cylinder kept in the Tower of London, but now it is defined in terms of
the kilogram as
1 Ibm - 0.453 592 37 kg
A related unit is the pound mole (lb mol), which is an amount of substance in pounds
mass numerically equal to the molecular weight of that substance. It is important to distin-
guish between a pound mole and a mole (gram mole).
In the English Engineering System of Units, the unit of force is the pound force
(Ibf), defined as the force with which the standard pound mass is attracted to the earth
under conditions of standard acceleration of gravity, which is that at 45° latitude and sea
level elevation, 9.806 65 m/s 2 or 32.1740 ft/s 2 . Thus, it follows from Newton's second
law, that
1 Ibf = 32.174 IbmfVs 2
which is a necessary factor for the purpose of units conversion and consistency. Note that
we must be careful to distinguish between a Ibm and a Ibf, and we do not use the term
pound alone.
The term weight is often used with respect to a body and is sometimes confused
with mass. Weight is really correctly used only as a force. When we say a body weighs so
much, we mean that this is the force with which it is attracted to the earth (or some other
body), that is, the product of its mass and the local gravitational acceleration. The mass of
a substance remains constant with elevation, but its weight varies with elevation.
Energy m 21
EXAMPLE 2.1 What is the weight of a one kg mass at an altitude where the local acceleration of gravity
is 9.75 m/s 2 ?
Solution
Weight is the force acting on the mass, which from Newton's second law is
F = mg = 1 kg X 9.75 m/s 2 X [1 N s 2 /kg m] = 9.75 N
EXAMPLE 2. IE What is the weight of a one Ibm mass at an altitude where the local acceleration of grav-
ity is 32.0 ft/s 2 ?
Solution
Weight is the force acting on the mass, which from Newton's second law is :
F = mg = 1 lbra X 32.0 ft/s 2 X [lbf s 2 /32.174 Ibm ft] === 0.9946 Ibf
2.6 ENERGY
One of the very important concepts in a study of thermodynamics is that of energy. En-
ergy is a fundamental concept, such as mass or force and, as is often the case with such
concepts, is very difficult to define. Energy has been defined as the capability to produce
an effect. Fortunately the word energy and the basic concept that this word represents are
familiar to us in everyday usage, and a precise definition is not essential at this point.
Energy can be stored within a system and can be transferred (as heat, for example)
from one system to another. In a study of statistical thermodynamics we would examine,
from a molecular view, the ways in which energy can be stored. Because it is helpful in a
study of classical thermodynamics to have some notion of how this energy is stored, a
brief introduction is presented here.
Consider as a system a certain gas at a given pressure and temperature contained
within a tank or pressure vessel. When considered from the molecular view, we identify
three general forms of energy:
/^Inter molecular p otential^ energy, which is associated with the forces between
molecules.
( 2AMolecular kineticenergy, which is associated with the translational velocity of indi-
vidual molecules.
( S ^ntramolecular energ y (that within the individual molecules), which is associated
with the molecular and atomic structure and related forces.
The first of these forms of energy, the intermolecular potential energy, depends on
the magnitude of the intermolecular forces and the position the molecules have relative to
each other at any instant of time. It is impossible to determine accurately the magnitude of
this energy because we do not know either the exact configuration and orientation of the
molecules at any time or the exact intermolecular potential function. However, there are
two situations for which we can make good approximations. The first situation is at low or
22 a Chapter Two Some Concepts and Definitions
FIGURE 2.4 The
coordinate system for a
diatomic molecule.
Vapor H 2
rs(steam). : -v-
A
Heat
FIGURE 2.5 Heat
transfer to water.
moderate densities. In this case the molecules are relatively widely spaced, so that only
two-molecule or two- and three-molecule interactions contribute to the potential energy.
At these low and moderate densities, techniques are available for" determining, with rea-
sonable accuracy, the potential energy of a system composed of reasonably simple mole-
cules. The second situation is at very low densities; under these conditions the average
intermolecular distance between molecules is so large that the potential energy may be as-
sumed to be zero. Consequently, we have in this case a system of independent particles
(an ideal gas) and, therefore, from a statistical point of view, we are able to concentrate
our efforts on evaluating the molecular translational and internal energies.
The translational energy, which depends only on the mass and velocities of the mol-
ecules, is determined by using the equations of mechanics—either quantum or classical.
The intramolecular internal energy is more difficult to evaluate because, in general, it
may result from a number of contributions. Consider a simple monatomic gas such as he-
Hum. Each molecule consists of a helium atom. Such an atom possesses electronic energy as
a result of both orbital angular momentum of the electrons about the nucleus and angular mo-
mentum of the electrons spinning on their axes. The electronic energy is commonly very
small compared with the translational energies. (Atoms also possess nuclear energy, which,
except in the case of nuclear reactions, is constant. We are not concerned with nuclear energy
at this time.) When we consider more complex molecules, such as those composed of two or
three atoms, additional factors must be considered. In addition to having electronic energy, a
molecule can rotate about its center of gravity and thus have rotational energy. Furthermore,
the atoms may vibrate with respect to each other and have vibrational energy. In some situa-
tions there may be an interaction between the rotational and vibrational modes of energy.
Consider a diatomic molecule, such as oxygen, as shown in Fig. 2.4. In addition to
translation of the molecule as a solid body, the molecule can rotate about its center of
mass in two normal directions, about the x axis and about the z axis (rotation about the y
axis is negligible), and the two atoms can also vibrate, that is, stretch the bond joining the
atoms along the y axis. A more rapid rotation increases the rotational energy, and a
stronger vibration results in an increase of vibrational energy of the molecule.
More complex molecules, such as typical polyatomic molecules, are usually three-
dimensional in structure and have multiple vibrational modes, each of which contributes
to the energy storage of the molecule. The more complicated the molecule is, the larger
the number of degrees of freedom that exist for energy storage. This subject of the modes
of energy storage and their evaluation is discussed in some detail in Appendix C, for those
interested in a further development of the quantitative effects from a molecular viewpoint.
This general discussion can be summarized by referring to Fig. 2.5. Let heat be trans-
ferred to the water. During this process the temperature of the liquid and vapor (steam) will
increase, and eventually all the liquid will become vapor. From the macroscopic view we
SPECIFIC VOLUME AND DENSITY H 23
are concerned only with the energy that is transferred as heat, the change in properties,
such as temperature and pressure, and the total amount of energy (relative to some base)
that the H 2 contains at any instant. Thus, questions about how energy is stored in the H 2
do not concern its. From a microscopic viewpoint we are concerned about the way in
which energy is stored in the molecules. We might be interested in developing a model of
the molecule so that we could predict the amount of energy required to change the temper-
ature a given amount. Although the focus in this book is on the macroscopic or classical
viewpoint, it is helpful to keep in mind the microscopic or statistical perspective as well, as
the relationship between the two helps us in understanding basic concepts such as energy.
2.7 Specific Volume and Density
The specific volume of a substance is defined as the volume per unit mass and is given the
symbol v. The density of a substance is defined as the mass per unit volume, and it is
therefore the reciprocal of the specific volume. Density is designated by the symbol p.
Specific volume and density are intensive properties.
The specific volume of a system in a gravitational field may vary from point to
point. For example, if the atmosphere is considered a system, the specific volume in-
creases as the elevation increases. Therefore, the definition of specific volume involves
the specific volume of a substance at a point in a system.
Consider a small volume 8V of a system, and let the mass be designated Sm. The
specific volume is defined by the relation
v = lim
sv-^sv om
where 5 V is the smallest volume for which the mass can be considered a continuum: Vol-
umes smaller than this will lead to the recognition that mass is not evenly distributed in
space but is concentrated in particles as molecules, atoms, electrons, etc. This is tentatively
indicated in Fig. 2.6, where in the limit of a zero volume the specific volume may be infi-
nite (the volume does not contain any mass) or very small (the volume is part of a nucleus).
FIGURE 2.6 The
continuum limit for the
specific volume.
24 H Chapter two Some Concepts and definitions
.Gases
Gas in
vacuum
Aim.
air
_l I i ' r 1 1 1 il
' i < i <
Solids
Fiber Wood
Cotton
Wool
A!
Ice
Rock
Lead
Ag Au
Propane Water
Hg
_l i ' T 1 I I I li
FIGURE 2.7 Density
of common substances.
10"
10°
10 1 10 2
Density [kg/m 3 ]
j i '
10 3
10 4
Thus, in a given system, we should speak of the specific volume or density at a
point in the system, and recognize that this may vary with elevation. However, most of
the systems that we consider are relatively small, and the change in specific volume with
elevation is not significant. Therefore, we can speak of one value of specific volume or
density for the entire system.
In this text, the specific volume and density will be given either on a mass or on a mole
basis. A bar over the symbol (lowercase) will be used to designate the property on a mole
basis. Thus, v will designate molal specific volume and p will designate the molal density. In
SI units, those for specific volume are m 3 /kg and mVmol (or mVkmol); for density the corre-
sponding units are kg/m 3 and mol/m 3 (or kmol/m 3 ). In English units, those for specific vol-
ume are ftVlbm and fWlb mol; the corresponding units for density are lbm/ft 3 and lb mol/ft 3 .
Although the SI unit for volume is the cubic meter, a commonly used volume unit
is the liter (L), which is a special name given to a volume of 0.001 cubic meters, that is,
1 L = 10~ 3 m 3 . The general ranges of density for some common solids, liquids, and
gases are shown in Fig. 2.7. Specific values for various solids, liquids and gases in SI
units are listed in Tables A.3, A.4, and A.5, respectively, and in English units in Tables
F.2, F.3, and F.4.
EXAMPLE 2.2 A 1 -m 3 container, Fig. 2.8, is filled with 0. 12 m 3 of granite, 0. 1 5 m 3 of sand, 0.2 m 3 of liq-
uid 25°C water, and the rest of the volume, 0.53 m 3 , is air with a density of 1.15 kg/m 3 .
Find the overall (average) specific volume and density.
Solution
From the definition of specific volume and density we have:
u = Vim and p = mfV= 1/v
We need to find the total mass taking density from Tables A.3 and A.4
'"granite = pfWte = 2750 kg/m 3 X 0. 12 m 3 = 330 kg
*H«nd = Pzmd Vs*a = 15{ W X 0.15 m 3 = 225 kg
»W = JtWr ^wito = 997 kg/m 3 X 0.2 m 3 = 199.4 kg
™& = Pair ^air = 1.15 kg/m 3 X 0.53 m 3 = 0.61 kg
pressure BO 25
Remark: It is misleading to include air in the
numbers for p and V, as the air is separate
from the rest of the mass.
FIGURE 2.8 Sketch for Example 2.2.
Now the total mass becomes
m M = "Igrante + ™sand + '"water + m air ~ 755 kg
and the specific volume and density can be calculated
v = VJm tot = 1 m 3 /755 kg = 0.001325 m 3 /kg
P = mJVn. = 755 kg/1 m 3 = 755 kg/m 3
2.8 Pressure
When dealing with liquids and gases, we ordinarily speak of pressure; for solids we speak
of stresses. The pressure in a fluid at rest at a given pointjsJhe same in all direction's, and
we define pressure as t,he ji~6rmalT:6^^ per unit^area\ More specifically, if
S A is a small area, 8 A r the smallest area over whichwe can consider the fluid a contin-
uum, and 8F n the component of force normal to SA, we define pressure, P, as
where the lower limit corresponds to sizes as mentioned for the specific volume, shown in
Fig. 2.6. The pressure P at a point in a fluid in equilibrium is the same in all directions. In
a viscous fluid in motion, the variation in the state of stress with orientation becomes an
important consideration. These considerations are beyond the scope of this book, and we
will consider pressure only in terms of a fluid in equilibrium.
The unit for pressure in the International System is the force of one newton acting
on a square meter area, which is called the pascal (Pa). That is,
Two other units, not part of the International System, continue to be widely used,
These are the bar, where
P = lim
SA-^A' SA
1 Pa = 1 N/m 2
1 bar= 10 s Pa
0.1 MPa
and the standard atmosphere, where
1 atm= 101 325 Pa
14.696 lbf/in 2
26 si Chapter two some concepts and definitions
FIGURE 2.9 The
balance of forces on a
movable boundary relates
to inside gas pressure.
Gas 7*
which is slightly larger than the bar. In this text, we will normally use the SI unit, the pas-
cal, and especially the multiples of kilopascal and megapascal. The bar will be utilized
often in the examples and problems, but the atmosphere will not be used, except in speci-
fying certain reference points.
Consider a gas contained in a cylinder fitted with a movable piston, as shown in
Fig. 2.9. The pressure exerted by the gas on all its boundaries is the same, assuming that
the gas is in an equilibrium state. This pressure is fixed by the external force acting on
the piston, since there must be a balance of forces for the piston to remain stationary.
Thus, the product of the pressure and the movable piston area must be equal to the exter-
nal force. If the external force is now changed, in either direction, the gas pressure inside
must accordingly adjust, with appropriate movement of the piston, to establish a force
balance at a new equilibrium state. As another example, if the gas in the cylinder is
heated by an outside body, which tends to increase the gas pressure, the piston will move
instead, such that the pressure remains equal to whatever value is required by the exter-
nal force.
EXAMPLE ^3 The hydraulic piston/cylinder system shown in Fig. 2.10 has a cylinder diameter of
D = 0.1 m with a piston and rod mass of 25 kg. The rod has a diameter of 0.01 m with
an outside atmospheric pressure of 101 kPa. The inside hydraulic fluid pressure is 250
kPa. How large a force can the rod push within the upward direction?
Solution
We will assume a static balance of forces on the piston (positive upward) so
F nit = ma =
= ^cy]4y[ ~ -fflC^cyl ~ ^rod) ~F~ Ttl p g
"rod
>'o O
tl l l 1 I ill
!l 1 1 1 1 Hi
FIGURE 2,10 Sketch for Example 2.3.
Pressure U 27
solve for F
F = P cyl A cyl - P Q (A cyl - A lo3 ) - m p g
The areas are:
A cyl = m 2 = ttDV4 = 1 0.1 2 m 2 - 0.007 854 m 2
^ rod = -irt 2 = ttD 2 IA = -0.0 l 2 m 2 = 0.000 078 54 m 2 \
So the force becomes
F - [250 X 0.007 854 - 101(0.007 854 - 0.000 078 54)] 1000 - 25 X 9.81
- 1963.5 - 785.32 - 245.25
- 932.9 N
Note that we must convert kPa to Pa to get units of N.
In most thermodynamic investigations we are concerned with absolute pressure
Most pressure and vacuum gauges, however, read the difference between the abso (CrtS)
pressure and the atmospheric pressure existing at the gauge. This is referred to as^gjtng
pressure. This is shown graphically in Fig. 2.1 1, and the following examples illustrate the
principles. Pressures below atmospheric and slightly above atmospheric, and pressure dif-
ferences (for example, across an orifice in a pipe), are frequently measured with a
manometer, which contains water, mercury, alcohol, oil, or other fluids.
Pahs !> - -
FIGURE 2.11
Illustration of terms used
in pressure measurement.
Ordinary pressure gauge
Ordinary vacuum gauge
Barometer reads
atmospheric pressure
28 H CHAPTER TWO SOME CONCEPTS AND DEFINITIONS
Fluid
P
FIGURE 2.12
Example of pressure
measurement using a
column of fluid.
1
Pufrn - Pa
—B
Consider the column of fluid of height H standing above point B in the manometer
shown in Fig. 2. 12. The force acting downward at the bottom of the column is
PoA + mg = PqA+ pAgH
where m is the mass of the fluid column, A is its cross-sectional area, and p is its density.
This force must be balanced by the upward force at the bottom of the column, which is
P&4. Therefore,
Pb-Pq = PgH
Since points A and B are at the same elevation in columns of the same fluid, their pres-
sures must be equal (the fluid being measured in the vessel has a much lower density,
such that its pressure P is equal to P A ). Overall,
"&P-P-P,^pgH J (2.2)
For distinguishing between absolute and gauge pressure in this text, the term pascal
will always refer to absolute pressure. Any gauge pressure will be indicated as such.
EXAMPLE A mercury (Hg) manometer is used to measure the pressure in a vessel as shown in Fig.
\/ 2.12. The mercury has a density of 13 590 kg/m 3 , and the height difference between the
two columns is measured to be 24 cm. We want to determine the pressure inside the vessel.
Solution
The manometer measures the gauge pressure as a pressure difference. From Eq. 2.2/
Ai 3 = F gsuge - pgH = 13 590 X 9.806 65 X 0.24
- 31 985 ^|^m = 31 985 Pa = 31.985 kPa
m s . —
= 0.316 atm
To get the absolute pressure inside the vessel we have
We need to know the atmospheric pressure measured by a barometer (absolute pres-
sure). Assume this pressure is known as 750 mm Hg, being measured with a setup sirni-
PRESSURE S 29
lar to the one in Fig. 2. 12 with one side open to the atmosphere and the other side closed
so there is mercury vapor with a very small pressure on top of the liquid column. The
absolute pressure in the vessel becomes
= AP + P tta = 31 985 + 13 590 X 0.750 X 9.806 65
- 31 985 + 99 954 - 131 940 Pa = 1.302 arm
EXAMPL:
A mercury (Hg) manometer is used to measure the pressure in a vessel as shown in Fig.
2.12. The mercury has a density of 848 lbm/ft 3 , and the height difference between the two
columns ts measured to be 9.5 in. We want to determine the pressure inside the vessel.
Solution
The manometer measures the gauge pressure as a pressure difference. FromEq. 2.2,
- 848 ^ X 32.174 ~ X 9,5 in. X
1 ft 3
1728 in 3
X
1 Ms 2
32.174 Ibmft
ft 3
- 4.66 Ibf/in 2
To get the absolute pressure inside the vessel we have
Pa ~ -^vessel = Pq~ + -Patm
We need to know the atmospheric pressure measured by a barometer (absolute pres-
sure). Assume this pressure is known as 29.5 in. Hg, being measured with a setup simi-
lar to the one above with one side open to the atmosphere and the other side closed so
there is mercury vapor with a very small pressure on top of the liquid column. The ab-
solute pressure in the vessel becomes
^vessd ~ AP + P ztta
1 " ' 1 1 + 4.66
= 848 X 32.174 X 29.5 X^X^4
= 19.14 Ibf/in 2
EXAMP^g^S What is the pressure at the bottom of the 7.5-m-tall storage tank of fluid at 25°C shown
in Fig. 2.13? Assume the fluid is gasoline with atmospheric pressure 101 kPa on the top
surface. Repeat the question 'for liquid refrigerant R-134a when the top surface pressure
is 1 MPa.
A Solution
The densities of the liquids are listed in Table A.4:
Pgasoii^e = 750 kg/m 3 ; p R . t34a = 1206 kg/m 3
The pressure difference due to the gravity is, from Eq. 2.2,
FIGURE 2.13 Sketch
for Example 2.5. AP = pgH
30 B chaptertwo some concepts and Definitions
The total pressure is
P = P top + AP
For the gasoline we get
AP = pgH = 750 kg/m 3 X 9.807 m/s 2 X 7.5 m = 55 164 Pa
Now convert all pressures to kPa
P= 101 + 55.164 = 156.2 kPa
For the R- 134a we get
AP = pgH = 1206 kg/m 3 X 9.807 m/s 2 X 7.5 m = 88 704 Pa
Now convert all pressures to kPa
P = 1000 + 88.704 = 1089 kPa
EXAMPLE 2.6 A piston/cylinder with cross-sectional area of 0.01 m 2 is connected with a hydraulic line
to another piston cylinder of cross-sectional area of 0.05 m 2 . Assume both chambers and
the line are filled with hydraulic fluid of density 900 kg/m 3 and the larger second pis-
ton/cylinder is 6 m higher up in elevation. The telescope arm and the buckets have hy-
draulic piston/cylinders moving them, as seen in Fig. 2.14. With an outside atmospheric
pressure of 100 kPa and a net force of 25 kN on the smallest piston, what is the balanc-
ing force on the second larger piston?
Solution
When the fluid is stagnant and at the same elevation we have the same pressure through-
out the fluid. The force balance on the smaller piston is then related to the pressure (we
neglect the rod area) as
F.+PoA^P.A, ^
from which the fluid pressure is
P, = P + F X IA\ = WO kPa + 25 kN/0.01 m 2 = 2600 kPa
J,
FIGURE 2,14 Sketch for Example 2.6.
The Zeroth Law of Thermodynamics B 31
The pressure at the higher elevation in piston/cylinder 2 is, from Eq. 2.2,
P 2 =P l ~ pgH = 2600 kPa - 900 kg/m 3 X 9.81 m/s 2 X 6 m/flOOO Pa/kPa)
where the second term is divided by 1000 to convert from Pa to kPa. Then the force bal-
ance on the second piston gives
Although temperature is a familiar property, defining it exactly is difficult. We are aware
of "temperature" first of all as a sense of hotness or coldness when we touch an object.
We also leam early that when a hot body and a cold body are brought into contact, the hot
body becomes cooler and the cold body becomes warmer. If these bodies remain in con-
tact for some time, they usually appear to have the same hotness or coldness. However,
we also realize that our sense of hotness or coldness is very unreliable. Sometimes very
cold bodies may seem hot, and bodies of different materials that are at the same tempera-
ture appear to be at different temperatures.
Because of these difficulties in defining temperature, we define equality of tempera-
ture- Consider two blocks of copper, one hot and the other cold, each of which" is in con-
tact with a mercury-in-glass thermometer. If these two blocks of copper are brought into
thermal communication, we observe that the electrical resistance of the hot block de-
creases with time and that of the cold block increases with time. After a period of time has
elapsed, however, no further changes in resistance are observed. Similarly, when the
blocks are first brought in thermal communication, the length of a side of the hot block
decreases with time, but the length of a side of the cold block increases with time. After a
period of time, no further change in length of either of the blocks is perceived. In addition,
the mercury column of the thermometer in the hot block drops at first and that in the cold
block rises, but after a period of time no further changes in height are observed. We may
say, therefore, that two bodies have equality of temperature if, when they are in thermal
communication, no change in any observable property occurs.
Now consider the same two blocks of copper and another thermometer. Let one block of
copper be brought into contact with the thermometer until equality of temperature is es-
tablished, and then remove it. Then let the second block of copper be brought into contact
with the thermometer. Suppose that no change in the mercury level of the thermometer
occurs during this operation with the second block. We then can say that both blocks are
in thermal equilibrium with the given thermometer.
The zeroth law of thermodynamics states that when two bodies have equality of
temperature with a third body, they in turn have equality of temperature with each other.
Pi = '(JPi ~ PiMi = (2547 - 100) kPa X 0.05 m 2 = 122.4 kN
2.9 Equality of temperature
2.10 THE ZEROTH LAW OF THERMODYNAMICS
32 M chapter Two Some Concepts and Definitions
This seems obvious to us because we are so familiar with this experiment. Because the
principle is not derivable from other taws, and because it precedes the first and second
laws of thermodynamics in the logical presentation of thermodynamics, it is called the ze-
roth law of thermodynamics. This law is really the basis of temperature measurement.
Every time a body has equality of temperature with the thermometer, we can say that the
body has the temperature we read on the thermometer. The problem remains of how to re-
late temperatures that we might read on different mercury thermometers or obtain from
different temperature-measuring devices, such as thermocouples and resistance ther-
mometers. This observation suggests the need for a standard scale for temperature mea-
surements.
2.11 Temperature Scales
Two scales are commonly used for measuring temperature, namely the Fahrenheit (after
Gabriel Fahrenheit, 1686-1736) and the Celsius. The Celsius scale was formerly called
the centigrade scale but is now designated the Celsius scale after Anders Celsius
(1701-1744), the Swedish astronomer who devised this scale.
The Fahrenheit temperature scale is used with the English Engineering system of
units, and the Celsius scale with the SI unit system. Until 1954 both of these scales were
based on two fixed, easily duplicated points — the ice point and the steam point. The tem-
perature of the ice point is defined as the temperature of a mixture of ice and water that is
in equilibrium with saturated air at a pressure of 1 atm. The temperature of the steam
point is the temperature of water and steam, which are in equilibrium at a pressure of 1
atm. On the Fahrenheit scale these two points are assigned the numbers 32 and 212, re-
spectively, and on the Celsius scale the points are and 100, respectively. Why Fahren-
heit chose these numbers is an interesting story. In searching for an easily reproducible
point, Fahrenheit selected the temperature of the human body and assigned it the number
96. He assigned the number to the temperature of a certain mixture of salt, ice, and salt
solution. On this scale the ice point was approximately 32. When this scale was slightly
revised and fixed in terms of the ice point and steam point, the normal temperature of the
human body was found to be 98.6 F.
In this text the symbols F and °C will denote the Fahrenheit and Celsius scales, re-
spectively. The symbol Twill refer to temperature on all temperature scales.
At the tenth CGPM in 1954, the Celsius scale was redefined in terms of a single
fixed point and the ideal-gas temperature scale. The single fixed point is the triple point
of water (the state in which the solid, liquid, and vapor phases of water exist together in
equilibrium). The magnitude of the degree is defined in terms of the ideal-gas tempera-
ture scale, which is discussed in Chapter 7. The essential features of this new scale are a
single fixed point and a definition of the magnitude of the degree. The triple point of
water is assigned the value of 0.0 1°C. On this scale the steam point is experimentally
found to be 100.00°C. Thus, there is essential agreement between the old and new tem-
perature scales.
We have not yet considered an absolute scale of temperature. The possibility of
such a scale comes from the second law of thermodynamics and is discussed in Chapter 7.
On the basis of the second law of thermodynamics, a temperature scale that is indepen-
dent of any thermometric substance can be defined. This absolute scale is usually referred
to as the thermodynamic scale of temperature. However, it is very complicated to use this
SUMMARY H 33
scale directly, and therefore, a more practical scale, the International Temperature Scale,
which closely represents the thermodynamic scale, has been adopted.
The absolute scale related to the Celsius scale is the Kelvin scale (after William
Thomson, 1824-1907, who is also known as Lord Kelvin), and is designated K (without
the degree symbol). The relation between these scales is
K = °C 4- 273.15 (2.3)
In 1967, the CGPM defined the kelvin as 1/273.16 of the temperature at the triple point of
water. The Celsius scale is now defined by this equation instead of by its earlier definition.
The absolute scale related to the Fahrenheit scale is the Rankine scale 1 ' and is desig-
nated R. The relation between these scales is
R - F + 459.67 (2.4)
A number of empirically based temperature scales, to standardize temperature mea-
surement and calibration, have been in use over the last 70 years. The most recent of these
is the International Temperature Scale of 1990, or ITS-90. It is based on a number of fixed
and easily reproducible points that are assigned definite numerical values of temperature,
and on specified formulas relating temperature to the readings on certain temperature-
measuring instruments for the purpose of interpolation between the defining fixed points.
Details of the ITS-90 are not considered further in this text. It is noted that this scale is a
practical means for establishing measurements that conform closely to the absolute ther-
modynamic temperature scale.
'UMMARY We j nrroc iuce a thermodynamic system as a control volume, which for a fixed mass is a
control mass. Such a system can be isolated, exchanging neither mass, momentum, or en-
ergy with its surroundings. A closed system versus an open system refers to the ability of
mass exchange with the surroundings. If properties for a substance change, the state
changes and a process occurs. When a substance has gone through several processes re-
turning to the same initial state it has completed a cycle.
Basic units for thermodynamic and physical properties are mentioned and most are
covered in Table A.l. Thermodynamic properties such as density p, specific volume v,
pressure P, and temperature Tare introduced together with units for these. Properties are
classified as intensive, independent of mass (like u), or extensive, proportional to mass
(like V). Students should already be familiar with other concepts from physics such as
force F, velocity V, and acceleration a. Application of Newton's law of motion leads to
the variation of the static pressure in a column of fluid and the measurements of pressure
(absolute and gauge) by barometers and manometers. The normal temperature scale and
the absolute temperature scale are introduced.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Define (choose) a control volume (C.V.) around some matter; sketch the content
and identify storage locations for mass; and identify mass and energy flows crossing
the C.V. surface.
• Know properties P,T,v, and p and their units.
• Know how to look up conversion of units in Table A.l.
• Know that energy is stored as kinetic, potential, or internal (in molecules).
34 H Chapter Two Some Concepts and Definitions
• Know that energy can be transferred.
• Know the difference between (v, p) and (V t m) intensive versus extensive.
• Apply a force balance to a given system and relate it to pressure P.
• Know the difference between a relative (gauge) and absolute pressure P.
• Understand the working of a manometer or a barometer and get AP or P from
heights.
• Know the difference between a relative and absolute temperature T.
• Have an idea about magnitudes (v, p, P, J).
Most of these concepts will be repeated and reinforced in the following chapters such as
properties in Chapter 3, energy transfer as heat and work in Chapter 4, and internal energy
in Chapter 5 together with their applications.
KEY CONCEPTS ControI v0 l U me everything inside a control surface
and Formulas '
Pressure definition P = — (mathematical limit for small A)
Specific volume
V
m
Density p = (Tables A.3, A.4, A.5, F.2, F.3, and F.4)
Static pressure variation AP = pgH (depth H in fluid of density p)
Absolute temperature 7[K] = T[°C] + 273. 15
T[R] = T[F) + 459.67
Units Table A, 1
Concepts from Physics
Newton's law of motion F = ma
Acceleration a =^ = ™
dt 2 dt
Velocity V = ^
dt
Concept-study Guide Problems
2.1 Make a control volume (CV.) around the turbine
in the steam power plant in Fig. 1.1 and list the
flows of mass and energy that are there.
2.2 Make a control volume around the whole power
plant in Fig. 1.2 and with the help of Fig. 1.1 list
what flows of mass and energy are in or out and
any storage of energy. Make sure you know what is
inside and what is outside your chosen CV.
2.3 Make a control volume that includes the steam
flow around in the main turbine loop in the nuclear
propulsion system in Fig. 1.3. Identify mass flows
(hot or cold) and energy transfers that enter or
leave the CV.
2.4 Take a control volume around your kitchen refrig-
erator and indicate where the components shown in
Fig. 1.6 are located and show all flows of energy
transfer.
2.5 An electric dip heater is put into a cup of water and
heats it from 20°C to 80°C. Show the energy
flow(s) and storage and explain what changes.
Homework Problems U 35
2.6 Separate the list P, F s V, v, p, T, a, m, Z, t, and V
into intensive, extensive, and nonproperttes.
2.7 An escalator brings four people of total mass 300
kg, 25 m up in a building. Explain what happens
with respect to energy transfer and stored energy.
2.8 Water in nature exists in different phases — solid,
liquid, and vapor (gas). Indicate the relative magni-
tude of density and specific volume for the three
phases.
2.9 Is density a unique measure of mass distribution in
a volume? Does it vary? If so, on what kind of
scale (distance)?
2.10 Density of fibers, rock wool insulation, foams and
cotton is fairly low. Why is that?
2.11 How much mass is there approximately in 1 L of
mercury (Hg)? Atmospheric air?
2.12 Can you carry 1 m 3 of liquid water?
2.13 A manometer shows a pressure difference of 1 m
of liquid mercury. Find AP in kPa.
2.14 You dive 5 m down in the ocean. What is the ab-
solute pressure there?
Homework Problems
Properties and Units
2.23 A steel cylinder of mass 2 kg contains 4 L of liquid
water at 25°C at 200 kPa. Find the total mass and
volume of the system. List two extensive and three
intensive properties of the water.
2.24 An apple "weighs" 80 g and has a volume of 100
cm 3 in a refrigerator at 8°C. What is the apple den-
sity? List three intensive and two extensive proper-
ties of the apple.
2.25 One kilopond (1 kp) is the weight of 1 kg in the
standard gravitational field. How many Newton's
(N) is that?
2.26 A pressurized steel bottle is charged with 5 kg of
oxygen gas and 7 kg of nitrogen gas. How many
kmoles are in the bottle?
Force and Energy
2.27 The "standard" acceleration (at sea level and 45°
latitude) due to gravity is 9.806 65 m/s a . What is
the force needed to hold a mass of 2 kg at rest in
2.15 What pressure difference does a 10-m column of
atmospheric air show?
2.16 The pressure at the bottom of a swimming pool is
evenly distributed. Suppose we look at a cast-iron
plate of 7272 kg lying on the ground with an area
of 100 m 2 . What is the average pressure below
that? Is it just as evenly distributed?
2.17 A laboratory room keeps a vacuum of 0.1 kPa.
What net force does that put on the. door of size
2 m by 1 m?
2.18 A tornado rips off a 100-m 2 roof with a mass of
1000 kg. What is the minimum vacuum pressure
needed to do that if we neglect the anchoring forces?
2.19 What is a temperature of -5°C tn degrees Kelvin?
2.20 What is the smallest temperature in degrees Celsius
you can have? Kelvin?
2.21 Density of liquid water is p = 1008 - 772 [kg/m 3 ]
with Tin °C. If the temperature increases 10°C how
much deeper does a 1-m layer of water become?
2.22 Convert the formula for water density in Problem
2.21 to be for Tin Kelvin.
this gravitational field? How much mass can a
force of 1 N support?
2.28 A force of 125 N is applied to a mass of 12 kg in
addition to the standard gravitation. If the direction
of the force is vertical up, find the acceleration of
the mass.
2.29 A model car rolls down an incline with a slope
such that the gravitational "pull" in the direction of
motion is one-third of the standard gravitational
force (see Problem 2.27). If the car has a mass of
0.45 kg, find the acceleration.
2.30 When you move up from the surface of the earth,
the gravitation is reduced as g = 9.807 — 3.32 X
10~ 6 z, with z as the elevation in meters. By how
many percent is the weight of an airplane reduced
when it cruises at 1 1 000 m?
2.31 A car is driven at 60 km/h and is brought to a full
stop with constant deceleration in 5 s. If the total
car and driver mass is 1 075 kg, find the necessary
force.
2.32 A car of mass 1775 kg travels with a velocity of
100 km/h. Find the kinetic energy. How high
36 H CHAPTER TWO SOME CONCEPTS AND DEFINITIONS
should the car be lifted in the standard gravitational
field to have a potential energy that equals the ki-
netic energy?
2.33 A 1200 kg car moving at 20 km/h is accelerated at
a constant rate of 4 m/s 2 up to a speed of 75 km/h.
What are the force and total time required?
2.34 A steel plate of 950 kg accelerates from rest at 3
m/s 2 for a period of 10 s. What force is needed and
what is the final velocity?
2.35 A 15-kg steel container has 1.75 kmole of liquid
propane inside. A force of 2 kN now accelerates
this system. What is the acceleration?
2.36 A bucket of concrete of total mass 200 kg is raised
by a crane with an acceleration of 2 m/s 2 relative to
the ground at a location where the local gravita-
tional acceleration is 9.5 m/s 2 . Find the required
force.
2.37 On the moon the gravitational acceleration is ap-
proximately one-sixth that on the surface of the
earth. A 5-kg mass is "weighed" with a beam bal-
ance on the surface of the moon. What is the ex-
pected reading? If this mass is weighed with a
spring scale that reads correctly for standard
gravity on earth (see Problem 2.27), what is the
reading?
Specific Volume
2.38 A 5-m 3 container is filled with 900 kg of granite
(density of 2400 kg/m 3 ) and the rest of the volume is
air with density equal to 1.15 kg/m 3 . Find the mass
of air and the overall (average) specific volume.
2.39 A tank has two rooms separated by a membrane.
Room A has 1 kg of air and a volume of 0.5 m 3 ;
room B has 0.75 m 3 of air with density 0.8 kg/m 3 .
The membrane is broken and the air comes to a
-uniform state. Find the final density of the air.
A 1-m 3 container is filled with 400 kg of granite
stone, 200 kg of dry sand, and 0.2 m 3 of liquid 25°C
water. Use properties from Tables A.3 and A.4.
Find the average specific volume and density of the
masses when you exclude air mass and volume.
A 1-m 3 container is filled with 400 kg of granite
stone, 200 kg of dry sand, and 0.2 m 3 of liquid
25°C water. Use properties from Tables A.3 and
A.4 and use an air density of 1.1 kg/m 3 . Find the
average specific volume and density of the 1-m 3
volume.
2.40
2.41
2.42 One kilogram of diatomic oxygen (0 2 , molecular
weight of 32) is contained in a 500-L tank. Find the
specific volume on both a mass and mole basis (u
and v).
2 A3 A 15-kg steel gas tank holds 300 L of liquid gaso-
line, having a density of 800 kg/m 3 . If the system is
decelerated at 6 m/s 2 what is the needed force?
Pressure
2.44 A hydraulic lift has a maximum fluid pressure of
500 kPa. What should the piston/cylinder diameter
be so it can lift a mass of 850 kg?
2.45 A piston/cylinder with a cross-sectional area of
0.01 m 2 has a piston mass of 100 kg resting on the
stops, as shown in Fig. P2.45. With an outside at-
mospheric pressure of 100 kPa, what should the
water pressure be to lift the piston?
FIGURE P2.45
2.46 A vertical hydraulic cylinder has a 125-mm-diame-
ter piston with hydraulic fluid inside the cylinder
and an ambient pressure of 1 bar. Assuming stan-
dard gravity, find the piston mass that will create a
pressure inside of 1500 kPa.
2.47 A valve in the cylinder shown in Fig. P2.47 has a
cross-sectional area of 1 1 cm 2 with a pressure of
735 kPa inside the cylinder and 99 kPa outside.
How large a force is needed to open the valve?
FIGURE P2.47
homework Problems M 37
2.48 A cannnonball of 5 kg acts as a piston in a cylinder
of 0.15 m diameter. As the gunpowder is burned, a
pressure of 7 MPa is created in the gas behind the
ball. What is the acceleration of the ball if the
cylinder (cannon) is pointing horizontally?
2.49 Repeat the previous problem for a cylinder (cannon)
pointing 40° up relative to the horizontal direction.
2.50 A large exhaust fan in a laboratory room keeps the
pressure inside at 10 cm of water relative vacuum
to the hallway. What is the net force on the door
measuring 1.9 m by 1.1 m?
2.51 What is the pressure at the bottom of a 5-m-talI
column of fluid with atmospheric pressure of 101
kPa on the top surface if the fluid is
a. water at 20°C
b. glycerine at 25°C
c. light oil
2.52 The hydraulic lift in an auto-repair shop has a
cylinder diameter of 0.2 m. To what pressure
should the hydraulic fluid be pumped to lift 40 kg
of piston/arms and 700 kg of a car?
2.53 A 2. 5 -m- tall steel cylinder has a cross-sectional
area of 1.5 m 2 . At the bottom with a height of 0.5
m is liquid water on top of which is a 1-m-high
layer of gasoline. This is shown in Fig. P2.53. The
gasoline surface is exposed to atmospheric air at
101 kPa. What is the highest pressure in the water?
1 m
0.5 m
1_
Air
Gasoline
2.5 m
FIGURE P2.53
2.54 At the beach, atmospheric pressure is 1025 mbar.
You dive 15 m down in the ocean and you later
climb a hill up to 250 m in elevation. Assume the
density of water is about 1000 kg/m 3 and the den-
sity of air is 1.18 kg/m 3 . What pressure do you feel
at each place?
2.55 A piston, m p = 5 kg, is fitted in a cylinder, A = \5
. cm 2 , that contains a gas. The setup is in a cen-
trifuge that creates an acceleration of 25 m/s 2 in the
direction of piston motion toward the gas. Assum-
ing standard atmospheric pressure outside the
cylinder, find the gas pressure.
2.56 A steel tank of cross-sectional area 3 m 2 and 1 6 m
tall weighs 10 000 kg and is open at the top as
shown in Fig. P2.56, We want to float it in the
ocean so it sticks 10 m straight down by pouring
concrete into the bottom of it. How much concrete
should we put in?
Air
Ocean
10 m
Concrete
FIGURE P2.56
2.57 Liquid water with density p is filled on top of a thin
piston in a cylinder with cross-sectional area A and
total height H, as shown in Fig. P2.57. Ah is let in
under the piston so it pushes up, spilling the water
over the edge. Deduce the formula for the air pres-
sure as a function of the piston elevation from the
bottom, h.
h i
-(g) Air
FIGURE P2.S7
Manometers and Barometers
2.58 The density of atmospheric air is about 1.15 kg/m 3 ,
which we assume is constant. How large an ab-
solute pressure will a pilot see when flying 1500 m
above ground level where the pressure is 101 kPa?
38 B Chapter Two Some Concepts and Definitions
2.59 A differential pressure gauge mounted on a vessel
shows 1.25 MPa and a local barometer gives at-
mospheric pressure as 0.96 bar. Find the absolute
pressure inside the vessel.
2.60 Two vertical cylindrical storage tanks are full of
liquid water (density = 1000 kg/m 3 ) with the top
open to the atmosphere. One is 10 m tall and 2 m in
diameter; the other is 2.5 m tall with diameter 4 m.
What is the total force form the bottom of each
tank to the water and what is the pressure at the
bottom of each tank?
2.61 Blue manometer fluid of density 925 kg/m 3 shows a
column height difference of 3 cm vacuum with one
end attached to a pipe and the other open to P Q =
101 kPa. What is the absolute pressure in the pipe?
2.62 The absolute pressure in a tank is 85 kPa and the
local ambient absolute pressure is 97 kPa. If a U-
tube with mercury (density = 13 550 kg/m 3 ) is at-
tached to the tank to measure the vacuum, what
column height difference would it show?
2.63 The pressure gauge on an air tank shows 75 kPa
when the diver is 10 m down in the ocean. At what
depth will the gauge pressure be zero? What does
that mean?
2.64 A submarine maintains 101 kPa inside it and dives
240 m down m the ocean having an average den-
sity of 1030 kg/m 3 . What is the pressure difference
between the inside and the outside of the subma-
rine hull?
2.65 A barometer to measure absolute pressure shows
a mercury column height of 725 mm. The tem-
perature is such that the density of the mercury is
13 550 kg/m 3 . Find the ambient pressure.
2.66 An absolute pressure gauge attached to a steel cylin-
der shows 135 kPa. We want to attach a manometer
using liquid water a day that P atm =101 kPa. How
high a fluid level difference must we plan for?
2.67 The difference in height between the columns of a
manometer is 200 mm with a fluid of density 900
kg/m 3 . What is the pressure difference? What is the
height difference if the same pressure difference is
measured using mercury (density = 13 600 kg/m 3 )
as manometer fluid?
2.68 An exploration submarine should be able to go
4000 m down in the ocean. If the ocean density is
1020 kg/m 3 what is the maximum pressure on the
submarine hull?
2.69 Assume we use a pressure gauge to measure the air
pressure at street level and at the roof of a tall
building. If the pressure difference can be deter-
mined with an accuracy of 1 mbar (0.001 bar),
what uncertainty in the height estimate does that
correspond to?
2.70 A U-tube manometer filled with water (density —
1000 kg/m 3 ) shows a height difference of 25 cm.
What is the gauge pressure? If the right branch is
tilted to make an angle of 30° with the horizontal, as
shown in Ftg. P2.70, what should the length of the
column in the tilted tube be relative to the U-tube?
FIGURE P2.70
2.71 A barometer measures 760 mm Hg at street level
and 735 mm Hg on top of a building. How tall is
the building if we assume air density of 1.15 kg/m 3 ?
2.72 A piece of experimental apparatus, Fig. P2.72, is
located where g = 9.5 m/s 2 and the temperature is
5°C. An air flow inside the apparatus is determined
by measuring the pressure drop across an orifice
with a mercury manometer (see Problem 2.77 for
density) showing a height difference of 200 mm.
What is the pressure drop in kPa?
Air
FIGURE P2.72
2.73 Two piston/cylinder arrangements, A and B, have
their gas chambers connected by a pipe, as shown
in Fig. P2.73. Cross-sectional areas are A A —
75 cm 2 and A B = 25 cm 2 , with the piston mass in
A being m A = 25 kg. Assume the outside pressure
is 100 kPa and standard gravitation. Find the
HOMEWORK PROBLEMS M 39
mass m B so that none of the pistons have to rest
on the bottom.
FIGURE P2.73
2.74 Two hydraulic piston/cylinders are of the same size
and setup as in Problem 2.73, but with negligible
piston masses. A single point force of 250 N
presses down on piston A. Find the needed extra
force on piston B so that none of the pistons have
to move.
2.75 A pipe flowing light oil has a manometer attached
as shown in Fig. P2.75. What is the absolute pres-
sure in the pipe flow?
P = tol kPa
0.7 m
0.1 m
FIGURE P2.75
2.76 Two cylinders are filled with liquid water, p =
1000 kg/m 3 , and connected by a line with a closed
valve, as shown in Fig. P2.76. A has 100 kg and B
FIGURE P2.76
has 500 kg of water, their cross-sectional areas
- are A A = 0. 1 m 2 and A B — 0.25 m 2 , and the height
h is 1 m. Find the pressure on each side of the
valve. The valve is opened and water flows to an
equilibrium. Find the final pressure at the valve
location.
Temperature
2.77 The density of mercury changes appr6ximately lin-
early with temperature as
p Hg = 13 595 - 2.5 Tkg/m 3 (Tin Celsius)
so the same pressure difference will result in a
manometer reading that is influenced by tempera-
ture. If a pressure difference of 100 kPa is mea-
sured in the summer at 35°C and in the winter at
~15 C } what is the difference in column height be-
tween the two measurements?
2.78 A mercury thermometer measures temperature by
measuring the volume expansion of a fixed mass
of liquid Hg due to a change in the density (see
Problem 2.77). Find the relative change (%) in
volume for a change in temperature from 10°C
to 20°C.
2.79 Using the freezing and boiling point temperatures
for water in both Celsius and Fahrenheit scales, de-
velop a conversion formula between the scales.
Find the conversion formula between Kelvin and
Rankine temperature scales.
2.80 The atmosphere becomes colder at higher eleva-
tion. As an average the standard atmospheric ab-
solute temperature can be expressed as r atm =
288 — 6.5 X 10~ 3 z, where z is the elevation
in meters. How cold is it outside an airplane
cruising at 12 000 m expressed in Kelvin and in
Celsius.
Review Problems
2.81 Repeat Problem 2.72 if the flow inside the appara-
tus is liquid water (p = 1000 kg/m 3 ) instead of air.
Find the pressure difference between the two holes
flush with the bottom of the channel. You cannot
neglect the two unequal water columns.
2.82 The main waterline into a tall building has a pres-
sure of 600 kPa at 5 m elevation below ground
level. The building is shown in Fig. P2.82. How
much extra pressure does a pump need to add to
40 B ChapterTwo Some Concepts and Definitions
ensure a water line pressure of 200 kPa at the top
floor 150 m above ground?
150 m
'''' "A
5 m
Top floor
Ground
with air at 125 kPa over the water surface. This is
illustrated in Fig. P2.84. Assuming the water den-
sity is 1000 kg/m 3 and standard gravity, find the
pressure required to pump more water in at ground
level.
2.85 Two cylinders are connected by a piston, as shown
in Fig. P2.85. Cylinder^ is used as a hydraulic lift
and pumped up to 500 kPa. The piston mass is 25
kg and there is standard gravity. What is the gas
pressure in cylinder B7
Water main
L_VL L-
Pump
FIGURE P2.82
2.83 A 5-kg piston in a cylinder with diameter of 100 mm
is loaded with a linear spring and the outside atmos-
pheric pressure of 100 kPa as shown in Fig. P2.83.
The spring exerts no force on the piston when it is at
the bottom of the cylinder, and for the state shown,
the pressure is 400 kPa with volume 0.4 L. The
valve is opened to let some air in, causing the piston
to rise 2 cm. Find the new pressure.
Air
supply
line
D< = 100 mm
FIGURE P2.85
2.86 A dam retains a lake 6 m deep, shown in Fig. P2.86.
To construct a gate in the dam we need to know the
net horizontal force on a 5-m-wide and 6-m-tall
port section that then replaces a 5-m section of the
dam. Find the net horizontal force from the water
on one side and air on the other side of the port.
FIGURE P2.83
2.84 In the city water tower, water is pumped up to a
level 25 m above ground in a pressurized tank
:-=h 6 m
Side view
=3 Lakec=3t3^
5 m
Top view
FIGURE P2.86
English unit Problems H 41
English Unit Problems
English Unit Concept Problems
2.87E A mass of 2 lbm has acceleration of 5 ft/s 2 .
What is the needed force in lbf?
2.88E How much mass is in 0.25 gal of liquid mercury
(Hg)? Atmospheric air?
2.89E Can you easily carry a 1-gal bar of solid gold?
2.90E What is the temperature of —5 F in degrees
Rankine?
2.91E What is the smallest temperature in degrees
Fahrenheit you can have? Rankine?
English Unit Problems
2.92E An apple weighs 0.2 lbm and has a volume of 6
in 3 in a refrigerator at 38 F. What is the apple
density? List three intensive and two extensive
properties for the apple.
2.93E A 2500- lbm car moving at 15 mi/h is acceler-
ated at a constant rate of 15 ft/s 2 up to a speed
of 50 mi/h. What are the force and total time
required?
2.94E Two pound moles of diatomic oxygen gas are
enclosed in a 20-lbm steel container. A force of
2000 lbf now accelerates this system. What is
the acceleration?
2.95E A valve in a cylinder has a cross-sectional area
of 2 in 2 with a pressure of 1 00 psia inside the
cylinder and 14.7 psia outside. How large a
force is needed to open the valve?
2.96E One pound mass of diatomic oxygen (0 2 molec-
ular weight 32) is contained in a 100-gal tank.
Find the specific volume on both a mass and
mole basis (v and v).
2.97E A 30-lbm steel gas tank holds 10 ft 3 of liquid
gasoline having a density of 50 lbm/ft 3 . What
force is needed to accelerate this combined sys-
tem at a rate of 15 ft/s 2 ?
2.98E A laboratory room keeps a vacuum of 4 in. of
water due to the exhaust fan. What is the net
force on a door of size 6 ft by 3 ft?
2.99E A 7-ft-m tall steel cylinder has a cross-sectional
area of 15 ft 2 . At the bottom, with a height of 2
ft, is liquid water on top of which is a 4-ft-high
layer of gasoline. The gasoline surface is ex-
posed to atmospheric air at 14.7 psia. What is
the highest pressure in the water?
2.100E A U-tube manometer filled with water, density
62.3 lbm/ft 3 , shows a height difference of 10 in.
What is the gauge pressure? If the right branch
is tilted to make an angle of 30° with the hori-
zontal, as shown in Fig. P2.70, what should the
length of the column in the tilted tube be rela-
tive to the U-tube?
2. 10 IE A piston/cylinder with cross-sectional area of
0.1 ft 2 has a piston mass of 200 lbm resting on
the stops, as shown in Fig. P2.45. With an out-
side atmospheric pressure of 1 atm, what should
the water pressure be to lift the piston?
2.102E The main waterline into a tall building has a
pressure of 90 psia at 16 ft elevation below
ground level. How much extra pressure does a
pump need to add to ensure a waterline pres-
sure of 30 psia at the top floor 450 ft above
ground?
2.103E A piston, m p = 10 lbm, is fitted in a cylinder,
A — 2.5 in 2 , that contains a gas. The setup is in a
centrifuge that creates an acceleration of 75
ft/s 2 . Assuming standard atmospheric pressure
outside the cylinder, find the gas pressure.
2.104E The atmosphere becomes colder at higher ele-
vation. As an average the standard atmospheric
absolute temperature can be expressed as r atm —
518 — 3.84 X 10~ 3 z, where z is the elevation
in feet. How cold is it outside an airplane cruis-
ing at 32 000 ft expressed in Rankine and in
Fahrenheit?
2.105E The density of mercury changes approximately
linearly with temperature as
Pn g - 851.5 - 0.086 T lbm/ft 3 (J" in degrees
Fahrenheit)
so the same pressure difference will result in a
manometer reading that is influenced by temper-
ature. If a pressure difference of 14.7 Ibf/in 2 is
measured in the summer at 95 F and in the win-
ter at 5 F, what is the difference in column
height between the two measurements?
42 H Chapter Two Some Concepts and Definitions
Computer, design and open-ended problems
2.106 Write a program to list corresponding tempera-
tures in °C, K, F, and R from -50°C to 100°C in
increments of 10 degrees.
2.107 Write a program that will input pressure in kPa or
atm or lbf/in 2 and write the pressure out in kPa,
atm, bar, and lbf/in 2 .
2.108 Write a program to do the temperature correction
on a mercury barometer reading (see Problem
2.62). Input reading and temperature and output
corrected reading at 20°C and pressure in kPa.
2.109 Make a list of different weights and scales that are
used to measure mass directly or indirectly. In-
vestigate the ranges of mass and the accuracy that
can be obtained.
2.110 Thermometers are based on several principles.
Expansion of a liquid with a rise in temperature is
used in many applications. Electrical resistance,
thermistors, and thermocouples are common in
instrumentation and remote probes. Investigate a
variety of thermometers and make a list of then-
range, accuracy, advantages, and disadvantages.
2.111 Collect information for a resistance-, thermistor-,
and thermocouple-based thermometer suitable for
the range of temperatures form 0°C to 200°C. For
each of the three types list the accuracy and re-
sponse of the transducer (output per degree
change). Is any calibration or corrections neces-
sary when it is used in an instrument?
2.112 A thermistor is used as a temperature transducer.
Its resistance changes with temperature approxi-
mately as
R=R exp[a(l/T - VT )]
where it has the resistance R at temperature
T . Select the constants as R = 3000 O and T =
298 K, and compute a so it has the resistance of
200 VL at 100°C. Write a program to convert a
measured resistance, R, into information about the
temperature. Find information for actual thermis-
tors and plot the calibration curves with the for-
mula given in this problem and the recommended
correction given by the manufacturer,
2.113 Investigate possible transducers for the measure-
ment of temperature in a flame with temperatures
near 1000 K. Are any available for a temperature
of 2000 K?
2.114 Devices to measure pressure are available as dif-
ferential or absolute pressure transducers. Make a
list of 5 different differential pressure transducers
to measure pressure differences in order of 100
kPa. Note their accuracy, response (linear or ?),
and price,
2.115 A micromanometer uses a fluid with density
1000 kg/m 3 and it is able to measure the height
difference with an accuracy of ±0.5 mm. Its
range is a maximum height difference of 0.5 m.
Investigate if any transducers are available to re-
place the micromanometer.
2.116 An experiment involves the measurements of
temperature and pressure of a gas flowing in a
pipe at 300°C and 250 kPa. Write a report with a
suggested set of transducers (at least two alterna-
tives for each) and give the expected accuracy
and cost.
Properties of a ^
Pure Substance j
In the previous chapter we considered three familiar properties of a substance — specific
volume, pressure, and temperature. We now turn our attention to pure substances and
consider some of the phases in which a pure substance may exist, the number of indepen-
dent properties a pure substance may have, and methods of presenting thermodynamic
properties.
Properties and the behavior of substances are very important for our studies of de-
vices and thermodynamic systems. The steam power plant in Fig. 1.1 and the nuclear
propulsion system in Fig. 1.3 have very similar processes, using water as the working
substance. Water vapor (steam) is made by boiling at high pressure in the steam generator
followed by an expansion in the turbine to a lower pressure, a cooling in the condenser,
and a return to the boiler by a pump that raises the pressure. We must know the water
properties to properly size the equipment such as the burners or heat exchangers, turbine,
and pump for the desired transfer of energy and the flow of water. As the water is brought
from liquid to vapor we need to know the temperature for the given pressure, and we
must know the density or specific volume so that the piping can be properly dimensioned
for the flow. If the pipes are too small, the expansion creates excessive velocities, leading
to pressure losses and increased friction, and thus demanding a larger pump and reducing
the turbine work output.
Another example is a refrigerator, shown in Fig. 1 .6, where we need a substance
that will boil from liquid to vapor at a low temperature, say — 20°C. This absorbs energy
from the cold space, keeping it cold. Inside the black grille in the back or at the bottom,
the now hot substance is cooled by air flowing around the grille, so it condenses from
vapor to liquid at a temperature slightly higher than room temperature. When such a sys-
tem is designed, we need to know the pressures at which these processes take place and
the amount of energy, covered in Chapter 5, that is involved. We also need to know how
much volume the substance occupies, the specific volume, so that the piping diameters
can be selected as mentioned for the steam power plant. The substance is selected so that
the pressure is reasonable during these processes; it should not be too high, due to leak-
age and safety concerns, and not too low either, as air might leak into the system.
A final example of a situation where we need to know the substance properties is
the gas turbine and a variation thereof, namely the jet engine shown in Fig. 1.11. In these
systems, the working substance is a gas (very similar to air) and no phase change takes
place. A combustion process burns fuel and air, freeing a large amount of energy, which
heats the gas so that it expands. We need to know how hot the gas gets and how much the
expansion is so that we can analyze the expansion process in the turbine and the exit noz-
zle of the jet engine. In this device, we do need large velocities inside the turbine section
43
44 m Chapter Three Properties of a Pure Substance
and for the exit of the jet engine. This high- velocity flow pushes on the blades in the tur-
bine to create shaft work or pushes on the jet engine (something called thrust) to move the
aircraft forward.
These are just a few examples of complete thermodynamic systems where a sub-
stance goes through several processes involving changes of its thermodynamic state and
therefore its properties. As your studies progress, many other examples will be used to il-
lustrate the general subjects.
3.1 The Pure Substance
A pure substance is one that has a homogeneous and invariable chemical composition. It
may exist in more than one phase, but the chemical composition is the same in all phases.
Thus, liquid water, a mixture of liquid water and water vapor (steam), and a mixture of ice
and liquid water are all pure substances; every phase has the same chemical composition.
In contrast, a mixture of liquid air and gaseous air is not a pure substance because the
composition of the liquid phase is different from that of the vapor phase.
Sometimes a mixture of gases, such as air, is considered a pure substance as long as
there is no change of phase. Strictly speaking, this is not true. As we will see later, we
should say that a mixture of gases such as air exhibits some of the characteristics of a pure
substance as long as there is no change of phase.
In this text the emphasis will be on simple compressible substances. This term des-
ignates substances whose surfagje^eff^ el ectrica l effects are in-
significant when dealing with the sub stanc es. But changes in volume, such as those
associated with the expansion of a gas in a cylinder, are very important. Reference will be
made, however, to other substances for which surface, magnetic, and electrical effects are
important. We will refer to a system consisting of a simple compressible substance as a
simple compressible system.
3.2 VAPOR-LIQUTO-SOLIB-PHASE
EQUILIBRIUM IN A PURE SUBSTANCE
Consider as a system 1 kg of water contained in the piston/cylinder arrangement shown in
™, -« - Fig. 3. la. Suppose that the piston and weight maintain a pressure of 0. 1 MPa in the cylin-
Jfli^Qnif der and that the initial temperature is 20°C. As heat is transferred to the water, the temper-
ature increases appreciably, the specific volume increases slightly, and the pressure
remains constant. When the temperature reaches 99.6°C, additional heat transfer results in
a change of phase, as indicated in Fig. 3.1&. That is, some of the liquid becomes vapor,
and during this process both the temperature and pressure remain constant, but the spe-
cific volume increases considerably. When the last drop of liquid has vaporized, further
transfer of heat results in an increase in both temperature and specific volume of the
vapor, as shown in Fig. 3.1c.
The term saturation temperature designates the temperature at which vaporization
takes place at a given pressure. This pressure is called the saturation pressure for the given
temperature. Thus, for water at 99.6°C the saturation pressure is 0.1 MPa, and for water at
0. 1 MPa the saturation temperature is 99.6°C, For a pure substance there is a definite rela-
VAPOR-LIQUJD-SOLID-PHASE EQUILIBRIUM IN A PURE SUBSTANCE B 45
FIGURE 3.1
Constant-pressure change
from liquid to vapor phase
for a pure substance.
M
(b)
-Water vapor-.
(c)
tion between saturation pressure and saturation temperature. A typical curve, called the
vapor-pressure curve, is shown in Fig. 3.2.
/If a substance exists as liquid at the saturation temperature and pressure, it is called
( saturated liquid. If the temperature of the liquid is lower than the saturation temperature
for the existing pressure, it is called either a subcooled liquid (implying that the tempera-
ture is lower than the saturation temperature for the given pressure) or a compressed liq-
uid (implying that the pressure is greater than the saturation pressure for the given
temperature). Either term may be used, but the latter term will be used in this text.
When a substance exists as part liquid and part vapor at the saturation temperature,
its quality is defined as the ratio of the mass of vapor to the total mass. Thus, in Fig. 3.16,
if the mass of the vapor is 0.2 kg and the mass of the liquid is 0.8 kg, the quality is 0.2 or
20%. The quality may be considered an mtensive^pjcjierty and has the symbol x. Quality
has meaning only when the substance is in a saturated state, that is, at saturation pressure
and temperature.
If a substance exists as vapor at the saturation temperature, it is called saturated
vapor. (Sometimes the term dry saturated vapor is used to emphasize that the quality is
/ 1 00%.) When the vapor is at a temperature greater than the saturation temperature, it is
/ said to exist as superheated vapor. The pressure and temperature of superheated vapor are
independent properties, since the temperature may increase while the pressure remains
constant. Actually, the substances we call gases are highly superheated vapors.
Consider Fig. 3.1 again. Let us plot on the temperature-volume diagram of Fig.
3.3 the constant -pressure line that represents the states through which the water passes
as it is heated from the initial state of 0.1 MPa and 20°C. Let state A represent the initial
state, B the saturated-liquid state (99. 6°C), and line AB the process in which the liquid
is heated from the initial temperature to the saturation temperature. Point C is the
FIGURE 3.2 Vapor-
pressure curve of a pure
substance.
Temperature
46 S Chapter Three properties of a Pure Substance
FIGURE 3.3
Temperature-volume
diagram for water
showing liquid and vapor
phases (not to scale).
D
MI {A
E
Volume
saturated- vapor state, and line BC is the constant-temperature process in which the
change of phase from liquid to vapor occurs. Line CD represents the process in which
the steam is superheated at constant pressure. Temperature and volume both increase
during this process.
Now let the process take place at a constant pressure of 1 MPa, starting from an ini-
tial temperature of 20°C. Point E represents the initial state, in which the specific volume
is slightly less than that at 0.1 MPa and 20°C. Vaporization begins at point F, where the
temperature is 179.9°C. Point G is the saturated-vapor state, and line GH is the constant-
pressure process in which the steam in superheated.
In a similar manner, a constant pressure of 10 MPa is represented by. line IJKL, for
which the saturation temperature is 311.1°C.
At a pressure of 22.09 MPa, represented by line MNO, we find, however, that there
is no constant-temperature vaporization process. Instead, point N is a point of inflection
with a zero slope. This point is called the critical point. Atjhe_^rjticj^point ^^aturated-
liquid and saturated - va p or states _are identical ., TQi£jeriH2erature^
volume atjhej^itical point aj;e ^alledjhe criticaiJempAra^
cal volume. The critical-point data for some substances are given in Table 3.1. More ex-
'teSive^ata are given in Table A.2 in the appendix.
Table 3.1
Some Critical-Point Data
Critical
Temperature,
Critical
Pressure,
MPa
Critical
Volume,
m 3 /kg
Water
Carbon dioxide
Oxygen
Hydrogen
374.14
31.05
-118.35
22.09
7.39
5.08
1.30
0.003 155
0.002 143
0.002 438
-239.85
0.032 192
Vapor-Liquid-Solid-Phase Equilibrium in a Pure Substance H 47
A constant-pressure process at a pressure greater than the critical pressure is repre-
sented by line PQ. If water at 40 MPa and 20°C is heated in a constant-pressure process in
a cylinder as shown in Fig. 3.1, there will never be two phases present and the state shown
in Fig. 3.16 will never exist. Instead, there will be a continuous change in density and at
all times there will be only one phase present. The question then is when do we have a liq-
uid and when do we have a vapor? The answer is that this is not a valid question at super-
critical pressures. We simply term the substance a fluid. However, rather arbitrarily, at
temperatures below the critical temperature we usually refer to it as a compressed liquid
and at temperatures above the critical temperature as a superheated vapor. , It should be
emphasized, however, that at pressures above the critical pressure we never have a liquid
and vapor phase of a pure substance existing in equilibrium.
In Fig. 3.3, line NJFB represents the saturated-liquid line and line NKGC represents
the saturated-vapor line.
By convention, the subscript / is u sed to designate a property of a s aturated liqui d
and the subscript g a property of a saturated vap or (subscript g being used to denote satu-
ration temperature and pressure). Thus, a saturation condition involving part liquid and
part vapor such as in Fig. 3.1b can be shown on J-u coordinates as in Fig. 3.4. All of the
liquid present is at state / with specific volume ty and all of the vapor present is at state g
with v g . The total volume is the sum of the liquid volume and the vapor volume, or
V= ?iiq + = m Viq v f + m^Vg
The average specific volume of the system v is then
„ s V '"liq m vap
^ v =£i = ljTVf + ^rv S =V-x)Vf+ (3-D
in terms of the definition of quality x = m np /m.
Using the definition
v /g = v g ~ v f
Equation 3.1 can also be mitten as
v = v f +xv /g (3.2)
Now, the quality x can he viewed as the fraction (v ~ vj)lv fg of the distance between satu-
rated liquid and saturated vapor, as indicated in Fig. 3.4.
Let us now consider another experiment with the piston/cylinder arrangement.
Suppose that the cylinder contains 1 kg of ice at — 20°C, 100 kPa. When heat is trans-
ferred to the ice, the pressure remains constant, the specific volume increases slightly,
FIGURE 3.4 T-v
diagram for the two-phase
liquid-vapor region to
show the quality specific
volume relation.
CR. Point
Sup.
vapor
48 H Chapter Three properties of a Pure Substance
Table 3.2
Some Solid-Ltqiiid-Vapor Triple-Point Data
Temperature,
°C
Pressure,
kPa
Hydrogen (normal)
-259
7.194
Oxygen
-219
0.15
Nitrogen
-210
12.53
Carbon dioxide
-56.4
520.8
Mercury
-39
0.000 000 13
Water
0.01
0.6113
Zinc
419
5.066
Silver
961
0.01
Copper
1083
0.000 079
and the temperature increases until it reaches 0°C, at which point the tee melts and the
temperature remains constant. In this state the ice is called a saturated solid. For most
substances the specific volume increases during this melting process, but for water the
specific volume of the liquid is less than the specific volume of the solid. When all the
ice has melted, a further heat transfer causes an increase in temperature of the liquid.
If the initial pressure of the ice at -20°C is 0.260 kPa, heat transfer to the ice results
in an increase in temperature to -10°C. At this point, however, the ice passes directly
jronijhejsr^^ injhl0ocj^sj^ Further heat
transfer results in superheating of the vapor.
Finally, consider an initial pressure of the ice of 0.6113 kPa and a temperature of
-20°C. Through heat transfer let the temperature increase until it reaches 0.0 1°C. At this '
point, however, further heat transfer may cause some of the ice to become vapor and some
to become liquid, for at this point it is possible to have the three phases in equilibrium.
This point is called the triple point, which is defined as thejjtetejnjw^^
may be present in eqmh^ium. The pressure and temperature at the triple point for a num-
ber~of substanceTare given in Table 3.2.
This whole matter is best summarized by the diagram of Fig. 3.5, which shows how
the solid, liquid, and vapor phases may exist together in equilibrium. Along the subhma-
tionlmejhe^so^ ^M^-^^J^^SlliSH?
liquid phases are injejm^riurn^ andjlong Jhe .vaporization line thejiqmd^nrljvarjor
phas^TaTe^equi^rium, The only point at which all three phases may exist in equilib-
rmlnTFmTtriple point. The vaporization line ends at the critical point because there is no
distinct change from the liquid phase to the vapor phase above the critical point.
Consider a solid in state A, as shown in Fig. 3.5. When the temperature increases
but the pressure (which is less than the triple-point pressure) is constant, the substance
passes directly from the solid to the vapor phase. Along the constant-pressure line EF, the
substance passes from the solid to the liquid phase at one temperature, and then from the
liquid to the vapor phase at a higher temperature. Constant-pressure line CD passes
through the triple point, and it is only at the triple point that the three phases may exist to-
gether in equilibrium. At a pressure above the critical pressure, such as GH t there is no
sharp distinction between the liquid and vapor phases.
VAFOR-LIQOID-SOLID-PHASE EQUILIBRIUM IN A PURE SUBSTANCE H 49
Although we have made these comments with rather specific reference to water
(only because of our familiarity with water), all pure substances exhibit the same general
behavior. However, the triple-point temperature and critical temperature vary greatly
from one substance to another. For example, the critical temperature of helium, as given
in Table A.2, is 5.3 K. Therefore, the absolute temperature of helium at ambient condi-
tions is over 50 times greater than the critical temperature. In contrast, water has a critical
temperature of 374.14°C (647.29 K), and at ambient conditions the temperature of water
is less than half the critical temperature. Most metals have a much higher critical tempera-
ture than water. When we consider the behavior of a substance in a given state, it is often
helpful to think of this state in relation to the critical state or triple point. For example, if
the pressure is greater than the critical pressure, it is impossible to have a liquid phase and
a vapor phase in equilibrium. Or, to consider another example, the states at which vac-
uum-melting a given metal is possible can be ascertained by a consideration of the proper-
ties at the triple point. Iron at a pressure just above 5 Pa (the triple-point pressure) would
melt at a temperature of about 1535°C (the triple-point temperature).
Figure 3.6 shows the three-phase diagram for carbon dioxide, in which it is seen
(see also Table 3.2) that the triple-point pressure is greater than normal atmospheric pres-
sure, which is very unusual. Therefore, the commonly observed phase transition under
conditions of atmospheric pressure of about 100 kPa is a sublimation from solid directly
to vapor, without passing through a liquid phase, which is why solid carbon dioxide is
commonly referred to as dry ice. We note from Fig. 3.6 that this phase transformation at
100 kPa occurs at a temperature below 200 K.
Finally, it should be pointed out that a pure substance can exist in a number of dif-
ferent solid phases. A transition from one solid phase to another is called an allotropic
transformation. Figure 3.7 shows a number of solid phases for water. A pure substance
can have a number of triple points, but only one triple point has a solid, liquid, and vapor
equilibrium. Other triple points for a pure substance can have two solid phases and a liq-
uid phase, two solid phases and a vapor phase, or three solid phases.
50 CHAPTER THREE PROPERTIES OF A PURE SUBSTANCE
FIGURE 3.7 Water phase diagram.
TABLES OF THERMODYNAMIC PROPERTIES Pi 51
3.3 Independent Properties
of a Pure Substance '
One important reason for introducing the concept of a pure substance is that the state of a
simple compressible pure substance (that is, a pure substance in the absence of motion,
- gravity, and surface, magnetic, or electrical effects) is defined by two indepen dent proper-
^ ties. For example, if the specific volume and temperature of superheated steamlife~sp^ct : *
fled, the state of the steam is determined.
To understand the significance of the term independent property, consider the
saturated-liquid and saturated- vapor states of a pure substance. These two states have
the same pressure and the same temperature, but they are definitely not the same state.
In a saturation state, therefore, pressure and temperature are not independent proper-
ties. Two independent properties such as pressure and specific volume or pressure and
quality are required to specify a saturation state of a pure substance.
The reason for mentioning previously that a mixture of gases, such as air, has the
same characteristics as a pure substance as long as only one phase is present, concerns
precisely this point. The state of air, which is a mixture of gases of definite composition,
is determined by specifying two properties as long as it remains in the gaseous phase. Air
then can be treated as a pure substance.
3,4 Tables of thermodynamic properties
Tables of thermodynamic properties of many substances are available, and in general, all
these tables have the same form. In this section we will refer to the steam tables. The
steam tables are selected both because they are a vehicle for presenting thermodynamic
tables and because steam is used extensively in power plants and industrial processes.
Once the steam tables are understood, other thermodynamic tables can be readily used.
Several different versions of steam tables have been published over the years. The
set included in Appendix B, Table B. 1 , is a summary based on a complicated lit to the be-
havior of water. It is very similar to the Steam Tables by Keenan, Keyes, Hill, and Moore,
published in 1969 and 1978. We will concentrate here on the three properties already dis-
cussed in Chapter 2 and in Section 3.2, namely T, P, and u i and note that the other proper-
ties listed in the set of Tables B. 1, n, h, and s, will be introduced later.
The steam tables in Appendix B consist of five separate tables, as indicated in Fig.
3.8. The region of superheated vapor in Fig. 3.5 is given in Table B.1.3, and that of com-
pressed liquid is given in Table B.I. 4. The compressed-solid region shown in Fig. 3.5 is
not listed in the appendix. The saturated-liquid and saturated-vapor region, as seen in the T
and v diagram of Fig. 3.3 (and as the vaporization line in Fig. 3.5), is listed according to the
values of Tin Table B.l.l and according to the values ofP (TandPare not independent in
the two-phase regions) in Table B.1.2. Similarly, the saturated-solid and saturated- vapor
region is listed according to T in Table B.1.5, but the saturated-solid and saturated- liquid
region, the third phase boundary line shown in Fig. 3.5 is not listed in the appendix.
In Table B.l.l, the first column after the temperature gives the corresponding sat-
uration pressure in kilopascals. The next three columns give specific volume in cubic
meters per kilogram. The first of these columns gives the specific volume of the satu-
rated liquid, v f ; the third column gives the specific volume of the saturated vapor v g ;
and the second column gives the difference between the two, iy g , as defined in Section
52 M Chapter Three Properties of a Pure Substance
FIGURE 3.8 Listing
of the steam tables. v T
3.2. Table B.1.2 lists the same information as Table B.l.l, but the data are listed ac-
cording to pressure, as mentioned earlier.
As an example, let us calculate the specific volume of saturated steam at 200°C hav-
ing a quality of 70%, Using Eq. 3.1 gives
v = 0.3(0.001 156) + 0.7(0.127 36)
= 0.0895 nrVkg
Table B.1.3 gives the properties of superheated vapor. In the superheated region,
pressure and temperature are independent properties; therefore, for each pressure a large
number of temperatures are given, and for each temperature four thermodynamic proper-
ties are listed, the first one being specific volume. Thus, the specific volume of steam at a
pressure of 0.5 MPa and 200°C is 0.4249 m 3 /kg.
Table B.1.4 gives the properties of the compressed liquid. To demonstrate the use ,
of this table, consider a piston and a cylinder (as shown in Fig. 3.9) that contains 1 kg of
saturated-liquid water at 100°C. Its properties are given hi Table B.l.l, and we note that
the pressure is 0.1013 MPa and the specific volume is 0.001 044 nrVkg. Suppose the pres-
sure is increased to 10 MPa while the temperature is held constant at 100°C by the neces- ■
sary transfer of heat, Q. Since water is slightly compressible, we would expect a slight
decrease in specific volume during this process. Table B.1.4 gives this specific volume as
0.001 039 nrVkg. This is only a slight decrease, and only a small error would be made if
one assumed that the volume of a compressed liquid is equal to the specific volume of the
saturated liquid at the same temperature. In many situations this is the most convenient
procedure, particularly when compressed-liquid data are not available. It is very important
to note, however, that the specific volume of saturated liquid at the given pressure, 10
FIGURE 3.9
Illustration of
compressed-liquid state.
Heat transfer
(in an amount
that will maintain *
constant temperature)
Tables of Thermodynamic properties H 53
MPa, does not give a good approximation. This value, from Table B.1.2, at a temperature
of 31 1.1 °C, is 0.001 452 m 3 /kg, which is in error by almost 40%.
Table B.1.5 of the steam tables gives the properties of saturated solid and saturated
vapor that are in equilibrium. The first column gives the temperature, and the second col-
umn gives the corresponding saturation pressure. As would be expected, all these pres-
sures are less than the triple-point pressure. The next two columns give the specific
volume of the saturated solid and saturated vapor.
Appendix B also includes thermodynamic tables for several other substances; refriger-
ant fluids ammonia, R-12, R-22, and R-134a and the cryogenic fluids nitrogen and methane.
In each case, only two tables are given — saturated liquid-vapor listed by temperature
(equivalent to Table B.l.l for water), and superheated vapor (equivalent to Table B.l .3).
Let us now consider a number of examples to illustrate the use of thermodynamic
tables for water and also the other substances listed in Appendix B.
EXAMPLE 3.1 Determine the phase for each of the following water states using the Appendix B tables
and indicate the relative position in the P-v, T-v, and P-T diagrams.
a. 1201C, 500 kPa
b. 120X, 0.5 mVkg
Solution
a. Enter Table B.l.l with 120°C. The saturation pressure is 198.5 kPa, so we have a
compressed liquid, point a in Fig. 3.10, That is above the saturation line for 120°C.
We could also have entered Table B.1.2 with 500 kPa and found the saturation tem-
perature as 151.86°C, so we would say it is subcooled liquid. That is to the left of the
saturation line for 500 kPa as seen in the P-T diagram.
b. Enter Table B.l.l with 120°C and notice
v f = 0.00106 < v < v g = 0.S9186 mVkg
so the state is a two phase mixture of liquid and vapor, point b in Fig. 3. 10. The state
is to the left of the saturated vapor state and to the right of the saturated liquid state
both seen in the T-v diagram.
120 T
FIGURE 3,10 DiagramforExamp!e3.1.
54 H chapter Three Properties of a Pure Substance
EXAMPLE 3.2 Determine the phase for each of the following states using the Appendix B tables and in-
dicate the relative position in the P-v, T-v and P-T diagrams, as in Figs. 3. 1 1 and 3.12.
a. Ammonia 30°C, 1000 kPa
b. R-22 200kPa, O.I5m 3 /kg
Solution
a. Enter Table B.2.1 with 30°C. The saturation pressure is 1167 kPa. As we have a
lower P, it is a superheated vapor state. We could also have entered with 1000 kPa
and found a saturation temperature of slightly less than 25°C, so we have a state that
is superheated about 5°C.
b. Enter Table B.4. 1 with 200 kPa and notice
u>u g ~0.1119m 3 /kg
so from the P-v diagram the state is superheated vapor. We can find the state in
Table B.4.2 between 40 and 50°C.
TABLES OF THERMODYNAMIC PROPERTIES H 55
EXAMPLE 3.3 Determine the temperature and quality (if defined) for water at a pressure of 300 kPa and
at each of these specific volumes:
a. 0.5 nrVkg
b. l.OmVkg
Solution
For each state, it is necessary to determine what phase or phases are present, in order to
know which table is the appropriate one to find the desired state information. That is, we
must compare the given information with the appropriate phase boundary values. Con-
sider a T-v diagram (or a P-u diagram) such as in Fig. 3.8. For the constant-pressure
line of 300 kPa shown in Fig. 3.13, the values for tyand v g shown there are found from
the saturation table, Table B.1.2.
a. By comparison with the values in Fig. 3.13, the state at which v is 0.5 nrVkg is seen
to be in the liquid-vapor two-phase region, at which T = I33.6°C, and the quality x is
found from Eq. 3.2 as
Note that if we did not have Table B.1.2 (as would be the case with the other sub-
stances listed in Appendix B), we could have interpolated in Table B. 1 . 1 between the
130°C and 135°C entries to get the lyand v g values for 300 kPa.
b. By comparison with the values in Fig. 3.13, the state at which vis 1 .0 nrVkg is seen to
be in the superheated vapor region, in which quality is undefined, and the temperature
for which is found from Table B. 1.3. In this case, Tis found by linear interpolation
between the 300 kPa specific-volume values at 300°C and 400°C, as shown in Fig.
3.14. This is an approximation for T, since the actual relation along the 300 kPa
constant-pressure line is not exactly linear.
From the figure we have
0.5 = 0.001 073 +x 0.604 75,
x = 0.825
slope =
T- 300
400 - 300
1.0 - 0.8753
1.0315 - 0.8753
solving this gives T = 379.8°C.
T
133.6°C
P = 300 kPa
FIGURE 3.13 A T-v diagram for
water at 300 kPa.
L_i
0.001073
0.60582
v
56 H Chapter Three properties of a Pure Substance
FIGURE 3.14 Tandw
values for superheated
vapor water at 300 kPa.
300
0.8753
1.0 1.0315
EXAMPLE 3.4 A closed vessel contains 0.1 m 3 of saturated liquid and 0.9 m 3 of saturated vapor R-134a
in equilibrium at 30°C. Determine the percent vapor on a mass basis.
Solution
Values of the saturation properties for R-134a are found from Table B.5.1. The
mass-volume relations then give
V ~ tn v
' vap '"vap v gj
mi
0.1
hq 0.000 843
0.9
= 118.6 kg
^ 0.026 71
= 33.7 kg
77i = 152.3 kg
™v*P _ 33.7
X —
0.221
™ 152.3
That is, the vessel contains 90% vapor by volume but only 22.1% vapor by mass.
EXAMPLE 3.4E A closed vessel contains 0.1 ft 3 of saturated liquid and 0.9 ft 3 of saturated vapor R-134a
in equilibrium at 90 F. Determine the percent vapor on a mass basis.
Solution
Values of the saturation properties for R-134a are found from Table F.10. The
mass-volume relations then give
m = 9.598 lbm
x = _ 2.245
0.1
iq 0.0136
v * p 0.4009
0.234
- 7.353 lbm
= 2.245 lbm
m 9.598
That is, the vessel contains 90% vapor by volume but only 23.4% vapor by mass.
Tables of thermodynamic Properties H 57
A rigid vessel contains saturated amrnonia vapor at 20°C. Heat is transferred to the sys-
tem nntil the temperature reaches 40°C. What is the final pressure?
Solution
Since the volume does not change during this process, the specific volume also remains
constant. From the ammonia tables, Table B.2. 1 , we have
vi = v 2 = 0.149 22 mVkg
Since v g at 40°C is less than 0. 149 22 mVkg, it is evident that in the final state the
ammonia is superheated vapor. By interpolating between the 800- and 1000-kPa
columns of Table B.2.2, we find that
P 2 = 945 kPa
EXAMPLE 3.5E A rigid vessel contains saturated ammonia vapor at 70 F. Heat is transferred to the sys-
tem until the temperature reaches 120 F. What is the final pressure?
Solution
Since the volume does not change during this process, the specific volume also remains
constant. From the ammonia tables, Table F.8,
v x = v 2 = 2.311 ftVlbm
Since v g at 120 F is less than 2.31 1 ftVlbm, it is evident that in the final state the
ammonia is superheated vapor. By interpolating between the 125- and 150-lbf/in 2
columns of Table F.8, we find that
P 2 = 145 lbfVin 2
EXAMPLE 3.6 Determine the missing property of P-v-T and x if applicable for the following states.
a. Nitrogen: -53.2°C, 600 kPa
b. Nitrogen: 10OK, 0.008 nrVkg
Solution
For nitrogen the properties are listed in Table B.6 with temperature in Kelvin.
a. Enter in Table B.6. 1 with T = 273.2 - 53.2 = 220 K, which is higher than the criti-
cal T in the last entry. Then proceed to the superheated vapor tables. We would also
have realized this by looking at the critical properties in Table A.2. From Table B.6,2
in the subsection for 600 kPa (T& = 96.37 K)
v - 0.10788 nrVkg
shown as point a in Fig. 3.15.
EXAMPLE 3.5
58 B chapter Three Properties of a Pure Substance
3400
600
CP.
3400
779
600
CP.
220
126
100
FIGURE 3.15 Diagram for Example 3.6.
b. Enter in Table B.6.I with T = 100 K, and we see
iy = 0.001 452 < v < v s = 0.0312 m 3 /kg
so we have a two-phase state with a pressure as the saturation pressure, shown as b in
Fig. 3.15
P sat = 779.2 kPa
and the quality from Eq. 3.2 becomes
x = (v- v$v fg = (0.008 - 0.001 452)70.029 75 = 0.2201
EXAMPLE 3.7 Determine the pressure for water at 200°C with v = 0.4 nvVkf
Solution
Start in Table B.U with 200°C and note that v>u g = 0.127 36 m 3 /kg so we have su-
perheated vapor. Proceed to Table B.1.3 at any subsection with 200°C; say we start at
200 kPa. There the v = 1.080 34, which is too large so the pressure must be higher. For
Thermodynamic Surfaces H 59
FIGURE 3.17 Linear 500
interpolation for Example
3.7.
0.35
0.4 0.42
The real constant- T curve is slightly curved and not linear,
but for manual interpolation we assume a linear variation.
500 kPa, u = 0.424 92, and for 600 kPa, v = 0.352 02, so it is bracketed. This is shown
in Fig. 3.16.
A linear interpolation, Fig. 3.17, between the two pressures is done to get P at
the desired v.
P = 500 + (600 - 500) ™n°-M%n = ^
3,5 THERMODYNAMIC SU&FACES
The matter discussed to this point can be well summarized by a consideration of a
pressure-specific volume-temperature surface. Two such surfaces are shown in Figs. 3.18
and 3.19. Figure 3.18 shows a substance such as water in which the specific volume in-
creases during freezing. Figure 3.19 shows a substance in which the specific volume de-
creases during freezing.
In these diagrams the pressure, specific volume, and temperature are plotted on
mutually perpendicular coordinates, and each possible equilibrium state is thus repre-
sented by a point on the surface. This follows directly from the fact that a pure substance
has only two independent intensive properties. All points along a quasi-equilibrium
process He on the P-v-T surface, since such a process always passes through equilibrium
states.
The regions of the surface that represent a single phase— the solid, liquid, and vapor
phases— are indicated. These surfaces are curved. The two-phase regions— the solid-liquid,
solid-vapor, and liquid-vapor regions— are ruled surfaces. By this we understand that they
are made up of straight Unes'parallel to the specific-volume axis. This, of course, follows
from the fact that in the two-phase region, lines of constant pressure are also lines of con-
stant temperature, although the specific volume may change. The triple point actually ap-
pears as the triple line on the P-u~T surface, since the pressure and temperature of the triple
point are fixed, but the specific volume may vary, depending on the proportion of each
phase.
It is also of interest to note the pressure-temperature and pressure-volume projec-
tions of these surfaces. We have already considered the pressure-temperature diagram
for a substance such as water. It is on this diagram that we observe the triple point. Var-
ious lines of constant temperature are shown on the pressure-volume diagram, and the
60 M chapter Three Properties of a Pure Substance
Vo[ume Temperature
FIGURE 3. IS Pressure-volume-temperature surface for a substance that
expands on freezing.
corresponding constant-temperature sections are lettered identically on the P-v~T sur-
face. The critical isotherm has a point of inflection at the critical point,
^ One notices that for a substance such as water, which expands on freezing, the
freezing temperature decreases with an increase in pressure. For a substance that contracts
on freezing, the freezing temperature increases as the pressure increases. Thus, as the
pressure of vapor is increased along the constant-temperature line abcdef'm Fig. 3.18, a
substance that expands on freezing first becomes solid and then liquid. For the substance
that contracts on freezing, the corresponding constant-temperature line (Fig. 3.19) indi-
cates that as the pressure on the vapor is increased, it first becomes liquid and then solid.
THE P-V-T BEHAVIOR QF LOW- AND MODE RATE-DENSITY GASES 9 61
Volume Temperature
FIGURE 3.19 Pressure-volume-temperature surface for a substance that
contracts on freezing.
3.6 the p-v-t behavior of low-
and Moderate-Density Gases
One form of energy possession by a system discussed in Section 2.6 was intennoiecular
(IM) potential energy, that associated with the forces between molecules. It was stated
there that at very low densities the average distances between molecules is so large that the
IM potential energy may effectively be neglected. In such a case, the particles would be in-
dependent of one another, a situation referred to as an ideal gas. Under this approximation,
62 B Chapter Three Properties of a Pure Substance
it has been observed experimentally that, to a close degree, a very low density gas behaves
according to the ideal gas equation of state
PV= nRT, Pv=RT j (3,3)
in which n is the number of kmol of gas, or
= m = kg
U M kg/kmol (3 ' 4)
In Eq. 3.3, R is the universal gas constant, the value of which is, for any gas,
R = 8.3145 J®*™ = 8.3145 kJ
kmol K ' kmol K
and T is the absolute (ideal gas scale) temperature in kelvins (i.e., J(K) = T(°C) +
273.15). It is important to note that Tmust always be the absolute temperature whenever
it is being used to multiply or divide in an equation. The ideal gas absolute temperature
scale will be discussed in more detail in Chapter 7. In the English Engineering system,
£= 1545 ftlbf
lb moli;
Substituting Eq. 3.4 into Eq. 3.3 and rearranging, we find that the ideal gas equation
of state can be written conveniently in the form
\_PV = mRT, Pv = RT^ (3.5)
where
= # CM)
in which R is a different constant for each particular gas. The value of R for a number of
substances is given in Table A. 5 of Appendix A, and also in English units in Table F.4.
EXAMPLE 3.8 What is the mass of air contained in a room 6 m X 10 m X 4 m if the pressure is 100
kPa and the temperature is 25X?
Solution
Assume air to be an ideal gas. By using Eq. 3.5 and the value of R from Table A. 5, we have
,„ _ PV _ 100 kN/m 2 X 240 m 3 .... ?fm , t
RT 0.287 kN m/kg K X 298.2 K ^ Kg
THE f-F-TBEHAVIOR OF LOW- AND MODERATE-DENSITY GASES 63
EXAMPLE 3.9 A tank has a volume of 0.5 m 3 and 'contains 10 kg of an ideal gas having a molecular
weight of 24. The temperature is 25°C. What is the pressure?
Solution
The gas constant is determined first:
^ R = 8.3145 kN m/kmol K
M 24kg/kmol
- 0.346 44 kN m/kg K
We now solve for P:
_ mRT _ 10 kg X 0.346 44 kN m/kg K X 298.2 K
P ~~ ~ 0.5 m 3
= 2066kPa
EXAMPLE 3.9E A tank has a volume of 15 ft 3 and contains 20 Ibm of an ideal gas having a molecular
weight of 24. The temperature is 80 F. What is the pressure?
Solution
The gas constant is determined first:
* = 1545 fUbf/lbmolR = ^ 4 ft {mhm R
M 24 lbm/lb mol
We now solve for P.
mRT = 20 Ibm X 64.4 ft Ibf/lbm R X 540 R = 32J M/in 2
V 144 in 2 /ft 2 X 15 ft 3
EXAMPLE 3,10 A gas-bell is submerged in liquid water with its mass counterbalanced with rope and
pulleys as shown in Fig. 3.20. The pressure inside is measured carefully to be 105 kPa,
and the temperature is 2 IT. A volume increase is measured to be 0.75 m 3 over a period
FIGURE 3.20 Sketch
for Example 3.10.
64 M CHAPTER THREE PROPERTIES OF A PURE SUBSTANCE
of 185 s. What is the volume flow rate and the mass flow rate of the flow into the hell as-
suming it is carbon dioxide gas?
Solution
The volume flow rate is
dt At 185 ' 4 111 /S
and the mass flow rate is m = pV ~ Vlv. At close to room conditions the carbon dioxide
is an ideal gas, so PV = mRT or v = RT/P, and from Table A.5 we have the ideal gas
constant P = 0. 1 889 kj/kg K. The mass flow rate becomes
• „ PV _ 105 X 0.040 54 kPa m 3 /s n n7 „ , ,
RT 0.1889(273.15 + 21) kJ/kg u - u/t,t,K «' s
Because of its simplicity, the ideal-gas equation of state is very convenient to use in
thermodynamic calculations. However, two questions are now appropriate. The ideal-gas
equation of state is a good approximation at low density. But what constitutes low den-
sity? Or, expressed in other words, over what range of density will the ideal-gas equation
of state hold with accuracy? The second question is, how much does an actual gas at a
given pressure and temperature deviate from ideal-gas behavior?
One specific example in response to these questions is shown in Fig. 3.21, a T-v dia-
gram for water that indicates the error in assuming ideal gas for saturated vapor and for su-
perheated vapor. As would be expected, at very low pressure or high temperature the error is
small, hut this becomes severe as the density increases. The same general trend would be the
600
400
300
200
100
"I
1 111 II lin[
i i jj h ii lit "T — r
1 11 -I:. 1
Ml I M 1 ll|
L
- 100%
w
\\
1%f
0.1%
\\
~ 270%
jfl7.6%
\\
I \ ■.
10MPa
1 MPa
deal /gas
f 0.2%
Error
<1%
1.5%
FIGURE 3.21 ' 1 — nil — ^ — i i 1 1 mil i — i i ii rriii i i , i i .TTTrrr-*
Temperature-specific 10 " 3 10~ 1 10° 10 1 10 2
volume diagram for water. Specific volume v [m 3 /kg]
the p-v-T Behavior of low- and moderate-density gases H 65
case in referring to Fig, 3. 18 or 3.19. As the state becomes further removed from the satura-
tion region (i.e., high Tor low P), the,gas behavior becomes closer to the ideal-gas model.
A more quantitative study of the question of the ideal-gas approximation can be
conducted by introducing the compressibility factor Z, defined as
\ Z=^ I
\ RT j
i /
I Pv=ZRT j ■■- (3.7)
Note that for an ideal gas Z = 1 , and the deviation of Z from unity is a measure of
the deviation of the actual relation from the ideal-gas equation of state.
Figure 3.22 shows a skeleton compressibility chart for nitrogen. From this chart we
make three observations. The first is that at all temperatures Z 1 as P -> 0. That is, as
the pressure approaches zero, the P-v-T behavior closely approaches that predicted by
the ideal-gas equation of state. Note also that at temperatures of 300 K and above (that is,
room temperature and above) the compressibility factor is near unity up to pressure of
about 10 MPa. This means that the ideal-gas equation of state can be used for nitrogen
(and, as it happens, air) over this range with considerable accuracy.
We further note that at lower temperatures or at very high pressures, the compressi-
bility factor deviates significantly from the ideal-gas value. Moderate-density forces of at-
traction tend to pull molecules together, resulting in a value of Z < 1, whereas very high
density forces of repulsion tend to have the opposite effect.
If we examine compressibility diagrams for other pure substances, we find that the di-
agrams are all similar in the characteristics described above for nitrogen, at least in a quali-
tative sense. Quantitatively the diagrams are all different, since the critical temperatures and
pressures of different substances vary over wide ranges, as evidenced from the values listed
66 B Chapter Three Properties of a Pure Substance
in Table A.2. Is there a way in which we can put all of these substances on a common
basis? To do so, we "reduce" the properties with respect to the values at the critical point.
The reduced properties are defined as
p
reduced pressure = P r — P c = critical pressure
" c
T
reduced temperature — T r = — , T c — critical temperature (3.8)
■* c
These equations state that the reduced property for a given state is the value of this
property in this state divided by the value of this same property at the critical point.
If lines of constant T r are plotted on a Z versus P r diagram, a plot such as that in Fig.
D.l is obtained. The striking fact is that when such Z versus P r diagrams are prepared for
a number of different substances, all of them very nearly coincide, especially when the
substances have simple, essentially spherical molecules. Correlations for substances with
more complicated molecules are reasonably close, except near or at saturation or at high
density. Thus, Fig. D.l is actually a generalized diagram for simple molecules, which
means that it represents the average behavior for a number of different simple substances.
When such a diagram is used for a particular substance, the results will generally be
somewhat in error. However, if P~v-T information is required for a substance in a region
where no experimental measurements have been made, this generalized compressibility
diagram will give reasonably accurate results. We need know only the critical pressure
and critical temperature to use this basic generalized chart.
In our study of thermodynamics, we will use Fig. D. 1 primarily to help us decide
whether, in a given circumstance, it is reasonable to assume ideal-gas behavior as a model.
For example, we note from the chart that if the pressure is very low (that is, <P C ), the ideal-
gas model can be assumed with good accuracy, regardless of the temperature. Furthermore,
at high temperatures (that is, greater than about twice T c % the ideal-gas model can be as-
sumed with good accuracy to pressures as high as four or five times P c . When the tempera-
ture is less than about twice the critical temperature and the pressure is not extremely low,
we are in a region, commonly termed superheated vapor, in which the deviation from ideal-
gas behavior may be considerable. In this region it is preferable to use tables of thermody-
namic properties or charts for a particular substance, as discussed in Section 3.4.
/
EXAMPLE 3,11 Is it reasonable to assume ideal-gas behavior at each of the given states?
a. Nitrogen at 20°C, 1.0 MPa
b. Carbon dioxide at 20°C, 1.0 MPa
c. Ammonia at 20°C, 1.0 MPa
Solution
In each case it is first necessary to check phase boundary and critical state data.
a. For nitrogen, the critical properties are, from Table A.2, 126.2 K, 3.39 MPa. Since
the given temperature, 293.2 K is more than twice T c and the reduced pressure is less
than 0.3, ideal gas behavior is a very good assumption.
THE P-V-T BEHAVIOR OF LOW- AND MODERATE-DENSITY GASES H
67
b. For carbon dioxide, the critical properties are 304,1 K, 7.38 MPa. Therefore, the reduced
properties are 0.96 and 0.136. From Appendix Fig D.l, C0 2 is a gas (although T < T c )
with a Zof about 0.95, so the ideal-gas model is accurate to within about 5% in this case.
c. The ammonia tables, Appendix B.2, give the most accurate information. From Table
B.2. 1 at 20°C, P g = 858 kPa. Since the given pressure of 1 MPa is greater than P g ,
this state is a compressed liquid, and not a gas.
EXAMPLE 3,12 Determine the specific volume for R- 1 34a at 1 00°C, 3.0 MPa, for the following models:
a. The R-134a tables, Table B.5
b. Ideal gas
c. The generalized chart, Fig, D.l
Solution
a. From Table B .5.2 at 1 00°C, 3 MPa,
v = 0.006 65 m 3 /kg (most accura te vah ie)
b. Assuming ideal gas, we have
-A = = o.osi 49 U
M 102.03- kgK
RT 0.081 49 X 373.2
P 3000
which is more'than-50% too large,
c. Using the generalized chart, Fig. D.l, we obtain
= 0.010 14m7kg
T ^M= LQ - p ^toir a74 ' z=0 - 67
!) = ZX^= 0.67 X 0.010 14 = 0.006 79 m 3 /kg
which is only 2% too large.
EXAMPLE 3,13 Propane in a steel bottle of volume 0. 1 m 3 has a quality of 1 0% at a temperature of 1 5°C.
Use the generalized compressibility chart to estimate the total propane mass and to find
the pressure.
Solution
To use the generalized chart we need the reduced pressure and temperature. From Table A.2
for propane, P c = 4250 kPa and T c ~ 369.8 K, The reduced temperature is, from Eq. 3.8,
T — ~ — 2-73 '^5 . + I 5 - 779 2 = 78
r T„ 369.8
68 M Chapter Three Properties of a Pure Substance
From Figure D.l, shown in Fig. 3,23, we can read for the saturated states
Sat. vapor
T r = 2.0 /
T r = 0.78
\ /T r =0.7
Sat. liquid
1
0.2
1 lnP r
Z f = 0.035,
Z, = 0.83
FIGURE 3,23 Diagram for
lixample 3.13.
For the two-phase state the pressure is the saturated pressure
P - P ri , x X P c - 0.2 X 4250 kPa - 850 kPa
The overall compressibility factor becomes, as Eq. 3.1 for v
Z=(\- x)Z f +xZ g = 0.9 X 0.035 + 0.1 X 0.83 = 0.1145
The gas constant from Table A.5 is R = 0.1886 kJ/kg K, so the gas law is Eq. 3.7.
PV=mZRT
850 X 0.1
kPa m 3
PV =
ZRT 0.1145 X 0.1886 X 288.15 kJ/kg
= 13.66 kg
Instead of the ideal-gas model to represent gas behavior, or even the generalized
compressibility chart, which is approximate, it is desirable to have an equation of state
that accurately represents the P-u~T behavior for a particular gas over the entire super-
heated vapor region. Such an equation is necessarily more complicated and conse-
quently more difficult to use. Many such equations have been proposed and used to
correlate the observed behavior of gases. To illustrate the nature and complexity of
these equations, we present one of the best known, the Benedict-Webb-Rubin equation
of state:
KT RTB -A -C^ RTb - a a* _g/ y) ^
y v 2 ' v 3 v 6 v'T 2 { v 2 ) { y)
This equation contains eight empirical constants and is accurate to densities of about
twice the critical density. The empirical constants for the Benedict-Webb-Rubin equation
for a number of substances are given in Appendix D.
An equation of state that accurately describes the relation among pressure, tempera-
ture, and specific volume is rather cumbersome and obtaining the solution requires con-
siderable time. When we use a digital computer, it is often most convenient to determine
the thermodynamic properties in a given state from such equations. However, in hand cal-
Computerized Tables H 69
dilations, it is much more convenient to tabulate values of pressure, temperature, specific
volume, and other thermodynamic properties for various substances. Such tables have
been presented in Appendix B.
3.7 COMPUTERIZED TABLES
Most of the tables in the appendix are supplied in a computer program on the disk accom-
panying this. book. The main program operates with a visual interface in the Windows en-
vironment on a PC-type computer and is generally self-explanatory.
The main program covers the full set of tables for water, refrigerants, and cryogenic
fluids, as in Tables B.l to B.7 including the compressed liquid region, which is only
printed for water. For these substances a small graph with the P~o diagram shows the re-
gion around the critical point down toward the triple line covering the compressed liquid,
two-phase liquid-vapor, dense fluid, and superheated vapor regions. As a state is selected
and the properties computed, a thin crosshair set of lines indicates the state in the diagram
so this can be seen with a visual impression of the state's location.
Ideal gases are covered corresponding to the Tables A.7 for air and A.8 or A.9 for
other ideal gases. You are able to select the substance and the units to work in for all the
various table sections giving a wider choice than the printed tables. Metric units (SI) or
standard English units for the properties can be used as well as a mass basis (kg or ibm) or
a mole basis, satisfying the need for the most common applications.
The generalized chart, Fig. D.l, with the compressibility factor, is included to allow
a more accurate value of Z to be obtained than can be read from the graph. This is particu-
larly useful for the case of a two-phase mixture where the saturated liquid and saturated
vapor values are needed. Besides the compressibility factor, this part of the program in-
cludes correction terms beyond ideal-gas approximations for changes in the other thermo-
dynamic properties.
The only mixture application that is included with the program is moist air.
EXAMPLE 3.14 Find the states in Examples 3.1 and 3.2 with the computer-aided thermodynamics tables,
CATT, and list the missing property of P-v-T and x if applicable.
Solution
Water states from Example 3.1: Click Water, click Calculator and then select Case 1
(T, P). Input (T, P) - (120, 0:5). The result is as shown in Fig. 3.24.
Compressed liquid v = 0.0106 mVkg (same as in Table B.1.4)
Click Calculator and then select Case 2 (T t v). Input (T, v) - (120, 0.5)
: ;> Two-phase x= 0.5601, P= 198.5 kPa -
Ammonia state from Example 3.2: Click Cryogenics; check that it is ammonia. Otherwise
select Ammonia, click Calculator, and then select Case 1 (T, P). Input (T, P) = (30, 1)
=> Superheated vapor v — 0.1321 m 3 /kg (same as in Table B.2,2)
70 a
Chapter Three Properties of a Pure Substance
FIGURE 3.24 CATT
Result for Example 3.1.
Ffe lahtes k°9 Wafer Help
P-VDiasram (log-log)
100
0.01 . 1
Specific Volume
~" \WaIerj(Refrigerants /Cryogenics /^Jr /Ideal Gases /Compressibility /Psychromelrics /
Input Typs -
** L T £ P
r 3. T S
C 4. T fcX
re. p fc h
rz.Pis
Specific LiiUisIpi' j 503-9 ~| U /!-.<! '. t
Specific Eniropj? ' (t-5273 | Wkg/K {
' Q»a5tp )0 "I Q <r= k.<= I j
I*/ OK j [~K Carted j j *? HuV j
R-22 state from Example 3.2: Click Refrigerants; check that it is R-22. Othenvise select
R-22 (Alt-R) click Calculator and then select Case 5 (P, v). Input (P, v) = (0.2, 0.15)
=> Superheated vapor T = 46.26°C
nummary
Thermodynamic properties of a pure substance and the phase boundaries for solid, liquid,
and vapor states are discussed. Phase equilibrium for vaporization (boiling liquid to
vapor), with the opposite direction being condensation (vapor to liquid); sublimation
(solid to vapor) or the opposite solidification (vapor to solid); and melting (solid to liquid)
or the opposite solidifying (liquid to solid) should be recognized. The three-dimensional
p„y_7' gur f ace and the tvvo-dimensional representations in the (P, T), (T, v) and (P, v) dia-
Key concepts and Formulas
m 71
Key Concepts
and formulas
grams, and the vaporization, sublimation, and fusion lines are related to the printed tables
in Appendix B. Properties from printed and computer tables covering a number of sub-
stances are introduced, including two-phase mixtures, for which we use the mass fraction
of vapor (quality). The ideal-gas law approximates the limiting behavior for low density.
An extension of the ideal-gas law is shown with the compressibility factor Z, and other
more complicated equations of state are mentioned.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Know phases and the nomenclature used for states and interphases.
• Identify a phase given a state (T, P).
• Locate states relative to the critical point and know Tables A.2 (F.l) and 3.2.
• Recognize phase diagrams and interphase locations.
• Locate states in the Appendix B tables with any entry: (T, P) > (T, v) or (P, v)
• Recognize how the tables show parts of the (T, P), (J, v) or {P, v) diagrams.
• Find properties in the two-phase regions; use quality x.
• Locate states using any combination of (T, P, v, x) including linear interpolation.
• Know when you have a liquid or solid and the properties in Tables A3, A. 4
(F.2, F.3).
- Know when a vapor is an ideal gas (or how to find out).
• Know the ideal-gas law and Table A.5 (F.4).
• Know the compressibility factor Z and the compressibility chart Fig. D.l.
• Know the existence of more general equations of state.
• Know how to get properties from the computer program.
Phases Solid, liquid, and vapor (gas)
Phase equilibrium T iZt , P sat , v fi v g , v {
Multiphase boundaries Vaporization, sublimation, and fusion lines;
Figs. 3.5 (general), 3.6 (C0 2 ) and 3.7 (water)
Critical point: Table 3.1, Table A.2 (F.l)
Triple point: Table 3.2
Equilibrium state Two independent properties (#1, #2)
Quality x = m np /m (vapor mass fraction)
1 — x = m lit Jm (liquid mass fraction)
Average specific volume v = (1 - x)v f + xu g (only two-phase mixture)
Equilibrium surface P—u-T Tables or equation of state
Ideal-gas law Py = RT PV= mRT - nRT
Universal gas constant R. = 8.3 145 kJ/kmol K
Gas constant R = RIM kJ/kg K, Table A.5 or M from Table A.2
ft lbtflbm R, Table F.4 or M from Table F. 1
Compressibility factor Z Pv = ZRT Chart for Z in Fig. D. 1
P T
Reduced properties P r = — T r = — Entr y t0 compressibility chart
" c * c
Equations of state Cubic, pressure explicit: Appendix D, Table D.l
B-W-R: Eq. 3.7 and Table D.2 for various substances
Lee Kesler: Appendix D, Table D.3, and Fig. D.l
72 B Chapter Three properties of a Pure Substance
Concept-Study Guide problems
3.1 What is the lowest temperature (approximately) at
which water can be liquid?
3.2 What is the percent change in volume as liquid
water freezes? Mention some effects in nature and
for our households the volume change can have.
3.3 When you skate on ice a thin liquid film forms
under the skate; how can that be?
3.4 An external water tap has the valve activated by a
long spindle so that the closing mechanism is lo-
cated well inside the wall. Why is that?
3.5 Some tools should be cleaned in water at a least
150°C. How high a P is needed?
3.6 Are the pressures in the tables absolute or gauge
pressures?
3.7 If I have 1 L of ammonia at room pressure and tem-
perature (100 kPa, 20°C) how much mass is that?
3.8 How much is the change in liquid specific volume
for water at 20°C as you move up from state i to-
ward state j in Fig. 3.18, reaching 15 000 kPa?
3.9 For water at 100 kPa with a quality of 10% find the
volume fraction of vapor.
3.10 Sketch two constant-pressure curves (500 kPa and
30 000 kPa) in a T~v diagram and indicate on the
curves where in the water tables you see the prop-
erties.
3.11 Locate the state of ammonia at 200 kPa, -10°C.
Indicate in both the P-v and the T-v diagrams the
Homework Problems
Phase Diagrams; Triple and Critical Points
/ 3.21 Modern extraction techniques can be based on dis-
solving material in supercritical fluids such as car-
bon dioxide. How high are the pressure and density
of carbon dioxide when the pressure and tempera-
ture are around the critical point? Repeat for ethyl
alcohol.
3.22 Find the lowest temperature at which it is possible
to have water in the liquid phase. At what pressure
must the liquid exist?
3.23 Water at 27°C can exist in different phases depen-
dent on the pressure. Give the approximate pres-
sure range in kPa for water being in each one of the
three phases, vapor, liquid, or solid.
location of the nearest states listed in the printed
Table B.2.
3.12 Why are most of the compressed liquid or solid re-
gions not included in the printed tables?
3.13 Water at 120°C with a quality of 25% has its tem-
perature raised 20°C in a constant volume process.
What is the new quality and pressure?
3.14 Water at 200 kPa with a quality of 25% has its tem-
perature raised 20°C in a constant pressure process.
What is the new quality and volume?
3.15 Why is it not typical to find tables for Ar, He, Ne
or air like an Appendix B table?
3.16 What is the relative (%) change in P if we double
the absolute temperature of an ideal gas keeping
mass and volume constant? Repeat if we double V
having m and T constant.
3.17 Calculate the ideal gas constant for argon and hy-
drogen based on Table A.2 and verify the value
with Table A. 5.
3.18 How close to ideal gas behavior (find Z) is ammo-
nia at saturated vapor, 100 kPa? How about satu-
rated vapor at 2000 kPa?
3.19 Find the volume of 2 kg of ethylene at 270 K, 2500
kPa using Z from Fig. D. 1 .
3.20 With T r = 0.85 and a quality of 0.6 find the com- "
pressibility factor using Fig. D.l.
i 3.24 What is the lowest temperature in Kelvins for
V_,^\vhich you can see metal as a liquid if the metal is
a. silver or b. copper?
3.25 If density of ice is 920 kg/m 3 , find the pressure at
the bottom of a 1000-m-thick ice cap on the North
Pole. What is the melting temperature at that
pressure?
3.26 Dry ice is the name of solid carbon dioxide. How
cold must it be at atmospheric (100 kPa) pres-
sure? If it is heated at 100 kPa what eventually
happens?
3.27 A substance is at 2 MPa and 17°C in a rigid tank.
Using only the critical properties, can the phase of
the mass be determined if the substance is nitrogen,
water, or propane?
[
Homework Problems M 73
3.28 Give the phase for the following states: - /
a. C0 2 at T = 267°C and P - 0.5 MPa
b. Air at T = 20°C and P - 200 kPa
c. NH 3 at T - 170°C and P = 600 kPa
General Tables
3.29 Determine the phase. of the substance at the given
state using Appendix B tables.
a. Water: 100°C, 500 kPa
b. Ammonia: -10°C, 150 kPa
c. R-12:0°C,350kPa
3.30 Determine whether water at each of the following
states is a compressed liquid, a superheated vapor,
or a mixture of saturated liquid and vapor.
a. 10 MPa, 0.003 m 3 /kg c. 200°C, 0.1 nrVkg
b. IMPa, 190°C d. lOkPa, 10°C
3.31 Give the phase for the following states:
a. H 2 at T — 275°C and P = 5 MPa
b. H 2 at T = -2°C and P = 100 kPa
3.32 Determine whether refrigerant R-22 in each of the
following states is a compressed liquid, a super-
heated vapor, or a mixture of saturated liquid and
vapor.
a. 50°C, 0.05 m 3 /kg c. 0.1 MPa, 0.1 m 3 /kg
b. 1.0MPa,20°C d. -20°C,200kPa
3.33 Fill out the following table for substance water:
PJkPa] 7[°C] u[m 3 /kg} x
a. 500
b. 500
c. 1400
d.
20
200
300
0.20
0.8
3.34 Place the four states a-d listed in Problem 3.33
as labeled dots in a sketch of the P-v and T-v
diagrams.
3.35 Determine the phase and the specific volume for
ammonia at these states using the Appendix B
table.
a. -10°C, 150 kPa
b. 20°C> lOOkPa
c. 60°C, quality 25%
3.36 Give the phase and the specific volume for the
^'following:
a. R-22 at T = -25°C and P = 100 kPa
b. R-22 at T = ~25°C and P = 300 kPa
c. R-12 at T - 5°C and P = 200 kPa
3.37 Fill out the following table for substance ammonia:
P[kPa]
JfC]
u[nrVkg]
50
50
0.IIS5
0.5
3.38 Place the two states a-b listed in Problem 3.37 as
labeled dots in a sketch of the P-v and T-v
^—,dia grams.
3.39 Calculate the following specific volumes:
_ a. R-134a: 50°C, 80% quality
b. Water: 4 MPa, 90% quality
c. Nitrogen: 120 K, 60% quality
3.40 Give the phase and the missing property of P, T, v,
and x.
a. R-134a,r= -20°C,P = 150 kPa
b. R-134a, P = 300 kPa, v = 0.072 m 3 /kg
c. CH 4 , T = 1 55 K, v = 0.04 nrVkg
d. CH 4l T = 350 K, v = 0.25 m 3 /kg
3.41 A sealed rigid vessel has volume of 1 m 3 and con-
tains 2 kg of water at 100°C. The vessel is now
heated. If a safety pressure valve is installed, at
what pressure should the valve be set to have a
maximum temperature of 200°C?
3.42 Saturated liquid water at 60°Cjs put under pressure
/to decrease the volume by 1% while keeping the
temperature constant. To what pressure should it
be compressed?
3.43 Saturated water vapor at 200 kPa is in a constant-
pressure piston/cylinder assembly. At this state the
piston is 0.1 m from the cylinder bottom. How much
is this distance if the temperature is changed to
a. 200°C
b. 100°C.
3.44 You want a pot of water to boil at 105°C. How
heavy a lid should you put on the 15-cm-diameter
pot when P atm = 101 kPa?
3.45 In your refrigerator the working substance evapo-
rates from liquid to vapor at — 20°C inside a pipe
74 n chapter Three Properties of a Pure Substance
3.47
around the cold section. Outside (on the back or
below) is a black grille, inside of which the work-
ing substance condenses form vapor to liquid at
+40°C. For each location find the pressure and the
change in specific volume (v) if
a. the substance is R-12
b. the substance is ammonia
3.46 Repeat the previous problem with the substances
a. R-134a-
b. R~22
A water storage tank contains liquid and vapor in
equilibrium at 110°C. The distance from the bot-
tom of the tank to the liquid level is 8 m, What is
the absolute pressure at the bottom of the tank?
Saturated water vapor at 200 kPa is in a constant-
pressure piston/cylinder assembly. At this state the
piston is 0.1 m from the cylinder bottom. How much
is this distance and what is the temperature if the
water is cooled to occupy half the original volume?
Two tanks are connected as shown in Fig. P3.49,
both containing water. Tank A is at 200 kPa, v
0.5 mVkg, V A = 1 m 3 , and tank B contains 3.5 kg at
0.5 MPa and 400°C. The valve is now opened and
the two come to a uniform state. Find the final spe-
cific volume.
3,53
3.49
FIGURE P3,49
3.50 Determine the mass of methane gas stored in a 2
m 3 tank at -30°C, 3 MPa. Estimate the percent
error in the mass determination if the ideal gas
\ model is used.
3.51 Saturated water vapor at 60°C has its pressure de-
creased to increase the volume by 10% while keep-
~" ing the temperature constant. To what pressure
should it be expanded?
3.52 Saturated water vapor at 200 kPa is in a constant-
pressure piston/cylinder device. At this state the
piston is 0.1 m from the cylinder bottom. How
much is this distance and what is the temperature if
the water is heated to occupy twice the original
volume?
3.54
3.55
3.56
_ 3.57
3.58
3.59
3.60
3.61
A boiler feed pump delivers 0.05 m 3 /s of water at
240°C, 20 MPa. What is the mass flow rate (kg/s)?
What would be the percent error if the properties of
saturated liquid at 240°C were used in the calcula-
tion? What if the properties of saturated liquid at
20 MPa were used?
Saturated vapor R~134a at 50°C changes volume at
constant temperature. Find the new pressure, and
quality if saturated, if the volume doubles. Repeat
the question for the case where the volume is re-
duced to half the original volume.
A storage tank holds methane at 120 K, with a
quality of 25%, and it warms up by 5°C per hour
due to a failure in the refrigeration system. How
much time will it take before the methane becomes
single phase and what is the pressure then?
A glass jar is filled with saturated water at 500 kPa
of quality 25%, and a tight lid is put on. Now it is
cooled to — 10°C. What is the mass fraction of
solid at this temperature?
Saturated (liquid + vapor) ammonia at 60°C is
contained in a rigid steel tank. It is used in an ex-
periment, where it should pass through the critical
point when the system is heated. What should the
initial mass fraction of liquid be?
A steel tank contains 6 kg of propane (liquid +
vapor) at 20°C with a volume of 0.015 m 3 . The
tank is now slowly heated. Will the liquid level in-
side eventually rise to the top or drop to the bottom
of the tank? What if the initial mass is 1 kg instead
of 6 kg?
A 400-m 3 storage tank is being constructed to hold
LNG, liquified natural gas, which may be assumed
to be essentially pure methane. If the tank is to
contain 90% liquid and 1 0% vapor, by volume, at
100 kPa, what mass of LNG (kg) will the tank
hold? What is the quality in the tank?
A sealed rigid vessel of 2 in 3 contains a saturated
' mixture of liquid and vapor R-134a at 10°C. If it is
heated to 50°C, the liquid phase disappears. Find the
pressure at 50°C and the initial mass of the liquid.
A pressure cooker (closed tank) contains water at
100°C with the liquid volume being 1/10 of the'
vapor volume. It is heated until the pressure
reaches 2.0 MPa. Find the final temperature. Has
the final state more or less vapor than the initial
state?
Homework Problems H 75
3.62 A pressure cooker has the lid screwed on tight. A
small opening with A — 5 mm 2 is covered with a
petcock that can be lifted to let steam escape. How
much mass should the petcock have to allow boiling
at 120°C with an outside atmosphere at 101.3 kPa?
Steam ■-
:-;CCVaf30(;, ; "-:-
ggg Liquid Eg!
FIGURE P3.62
r
3.63 Ammonia at 10°C with a mass of 10 kg is in a
\^_y piston/cylinder assembly with an initial volume of
1 m 3 . The piston initially resting on the stops has a
mass such that a pressure of 900 kPa will float it.
Now the ammonia is slowly heated to 50°C. Find
the final pressure and volume.
Ideal Gas
3.64 A cylinder fitted with a frictionless piston contains
butane at 25°C, 500 kPa. Can the butane reason-
ably be assumed to behave as an ideal gas at this
state?
3.65 A spherical helium balloon 10 m in diameter is at
ambient T and P, 15°C and 100 kPa. How much
helium does it contain? It can lift a total mass that
equals the mass of displaced atmospheric air. How
much mass of the balloon fabric and cage can then
be lifted?
3.66 Is it reasonable to assume that at the given states
the substance behaves as an ideal gas?
a. Oxygen at 30°C, 3 MPa
b. Methane at 30°C, 3 MPa
c. Water at 30°C, 3 MPa
d. R-134aat30°C J 3MPa
e. R-134aat30°C ) 100 kPa
3.67 A 1-m 3 tank is filled with a gas at room temperature
(20°C) and pressure (100 kPa). How much mass is
there if the gas is a. air, b. neon, or c. propane? ^
3.68 A rigid tank of 1 m 3 contains nitrogen gas at 600
kPa, 400 K. By mistake someone lets 0.5 kg flow
out. If the final temperature is 375 K, what is then
\the final pressure?
§9/ A cylindrical gas tank 1 m long, with inside diameter
of 20 cm, is evacuated and then filled with carbon
dioxide gas at 25°C. To what pressure should it be
charged if there should be 1 .2 kg of carbon dioxide?
1.70 A glass is cleaned in 45°C hot water and placed on
the table bottom up. The room air at 20°C that was
trapped in the glass gets heated up to 40°C and some
of it leaks out so that the net resulting pressure inside
is 2 kPa above the ambient pressure of 101 kPa. Now
the glass and the air inside cool do™ to room tem-
perature. What is the pressure inside the glass?
',71 A hollow metal sphere of 150 mm inside diame-
ter is weighed on a precision beam balance when
evacuated and again after being filled to 875 kPa
with an unknown gas. The difference in mass is
0.0025 kg, and the temperature is 25°C. What is
the gas, assuming it is a pure substance listed in
\TabIe A.5?
.72 A vacuum pump is used to evacuate a chamber
; where some specimens are dried at 50°C. The
- pump rate of volume displacement is 0.5 m 3 /s with
an inlet pressure of 0.1 kPa and temperature 50°C.
How much water vapor has been removed over a
30-min period?
.73 A 1-m 3 rigid tank has propane at 100 kPa, 300 K
and connected by a valve to another tank of 0,5 m 3
with propane at 250 kPa, 400 K. The valve is
opened and the two tanks come to a uniform state
at 325 K. What is the final pressure?
FIGURE P3.73
3.74 Verify the accuracy of the ideal-gas model when it
is used to calculate specific volume for saturated
water vapor as shown in Fig. 3.21. Do the calcula-
• . tion for 10 kPa and 1 MPa.
3.75 Assume we have three states of saturated vapor
R-134a at +40°C, 0°C, and -40°C. Calculate the
76 a CHAPTER THREE PROPERTIES OF A PURE SUBSTANCE
3.76
^3,78
specific volume at the set of temperatures and cor-
responding saturated pressure assuming ideal-
gas behavior. Find the percent relative error =
100(u — v g )/u g with u g from the saturated R-134a
table.
Do Problem 3.75, but for the substance R-12.
Do Problem 3.75, but for the substance ammonia,
ir in an automobile tire is initially at -I0°C and
90 kPa. After the automobile is driven awhile, the
temperature gets up to 10°C. Find the new pres-
sure. You must make one assumption on your own.
Air
FIGURE P3.78
3.79 An initially deflated and flat balloon is connected
by a valve to a 12-m 3 storage tank contahiing he-
lium gas at 2 MPa and ambient temperature, 20°C.
The valve is opened and the balloon is inflated at
constant pressure, P Q = 100 kPa, equal to ambient
pressure, until it becomes spherical at = 1 m. If
the balloon is larger than this, the balloon material
is stretched giving a pressure inside as
DJD
P = P n + C I-
Compressibility Factor
3.80
3.82
3.83
Argon is kept in a rigid 5-m 3 tank at ™30°C and 3
MPa. Determine the mass using the compressibil-
ity factor. What is the error (%) if the ideal-gas
model is used?
yWhat is the percent error in. specific volume if the
deal-gas model is used to represent the behavior of
'superheated ammonia at 40°C and 500 kPa? What
if the generalized compressibility chart, Fig. D.l, is
used instead?
A new refrigerant R-125 is stored as a liquid at
-20°C with a small amount of vapor. For a total of
1.5 kg R-125 find the pressure and the volume.
Many substances that normally do not mix well do
so easily under supercritical pressures. A mass of
125 kg ethylene at 7.5 MPa and 296.5 K is stored for
—>~-^uch a process. How much volume does it occupy?
3.84 Cajbon dioxide at 330 K is pumped at a very high
Vjpressure, 10 MPa, into an oil well. As it penetrates
. the rock/oil, the oil viscosity is lowered so it flows
out easily. For this process we need to know the
density of the carbon dioxide being pumped.
3.85 To plan a commercial refrigeration system using
R-123 we would like to know how much more vol-
ume saturated vapor R-123 occupies per kg at
— 30°C compared to the saturated liquid state.
3.86 A bottle with a volume of 0.1 m 3 contains butane
with a quality of 75% and a temperature of 300 K.
Estimate the total butane mass in the bottle using
/^~~^.the generalized compressibility chart.
3.87/Refrigerant R-32 is at -10°C with a quality of
15%. Find the pressure and specific volume.
3.88 A mass of 2 kg of acetylene is in a 0,045-m 3 rigid
container at a pressure of 4.3 MPa. Use the com-
pressibility chart to estimate the temperature by
trial and error.
3.89 A substance is at 2 MPa and 17°C in a 0.25 m 3
rigid tank. Estimate the mass using the compressi-
bility factor if the substance is a. air, b. butane, or
c. propane.
Review Problems
3.90 Tjetermine the quality (if saturated) or tempera-
V_^/ture (if superheated) of the following substances
at the given two states;
a. Water at
1: 120°C, 1 m 3 /kg; 2: 10 MPa, 0.01 m 3 /kg
b. Nitrogen at
1: 1 MPa, 0.03 rnVkg 2: 100 K, 0.03 mVkg
3.91 Fill out the following table for substance ammonia:
PfkPa] T[°C] ulm 3 /kg] x
400
10
0.15
1.0
3.92 Find the phase, quality x if applicable, and the
missing property P or T.
a. H 2 at T = 120°C with v = 0.5 m 3 /kg
b. H 2 atP = 100 kPa with v = 1.8 m 3 /kg
c. H 2 at T = 263 K with v = 200 m 3 /kg
homework problems M 77
, 3.93 Find the phase, quality x if applicable, and the
missing property P or T.
a. NH 3 at P = 800 kPa with u = 0.2 m 3 /kg
b. NH 3 at T = 20°C with y = 0. 1 m 3 /kg
3.94 Give the phase and the missing properties of P, T,
v s andx.
a. R-22 at T = 10°C with v = 0.01 m 3 /kg
b. H 2 at T = 350°C with y = 0.2 m 3 /kg
c. R-12 at T = -5°C and P = 200 kPa
d. R-134a at 294 kPa and v = 0.05 m 3 /kg
3.95 Give the phase and the missing properties of P,
T, v, and x. These may be a little more difficult
if the appendix tables are used instead of the
software.
a. R-22, T= 10°C, v = 0.036 m 3 /kg
b. H 2 0, v = 0.2 m 3 /kg, x = 0.5
c. H 2 0, T= 60°C, v = 0.001016 m 3 /kg
_d. NH 3 , T = 30°C, P = 60 kPa
"~e^R-134a, v = 0.005 m 3 /kg, x = 0.5
3.96 Ai5-m-long vertical tube of cross-sectional area
200 cm 2 is placed in a water fountain. It is filled
y ywith 15°C water; the bottom is closed and the top
is open to the 100 kPa atmosphere.
a. How much water is in the tube?
b. What is the pressure at the bottom of the
tube?
3.97 Consider two tanks, A and B, connected by a
valve, as shown in Fig. P3.97. Each has a volume
of 200 L, and tank A has R-12 at 25 C, 10% liq-
uid and 90% vapor by volume, while tank B is
evacuated. The valve is now opened and saturated
vapor flows from A to B until the pressure in B
has reached that in A, at which point the valve is
closed. This process occurs slowly such that all
temperatures stay at 25°C throughout the process.
How much has the quality changed in tank A dur-
ing the process?
3.98 A spring-loaded piston/cylinder assembly con-
, tains water at 500°C and 3 MPa. The setup is such
that pressure is proportional to volume, P = CV.
It is now cooled until the water becomes saturated
vapor. Sketch the P-v diagram and find the final
pressure.
( 3.99 A 1-m 3 rigid tank has air at 1500 kPa and ambient
\ v 300 K connected by a valve to a piston/cylinder,
V__^,Fig. P3.99. The piston of area 0.1 m 2 . .requires 250
kPa below it to float. The valve is opened and the
piston moves slowly 2 m up and the valve is
closed. During the process air temperature remains
at 300 K. What is the final pressure in the tank?
Valve
FIGURE P3.99
••100 A tank contains 2 kg of nitrogen at 100 K with a
quality of 50%. Through a volume flowmeter and
valve, 0.5 kg is now removed while the tempera-
ture remains constant. Find the final state inside
the tank and the volume of nitrogen removed if
the valve/meter is located at
a. The top of the tank
b. The bottom of the tank
.101 A piston/cylinder arrangement is loaded with a lin-
ear spring and the outside atmosphere. It contains
water at 5 MPa, 400°C with the volume being 0.1
m 3 , as shown in Fig. P3.101. If the piston is at the
FIGURE P3.97
FIGURE P3.101
78 M chapter Three Properties of a Pure Substance
bottom, the spring exerts a force such that P m =
200 kPa, The system now cools until the pressure
reaches 1200 kPa. Find the mass of water, the final
state (T 2 , v 2 ) and plot the P-u diagram for the
^-—--process.
i r 3.102 Water in a piston/cylinder is at 90°C, 100 kPa, and
,the piston loading is such that pressure is propor-
tional to volume, P = CV. Heat is now added until
the temperature reaches 200°C. Find the final pres-
sure and also the quality if in the two-phase region.
3.103 A container with liquid nitrogen at 100 K has a
cross-sectional area of 0.5 m 2 as shown in Fig.
P3.103. Due to heat transfer, some of the liquid
evaporates and in one hour the liquid level drops
30 mm. The vapor leaving the container passes
through a valve and a heater and exits at 500 kPa,
260 K. Calculate the volume rate of flow of nitro-
gen gas exiting the heater.
— (— *
Heater
FIGURE P3.I03
3.104 A cylinder containing ammonia is fitted with a
piston restrained by an external force that is pro-
portional to cylinder volume squared. Initial con-
ditions are 10°C f 90% quality, and a volume of 5
L. A valve on the cylinder is opened and addi-
tional ammonia flows into the cylinder until the
. mass inside has doubled. If at this point the pres-
\ sure is 1 .2 MPa, what is the final temperature?
3.105 A cylinder/piston arrangement contains water at
J 105°C, 85% quality, with a volume of 1 L. The
jl
system is heated, causing the piston to rise and
encounter a linear spring, as shown in Fig.
P3.105. At this point the volume is 1.5 L, the
piston diameter is 150 mm, and the spring con-
stant is 100 N/mm. The heating continues, so
the piston compresses the spring. What is the
cylinder temperature when the pressure reaches
200 kPa?
3.106 Refrigerant- 12 in a piston/cylinder arrangement is
initially at 50°C with.r = 1. It is then expanded in
a process so that P — Cu~ l to a pressure of 100
kPa. Find the final temperature and specific vol-
ume.
3.107 A 1-m 3 rigid tank with air at 1 MPa and 400 K is
connected to an air line as shown in Fig. P3.107,
The vaive is opened and air flows into the tank
until the pressure reaches 5 MPa, at which point
the valve is closed and the temperature inside is
450 K.
a. What is the mass of air in the tank before and
after the process?
b. The tank eventually cools to room temperature,
300 K. What is the pressure inside the tank
then?
Air
line
Tank
FIGURE P3.105
FIGURE P3.107
3.108 Ammonia in a piston/cylinder arrangement is at
700 kPa and 80°C. It is now cooled at constant
pressure to saturated vapor (state 2) at which
point the piston is locked with a pin. The cooling
continues to — 10°C (state 3). Show the processes
1 to 2 and 2 to 3 on both a P-u and a T-v dia-
gram.
3.109 A cylinder has a thick piston initially held by a
pin as shown in Fig. P. 3. 109. The cylinder con-
tains carbon dioxide at 200 kPa and ambient tem-
perature of 290 K. The metal piston has a density
of 8000 kg/m 3 and the atmospheric pressure is
Homework problems M 79
101 kPa. The pin is now removed, allowing the pis-
ton to move and after a while the gas returns to am-
bient temperature. Is the piston against the stops?
50 mm
100 mm
100 mm
CO,
Pin
-100 mm
FIGURE P3.109
J
3.110
For a certain experiment, R-22 vapor is contained
in a sealed glass tube at 20°C It is desired to
know the pressure at this condition, but there is rto
means of measuring it, since the tube is sealed.
However, if the tube is cooled to — 20°C small
droplets of liquid are observed on the glass walls.
jWhat is the initial pressure?
A piston/cylinder arrangement, shown in Fig.
P3.1I1, contains air at 250 kPa and 3Q0°C. The
50-kg piston has a diameter of 0.1 m and initially
pushes against the stops. The atmosphere is at 100
kPa and 20°C. The cylinder now cools as heat is
transferred to the ambient surroundings.
a. At what temperature does the piston begin to
move down?
b. How far has the piston dropped when the tem-
perature reaches ambient?
Air
25 cm
3.112 Air in a tank is at 1 MPa and room temperature of
. 20°C, It is used to fill an initially empty balloon to
a pressure of 200 kPa, at which point the diameter
is 2 m and the temperature is 20°C. Assume the
pressure in the balloon is linearly proportional to
its diameter and that the air in the tank also re-
mains at 20°C throughout the process. Find the
mass of air in the balloon and the minimum re-
quired volume of the tank.
3.113 A cylinder is fitted with a 10-cm-diameter pis-
ton that is restrained by a linear spring (force
proportional to distance), as shown in Fig.
P3.113. The spring force is 80 kN/m and the
piston initially rests on the stops, with a cylin-
der volume of 1 L. The valve to the air line is
opened and the piston begins to rise when the
cylinder pressure is 150 kPa. When the valve is
closed, the cylinder volume is 1.5 L and the
temperature is 80°C. What mass of air is inside
the cylinder?
Air suppfy line
FIGURE P3.111
FIGURE P3.113
114 A^SOO-L tank stores 100 kg of nitrogen gas at 150
,K. To design the tank the pressure must be estt-
— " mated and three different methods are suggested.
Which is the most accurate, and how different in
percent are the other two?
a. Nitrogen tables, Table B.6
b. Ideal gas
c. Generalized compressibility chart, Fig. D.l
3.115 What is the percent error in pressure if the ideal-
gas model is used to represent the behavior of su-
perheated vapor R-22 at 50°C, 0.03082 nrVkg?
What if the generalized compressibility chart, Fig.
D.l, is used instead? (Note that iterations are
needed.)
80 I! Chapter Three Properties of a Pure Substance
Linear Interpolation
3.116 Find the pressure and temperature for saturated
vapor R-12 with u ~ 0. 1 mVkg
3.117 YUse a linear interpolation to estimate properties of
, ammonia to fill out the table below.
P [kPa] T [°C] v [m 3 /kg] x
a. 550 0.75
b. 80 20
c. 10 0.4
3.118 Use a linear interpolation to estimate 7^ at 900
kPa for nitrogen. Sketch by hand the curve P $zt (T)
by using a few table entries around 900 kPa from
Table B.6.1. Is your linear interpolation above or
below the actual curve?
3.119 Use a double linear interpolation to find the pres-
sure for superheated R-134a at 13°C with v = 0.3
m 3 /kg.
3.120 Find the specific volume of ammonia at 140 kPa
' and 0°C.
3.121 Find the pressure of water at 200°C and specific
volume of 1.5 nrVkg.
Computer Tables
3.122 Use the computer software to find the properties
for water at the four states in Problem 3.33.
3.123 Use the computer software to find the properties for
ammonia at the four states listed in Problem 3.37.
3.124 Use the computer software to find the properties for
ammonia at the three states listed in Problem 3.1 17.
3.125 Find the value of the saturated temperature for ni-
trogen by linear interpolation in Table B.6.1 for a
pressure of 900 kPa. Compare this to the value
given by the computer software.
3.126 Write a computer program that lists the states P, T,
and v along the process curves in Problem 3.111.
3.127 Use the computer software to sketch the variation
of pressure with temperature in Problem 3.41. Ex-
tend the curve a little into the single-phase region.
ENGLISH UNIT PROBLEMS
English Unit Concept Problems
3.128E Cabbage needs to be cooked (boiled) at 250 F.
What pressure should the pressure cooker be set
for?
3.129E If I have 1 ft 3 of ammonia at 15 psia, 60 F how
much mass is that?
3.130E For water at 1 atm with a quality of 10% find the
volume fraction of vapor.
3. 13 IE Locate the state of R- 134a at 30 psia, 20 F. Indi-
cate in both the P-v and the T-v diagrams the
location of the nearest states listed in the printed
Table F.S.
3.132E Calculate the ideal gas constant for argon and
hydrogen based on Table F.l and verify the
value with Table F.4.
English Unit Problems
3.133E Water at 80 F can exist in different phases de-
pendent on the pressure. Give the approximate
pressure range in lbf/in 2 for water being in each
one of the three phases, vapor, liquid, or solid.
3.134E A substance is at 300 lbf/in 2 , 65 F in a rigid
tank. Using only the critical properties, can the
phase of the mass be determined if the substance
is nitrogen, water, or propane?
3.135E Determine whether water at each of the follow-
ing states is a compressed liquid, a superheated
vapor, or a mixture of saturated liquid and vapor.
a. 1800 lbiVin 2 , 0.03 ftVlbm
b. 150 lbf/in 2 , 320 F
c. 380 F, 3 ftVlbm
3.136E Determine whether water at each of the following
states is a compressed liquid, a superheated vapor,
or a mixture of saturated liquid and vapor.
a. 2 lbf/in 2 , 50 F
b. 270 F, 30 lbf/in 2
c. 160 F, 10 ftVlbm
3.137E Give the phase and the missing property of P, T,
v and x.
a. R-134a, T~ -lOF.J^ 18 psia .
b. R-134a,P = 50 psia, u = 1.3 ftVlbm
c. NH 3) T = 120 F, v = 0.9 ftVlbm
d. NH 3) T = 200 F, v = 11 ftVlbm
English UNrr problems H 81
3.138E Give the phase and the specific volume.
a. R-22, T = - 10 F, P = 30 lbf/in 2
b. R-22, T = - 10 F, P - 40 lbf/in 2
c. H 2 0, r = 280 F, p = 35 lbf/in 2
d. NH 3 , T = 60 F, P = 15 lbf/in 2
3.139E A water storage tank contains liquid and vapor
in equilibrium at 220 F. The distance from the
bottom of the tank to the liquid level is 25 ft.
What is the absolute pressure at the bottom of
the tank?
3.140E A sealed rigid vessel has volume of 35 ft 3 and
contains 2 Ibm of water at 200 F. The vessel is
now heated. If a safety pressure valve is in-
stalled, at what pressure should the valve be set
to have a maximum temperature of 400 F?
3.141E You want a pot of water to boil at 220 F. How
heavy a lid should you put on the 6-in.-diameter
pot wheni 5 ^ = 14.7 psia?
3.142E Saturated water vapor at 200 F has its pressure
decreased to increase the volume by 10%, keep-
ing the temperature constant. To what pressure
should it be expanded?
3.143E A hoiler feed pump delivers 100 ftVmin of
water at 400 F, 3000 lbf/in 2 . What is the mass
flowrate (Ibm/s)? What would be the percent
error if the properties of saturated liquid at 400
F were used in the calculation? What if the
properties of saturated liquid at 3000 lbf/in 2
were used?
3.144E A pressure cooker has the lid screwed on tight.
A small opening with A = 0.0075 in 2 is covered
with a petcock that can be lifted to let steam es-
cape. How much mass should the petcock have
to allow boiling at 250 F with an outside atmos-
phere at 15 psia?
3.145E A steel tank contains 14 Ibm of propane (liquid +"
vapor) at 70 F with a volume of 0.25 ft 3 . The
tank is now slowly heated. Will the liquid level
inside eventually rise to the top or drop to the
bottom of the tank? What if the initial mass is 2
Ibm instead of 14 Ibm?
3.146E A cylindrical gas tank 3 ft long, inside diameter
of 8 in., is evacuated and then filled with carbon
dioxide gas at 77 F. To what pressure should it
be charged if there should be 2.6 Ibm of carbon
dioxide?
3.147E A spherical helium balloon of 30 ft in diameter
is at ambient T and P, 60 F and 14.69 psia. How
much helium does it contain? It can lift a total
mass that equals the mass of displaced atmos-
pheric air. How much mass of the balloon fabric
and cage can then be lifted?
3.148E Give the phase and the specific volume.
a. C0 2 , T = 510 F, P = 75 lbf/in 2
b. Air, J= 68 F, P = 2 arm
c. Ar, T= 300 F, P = 30 lbf/in 2
3.149E What is the percent error in specific volume if
the ideal-gas model is used to represent the be-
havior of superheated ammonia at 100 F, 80
lbf/in 2 ? What if the generalized compressibility
chart, Fig. D.l, is used instead?
3.150E A cylinder is fitted with a 4-in.-diameter piston
that is restrained by a linear spring (force pro-
portional to distance) as shown in Fig. P3.113.
The spring force constant is 400 lbf/in. and the
piston initially rests on the stops, with a cylinder
volume of 60 in 3 . The valve to the air line is
opened and the piston begins to rise when the
cylinder pressure is 22 lbf/in 2 . When the valve is
closed, the cylinder volume is 90 in 3 and the
temperature is 180 F. What mass of air is inside
the cylinder?
3.151E A 35 -ft 3 rigid tank has propane at 15 psia,
540 R and connected by a valve to another
tank of 20 ft 3 with propane at 40 psia, 720 R.
The valve is opened and the two tanks come to
a uniform state at 600 R. What is the final
pressure?
3.152E Two tanks are connected together as shown
in Fig. P3.49, both containing water. Tank A
is at 30 lbf/in 2 , v = 8 ftVlbm, V = 40 ft 3 ,
and tank B contains 8 Ibm at 80 lbf/in 2 ,
750 F. The valve is now opened and the two
come to a uniform state. Find the final specific
volume.
3.153E A 35-ft 3 rigid tank has air at 225 psia and ambi-
ent 600 R connected by a valve to a piston/
cylinder. The piston of area 1 ft 2 requires 40
psia below it to float (see Fig. P3.99). The valve
is opened and the piston moves slowly 7 ft up
and the valve is closed. During the process air
temperature remains at 600 R. What is the final
pressure in the tank?
82 M chapter Three Properties of a pure Substance
3.154E Give the phase and the missing properties of P,
T, v, and x. These may be a little more difficult
if the appendix tables are used instead of the
software.
a. R-22, T—50F,v — 0.6 ftVlbm
b. H 2 0, v = 2 ftVlbm,* = 0.5
c. H 2 0, T = 150 F, v = 0.01632 ftVlbm
d. NH 3 , T — 80 F, P — 13ibf/in 2
e. R-134a, v = 0.08 ftVlbm, .x = 0.5
3.155E A pressure cooker (closed tank) contains water
at 200 F with the liquid volume being 1/10 of
the vapor volume. It is heated until the pressure
reaches 300 lbf/in 2 . Find the final temperature.
Has the final state more or less vapor than the
initial state?
3.156E Refrigerant-22 in a piston/cylinder arrangement
is initially at 120 F, x = 1 . It is then expanded in
a process so that P = Cv~ l to a pressure of 30
lbf/in 2 . Find the final temperature and specific
volume.
3.157E A substance is at 70 F, 300 lbf/in 2 in a 10-ft 3
tank. Estimate the mass from the compressibil-
ity chart if the substance is (a) air, (b) butane, or
(c) propane.
3.158E Determine the mass of an ethane gas stored in a
25 -ft 3 tank at 250 F, 440 lbf/in 2 using the com-
pressibility chart. Estimate the error (%) if the
ideal gas model is used.
Computer, design and Open-Ended Pr<
3.159 Make a spreadsheet that will tabulate and plot the
saturated pressure versus temperature for ammo-
nia starting with T = -40°C ending at the critical
point in steps of 1 0°C.
3.160 Make a spreadsheet that will tabulate and plot
values of P and T along a constant specific vol-
ume line for water. Starting state is 100 kPa, qual-
ity of 50% and the ending state is 800 kPa.
3.161 Write a computer program that lists the states P,
T, and v along with the process curves in Problem
3.111.
3.162 Use the computer software to sketch the variation
of pressure with temperature in Problem 3.55, Ex-
tend the curve a little into the single phase region.
v 3.163 By the use of the computer software find a few
of the states between the beginning and end
states and show the variation of pressure and
temperature as a function of volume for Prob-
lem 3.102.
3.164 In Problem 3.106 we wish to follow the path of
the process for the R-12 for any state between the
initial and final states inside the cylinder.
3.165 For any specified substance in Tables B.1-B.7, fit
a polynomial equation of degree n to tabular data
for pressure as a function of density along any
given isotherm in the superheated vapor region.
3.166 The refrigerant fluid in a household refrigerator
changes phase from liquid to vapor at the low
temperature in the refrigerator. It changes phase
from vapor to liquid at the higher temperature in
the heat exchanger that gives the energy to the
room air. Measure or otherwise estimate these
temperatures. Based on these temperatures make
a table with the refrigerant pressures for the re-
frigerants for which tables are available in Appen-
dix B. Discuss the results and the requirements
for a substance to be a potential refrigerant.
3.167 Repeat the previous problem for refrigerants that
are listed in Table A.2 and use the compressibility
chart Fig. D.l to estimate the pressures.
3.168 The saturated pressure as a function of tempera-
ture follows the correlation developed by Wagner
as
ltt P r = [WjT + w 2 r u + W3T 3 + vf 4 T 6 ]/r r
where the reduced pressure and temperature are
P r - P!P C and T r - T/T c . The temperature vari-
able is t — 1 T r , The parameters are found for
R-12andR-134a as
R-12 -6.91826 1.49560 -2.65015 -0.63170
R-134a -7.59884 1.48886 -3.79873 1.81379
Compare these correlations to the tables in Ap-
pendix B.
3.169 Find the constants in the curve fit for the satura-
tion pressure using Wagner's correlation as
Computer, design and open-ended problems B 83
shown in the previous problem for water and
methane. Find other correlations in the literature
and compare them to the tables and give the max-
imum deviation.
3.170 The specific volume of saturated liquid can be ap-
proximated by the Rackett equation as
with the reduced temperature, T r ~ T/T c> and the
compressibility factor, Z c = P c vJRT c . Using val-
ues from Table A.2 with the critical constants,
compare the formula to the tables for substances
where the saturated specific volume is available.
ilgg
W -
:
Work and Heat
In this chapter we consider work and heat. It is essential for the student of thermodynam-
ics to understand clearly the definitions of both work and heat, because the correct analy-
sis of many thermodynamic problems depends on distinguishing between them.
Work and heat are energy in transfer from one system to another and thus play a
crucial role in most thermodynamic systems or devices. As we want to analyze such sys-
tems, we need to model the heat and work as functions of properties and parameters char-
acteristic of the system or how it functions. An understanding of the physics involved
allows us to construct a model for the heat and work and use the result in our analysis of
the energy transfers and changes, which we will do with the first law of thermodynamics
in Chapter 5.
To facilitate an understanding of the basic concepts we present a number of physi-
cal arrangements that will enable us to express the work done from changes in the system
during a process. We will also examine work that is the result of a given process without
going into details about how the process physically can be made to occur. This is done
because such a description will be too complex and involve concepts that are not covered
so far, but at least we can examine the result of the process.
A general description of heat transfer in different situations is a subject that usually
is studied separately. However, a very simple introduction is beneficial so that the heat
transfer does not become too abstract and we can relate it to the processes we examine.
Heat transfer by conduction, convection (flow), and radiation is presented in terms of
very simple models, emphasizing that it is driven by a temperature difference.
4.1 DEFINITION OF WORK
Work is usually defined as a force F acting through a displacement ,v, where the displace-
ment is in the direction of the force. That is
(4.1)
This is a very useful relationship becausertenables us to find the work required to raise a
weight, to stretch a wire, or to move a charged particle through a magnetic field.
However, when treating thermodynamics from a macroscopic point of view, it is
advantageous to tie in the definition of work with the concepts of systems, properties, and
processes. We therefore define work as follows: Work is done by a system if the sole ef-
Hl^^^g^ feet on the surroundings (everything external to the system) could be the raising of a
"~ weight. Notice that the raising of a weight is in effect a force acting through a distance.
Notice also that our definition does not state that a weight was actually raised or that a
84
Definition of Work H 85
FIGURE 4.1 Example
of work crossing the
boundary of a system.
boundary
force actually acted through a given distance, but that the sole effect external to the sys-
tem could be the raising of a weight. Work done by a system is considered positive and
work done on a system is considered negative. The symbol W designates the work done
by a system.
In general, work is a form of energy in transit, that is, energy being transferred
across a system boundary. The concept of energy and energy storage or possession has
been discussed in some detail in Section 2.6. Work is the form of energy that fulfills the
definition given in the preceding paragraph.
Let us illustrate this definition of work with a few examples. Consider as a system
the battery and motor of Fig. 4.1c and let the motor drive a fan. Does work cross the
boundary of the system? To answer this question using the definition of work given ear-
lier, replace the fan with the pulley and weight arrangement shown in Fig. 4.1b. As the
motor turns, the weight is raised, and the sole effect external to the system is the raising of
a weight. Thus, for our original system of Fig. 4.1a, we conclude that work is crossing the
boundary of the system, since the sole effect external to the system could be the raising of
a weight.
Let the boundaries of the system be changed now to include only the battery shown
in Fig. 4.2. Again we ask the question, does work cross the boundary of the system? To
answer this question, we need to ask a more general question; Does the flow of electrical
energy across the boundary of a system constitute work?
The only limiting factor in having the sole external effect be the raising of a weight
is the inefficiency of the motor. However, as we design a more efficient motor, with lower
bearing and electrical losses, we recognize that we can approach a certain limit that meets
the requirement of having the only external effect be the raising of a weight. Therefore,
Pulley
FIGURE 4.2 Example
of work crossing the
boundary of a system
because of a flow of an
electric current across the
system boundary.
86 H Chapter Four Work and Heat
we can conclude that when there is a flow of electricity across the boundary of a system,
as in Fig. 4.2, it is work.
42 Units for Work
As already noted, work done by a system, such as that done by a gas expanding against a
piston, is positive, and work done on a system, such as that done by a piston compressing
a gas, is negative. Thus, positive work means that energy leaves the system, and negative
work means that energy is added to the system.
Our definition of work involves raising of a weight, that is, the product of a unit
force (one newton) acting through a unit distance (one meter). This unit for work in SI
units is called the joule (J)
1 J - INm
Power is the time rate of doing work and is designated by the symbol W:
dt
The unit for power is a rate of work of one joule per second, which is a watt ( W):
I W = 1 J/s
A familiar unit for power in English units is the horsepower (hp), where
1 hp - 550 ft lbf/s
Note that the work crossing the boundary of the system in Fig. 4.1 is that associated
with a rotating shaft. To get the expression for power we use the differential work from
Eq. 4.1 as
BW=Fdx = Frd6 = TdB
that is, force acting through a distance dx or a torque (T = Ft') acting through an angle of
rotation as shown in Fig. 4.3. Now the power becomes
W=§E = F 4!L = F yf = Fr d9. = Tco (4,2)
dt dt dt v ;
that is, a force times rate of displacement (velocity) or a torque times angular velocity.
F
FIGURE 4.3 Force
acting at radius r gives a
torque T = Fr,
Work done at the moving boundary of a simple compressible system
■ 87
It is often convenient to speak of the work per unit mass of the system, often termed
"specific work." This quantity is designated w and is defined
_ W
yy = - —
m
4.3 work done at the moving boundary
oe a simple Compressible System
We have already noted that there are a variety of ways in which work can be done on
or by a system. These include work done by a rotating shaft, electrical work, and the
work done by the movement of the system boundary, such as the work done in moving
the piston in a cylinder. In this section we will consider in some detail the work done
at the moving boundary of a simple compressible system during a quasi- equilibrium
process.
Consider as a system the gas contained in a cylinder and piston, as in Fig. 4.4. Let
one of the small weights be removed from the piston, which will cause the piston to
move upward a distance dL. We can consider this quasi-equilibrium process and calcu-
late the amount of work W done by the system during this process. The total force on
the piston is PA, where P is the pressure of the gas and A is the area of the piston.
Therefore, the work S W is
SW=PAdL
But A dL ~ dV, the change in volume of the gas. Therefore,
1
A
FIGURE 4.4 Example
of work done at the
moving boundary of a
system in a quasi-
equilibrium process.
8W=PdV~)
(4.3)
The work done at the moving boundary during a given quasi-equilibrium process can be
found by integrating Eq. 4.3. However, this integration can be performed only if we know
the relationship between P and V during this process. This relationship may be expressed
in the form of an equation, or it may be shown in the form of a graph.
Let us consider a graphical solution first. We use as an example a compression
process such as occurs during the compression of air in a cylinder, Fig. 4.5. At the begin-
ning of the process the piston is at position 1, and the pressure is relatively low. This state
is represented on a pressure-volume diagram (usually referred to as a P—V diagram). At
the conclusion of the process the piston is in position 2, and the corresponding state of the
gas is shown at point 2 on the P—V diagram. Let us assume that this compression was a
quasi-equilibrium process and that during the process the system passed through the states
shown by the line connecting states 1 and 2 on the P—V diagram. The assumption of a
quasi-equilibrium process is essential here because each point on line 1-2 represents a
definite state, and these states will correspond to the actual state of the system only if the
deviation from equilibrium is infinitesimal. The work done on the air during this compres-
sion process can be found by integrating Eq. 4.3:
=/>=/;
PdV
(4.4)
88 H CHAPTER FOUR WORK AND HEAT
FIGURE 4.5 Use of
pressure-volume diagram
to show work done at the
moving boundary of a
system in a quasi-
equilibrium process.
1
b
i
a
V
2
I
I
The symbol t W 2 is to be interpreted as the work done during the process from state 1 to
state 2. It is clear from examining the P-V diagram that the work done during this
process,
\\pdv
is represented by the area under the curve 1-2, area a-\-2-b-a. In this example the vol-
ume decreased, and the area a -\-2-b-a represents work done on the system. If the
process bad proceeded from state 2 to state 1 along the same path, the same area would
represent work done by the system.
Further consideration of a P-V diagram, such as Fig. 4.6, leads to another important
conclusion. It is possible to go from state 1 to state 2 along many different quasi-equilib-
rium paths, such as A, B, or C. Since the area underneath each curve represents the work
for each process, the amount of work done during each process not only is a function of
the end states of the process but depends on the path that is followed in going from one
state to another. For this reason work is called a path function or, in mathematical par-
lance, 8Wis an inexact differential.
This concept leads to a brief consideration of point and path functions or, to use an-
other term, exact and inexact differentials. Thermodynamic properties are point functions,
a name that comes from the fact that for a given point on a diagram (such as Fig. 4.6) or
surface (such as Fig. 3.18), the state is fixed, and thus there is a definite value of each
p
FIGURE 4.6 Various
quasi-equilibrium
processes between two
given states, indicating
that work is a path
function. b a V
r
work done at the moving boundary of a Simple compressible system H 89
property corresponding to this point. The differentials of point functions are exact differ-
entials, and the integration is simply
-2
dv= r 2 - v x
i
Thus, we can speak of the volume in state 2 and the volume in state 1 , and the
change in volume depends only on the initial and final states.
Work, however, is a path function, for, as has been indicated, the work done in a quasi-
equilibrium process between two given states depends on the path followed. The differentials
of path functions are inexact differentials, and the symbol 8 will be used in this text to desig-
nate inexact differentials (in contrast to d for exact differentials). Thus, for work, we write
-2
1
It would be more precise to use the notation [ W Ui which would indicate the work
done during the change from state 1 to state 2 along path ^, However, it is implied in the
notation , W 2 that the process between states 1 and 2 has been specified. It should be noted
that we never speak about the work in the system in state 1 or state 2, and thus we would
never write W 2 ~ W { .
In evaluating the integral of Eq. 4.4, we should always keep in mind that we wish to
determine the area under the curve in Fig. 4.6. In connection with this point, we identify
the following two classes of problems.
1. The relationship between P and V is given in terms of experimental data or in
graphical form (as, for example, the trace on an oscilloscope). Therefore, we may
evaluate the integral, Eq. 4.4, by graphical or numerical integration.
2. The relationship between P and V makes it possible to fit an analytical relationship
between them. We may then integrate directly.
One common example of this second type of functional relationship is a process
called a polytropic process, one in which
PV n - constant
throughout the process. The exponent n may possibly be any value from — co to +«>, depend-
ing on the particular process. For this type of process, we can integrate Eq. 4.4 as follows:
PV*= constant - PiH= p 2 V l
„ _ constant _ P^i _ ^tV\
■yn yn yn
2 f 2 AV f y~n+l
f 2 f 2 dV
PdV = constant ™ = constant , , ,
J i J i V \~n + 1
2 p , v _ constant (yl - n _ ^ _ P 2 V n 2 V 1 f -
_P 2 V 2 -P l V l
Note that the resulting Eq. 4.5 is valid for any exponent ?i, except n = 1. Where n = 1,
PK= constant = P X V X = P 2 V 2
(4.5)
90 @ Chapter Four Work and Heat
and
f*Pdr= Pfr Hf-P^ ln^ (4.6)
Note that in Eqs. 4.5 and 4.6 we did not say that the work is equal to the expressions
given in these equations. These expressions give us the value of a certain integral, that is,
a mathematical result. Whether or not that integral equals the work in a particular process
depends on the result of a thermodynamic analysis of that process. It is important to keep
the mathematical result separate from the thermodynamic analysis, for there are many sit-
uations in which work is not given by Eq. 4.4.
The polytropic process as described demonstrates one special functional relation-
ship between P and V during a process. There are many other possible relations, some of
which will be examined in the problems at the end of this chapter.
EXAMPLE 4.1 Consider as a system the gas in the cylinder shown in Fig. 4.7; the cylinder is fitted with
a piston on which a number of small weights are placed. The initial pressure is 200 kPa,
and the initial volume of the gas is 0.04 m 3 .
a. Let a Bunsen burner be placed under the cylinder, and let the volume of the gas in-
crease to 0.1 m 3 while the pressure remains constant. Calculate the work done by the
system during this process.
x W 2 = ^PdV
Since the pressure is constant, we conclude from Eq. 4.4 that
1 W 2 =p\\v = P{V 2 -V 1 )
J¥ 2 = 200 kPa X (0,1 - 0.04)m 3 = 12.0 kJ
b. Consider the same system and initial conditions, but at the same time as the Bunsen
burner is under the cylinder and the piston is rising, let weights be removed from
the piston at such a rate that, during the process, the temperature of the gas remains
constant.
If we assume that the ideal-gas model is valid, then, from Eq. 3.5,
PV^mRT
We note that this is a polytropic process with exponent n = 1. From our analysis, we
conclude that the work is given by Eq. 4.4 and that the integral in this equation is
given by Eq. 4.6. Therefore,
f 2 v 2
{ W 2 = } PdV^Pftbi^
200 kPa X 0.04 m 3 X In = 7.33 kJ
Gas
FIGURE 4.7 Sketch
for Example 4. 1.
Work Done at the Moving boundary of a Simple compressible System B 91
c. Consider the same system, but during the heat transfer let the weights be removed at
such a rate that the expression PV 13 = constant describes the relation between pres-
sure and volume during the process. Again the final volume is 0.1 m 3 . Calculate the
work.
This is a polytropic process in which n — 1.3. Analyzing the process, we con-
clude again that the work is given by Eq. 4.4 and that the integral is given by Eq. 4.5.
Therefore,
d. Consider the system and initial state given in the first three examples, but let the
piston be held by a pin so that the volume remains constant. In addition, let heat
be transferred from the system until the pressure drops to 100 kPa. Calculate the
work-
Since SW = P dV for a quasi-equilibrium process, the work is zero, because
there is no change in volume.
The process for each of the four examples is shown on the P-V diagram of Fig'. 4.8,
Process l-2a is a constant-pressure process, and area \-2a-f—e—\ represents the work.
Similarly, line 1-26 represents the process in which PV = constant, line \-2c the process
in which PV 1 ' 3 = constant, and line \-2d the constant-volume process. The student
should compare the relative areas under each curve with the numerical results obtained
for the amounts of work done.
6.41 kJ
p
FIGURE 4.8
Pressure-volume diagram
showing work done in the
various processes of
Example 4.1.
2d
2b
2c
e
f
V
92 m
Chapter Four Work and heat
EXAMPLE 4.2 Consider a slightly different piston/cylinder arrangement as shown in Fig. 4.9. In this
example the piston is loaded with a mass, m p , the outside atmosphere P Qj a linear spring,
and a single point force F lt The piston traps the gas inside with a pressure P. A force
balance on the piston in the direction of motion yields
with a zero acceleration in a quasi- equilibrium process. The forces, when the spring is in
contact with the piston, are
2F t =A4, ^ l F [ =m p g + P Q A + k s (x-x ) + F 1
with the linear spring constant, k s . The piston position for a relaxed spring is ,r 0) which
depends on how the spring is installed. The force balance then gives the gas pressure by
division with the area A as
P = Po + [m p g + F l + k s (x~x )]/A
To illustrate the process in a P-V diagram, the distance x is converted to volume
by division and multiplication with A :
This relation gives the pressure as a linear function of the volume, with the line
having a slope of C 2 = kJA 2 . Possible values of P and Fare as shown in Fig. 4.10 for
an expansion. Regardless of what substance is inside, any process must proceed along
the line in the P-V diagram. The work term in a quasi-equilibrium process then fol-
lows as
tW 2 = j P dV ~ area under the process curve
For a contraction instead of expansion, the process would proceed in the opposite direc-
tion from the initial point 1 along a line of the same slope shown in Fig. 4.10.
FIGURE 4,9 Sketch of physical system for Example 4.2.
work done at the moving boundary of a simple Compressible system
a 93
FIGURE 4.10 Hie
process curve showing
possible P-V combinations
for Example 4.2.
EXAMPLE 4.3 The cylinder/piston setup of Example 4.2 contains 0.5 kg ammonia at -20°C with a
quality of 25%. The ammonia is now heated to +20°C, at which state the volume is
observed to be 1.41 times larger. Find the final pressure and the work the ammonia
produced.
Solution
The forces acting on the piston, gravitation constant, external atmosphere at constant
pressure and the linear spring give a linear relation between P and v( V).
State 1: (T u *,) from Table B.2.1
Pi = = 190.2 kPa
Vl = v f +xiv A = 0.001 504 + 0.25 X 0.621 84 = 0.156 96 rnVkg
State 2: (T 2i v 2 = 1.41 Vt = 1.41 X 0.156 96 = 0.2213 nvVkg)
Table B.2.2 state very close to P 2 - 600 kPa.
Process: P = C x + C 2 v
The work term can now be integrated knowing P versus v and can be seen as the area in
the P-v diagram, shown in Fig. 4.11.
l W 2 = fpDV= fpmdv= area = m~ (Pj + P 2 )(v 2 - Vj)
- 0.5 kg j (190.2 + 600) kPa (0.2213 - 0.156 96) mVkg
- 12.71 kJ
94 M CHAPTER FOUR WORK AND HEAT
FIGURE 4.11 Diagrams for Example 4.3.
EXAMPLE 4.4 The piston/cylinder setup shown in Figure 4.12 contains 0.1 kg of water at 1000 kPa,
500°C. The water is now cooled with a constant force on the piston until it reaches half
the initial volume. After this it cools to 25°C while the piston is against the stops. Find
the final water pressure and the work in the overall process, and show the process in a
P-v diagram.
Solution
We recognize this is a two-step process, one of constant P and one of constant V. This
behavior is dictated by the construction of the device.
State 1
Process 1-la
la-2
State 2
(P, T) From Table B.1.3; v x = 0.354 11 m 3 /kg.
P = constant = F/A
v — constant -- v la = v 2 = v x !2
(T t v 2 = v x l2 - 0.177 06 m 3 /kg)
From Table B.l.l, v 2 < v g> so the process is two phase andP 2 = P^ t = 3.169 kPa.
i W 2 = j* PdV= m j 2 Pdv^ mP x (v la - v,) +
= 0.1 kg X 1000 kPa (0.177 06 - 0.354 11) mVkg - -17.7 kj
Note that the work done from lo to 2 is zero (no change in volume) as shown in
Fig. 4.13.
FIGURE 4.12 Sketch
for Example 4.4.
WORK DONE AT THE MOVING BOUNDARY OF A SIMPLE COMPRESSIBLE SYSTEM M 95
In this section we have discussed boundary movement work in a quasi-equilibrium
process. We shouid also realize that there may very well be boundary movement work in
a nonequiiibrium process. Then the total force exerted on the piston by the gas inside the
cylinder, PA, does not equal the external force, F ext , and the work is not given by Eq. 4.3.
The work can, however, be evaluated in terms of F Mt or, dividing by area, an equivalent
external pressure, P ext . The work done at the moving boundary in this case is
8W = F tM t!L = P at dV (4.7)
Evaluation of Eq. 4.7 in any particular instance requires a knowledge of how the external
force or pressure changes during the process.
EXAMPLE 4.5
FIGURE 4.14
Example of a
nonequiiibrium process.
Consider the system shown in Fig. 4.14 in which the piston of mass m p is initially held
in place by a pin. The gas inside the cylinder is initially at pressure P l and volume V lt
When the pin is now released, the external force per unit area acting on the system (gas)
boundary is comprised of two parts:
P a t = F*xlA = P + m p g!A
Calculate the work done by the system when the piston has come to rest.
After the piston is released, the system is exposed to the boundary pressure equal
to P m , which dictates the pressure inside the system, as discussed in Section 2.8 in con-
nection with Fig. 2.9. We further note that neither of the two components of this external
force will change with a boundary movement, since the cylinder is vertical (gravitational
force) and the top is open to the ambient surroundings (movement upward merely
pushes the air out of the way). If the initial pressure P x is greater than that resisting the
boundary, the piston will move upward at a finite rate, that is, in a nonequiiibrium
process, with the cylinder pressure eventually coming to equilibrium at the value P est . If
we were able to trace the average cylinder pressure as a function of time, it would typi-
cally behave as shown in Fig. 4.15. However, the work done by the system during this
process is done against the force resisting the boundary movement and is therefore given
by Eq. 4.7. Also, since the external force is constant during this process, the result is
iW 2
P^V=P K JV 2 -V 1 )
96 M CHAPTER FOUR work and heat
FIGURE 4.15
Cylinder pressure as a I —
function of time. Time
where V 2 is greater than V h and the work by the system is positive. If the initial pressure
had been less than the boundary pressure, the piston would have moved downward,
compressing the gas, with the system eventually coming to equilibrium at P extJ at a vol-
ume less than the initial volume, and the work would be negative, that is, done on the
system by its surroundings.
44 Other Systems That Involve work
In the preceding section we considered the work done at the moving boundary of a simple
compressible system during a quasi-equilibrium process and also during a nonequilibrium
process. There are other types of systems in which work is done at a moving boundary. In
this section we briefly consider three such systems, a stretched wire, a surface film, and
electrical work.
Consider as a system a stretched wire that is under a given tension 2T. When the
length of the wire changes by the amount dL, the work done by the system is
W=-VdL (4.8)
The minus sign is necessary because work is done by the system when dL is negative.
This equation can be integrated to have
,JF 2 = - J Vdt (4.9)
The integration can be performed either graphically or analytically if the relation between
9" and L is known. The stretched wire is a simple example of the type of problem in solid-
body mechanics that involves the calculation of work.
EXAMPLE 4,6 A metallic wire of initial length L is stretched. Assuming elastic behavior, determine
the work done in terms of the modulus of elasticity and the strain.
Let a = stress, e = strain, and E = the modulus of elasticity.
A
Other Systems That Involve Work
a 97
Therefore,
From the definition of strain,
de — —
Therefore,
SW=-VdL = ~AEeL a de
J e =o I
Now consider a system that consists of a liquid film having a surface tension SP. A
schematic arrangement of such a film, maintained on a wire frame, one side of which can
be moved, is shown in Fig. 4.16. When the area of the film is changed, for example, by
sliding the movable wire along the frame, work is done on or by the film. When the area
changes by an amount dA, the work done by the system is
SW= -VdA (4,10)
For finite changes,
(4.11)
We have already noted that electrical energy flowing across the boundary of a sys-
tem is work. We can gain further insight into such a process by considering a system in
which the only work mode is electrical. As examples of such a system, we can think of a
charged condenser, an electrolytic cell, and the type of fuel cell described in Chapter 1.
Consider a quasi-equnibrium process for such a system, and during this process let the
potential difference be % and the amount of electrical charge that flows into the system be
dZ. For this quasi-equilibrium process the work is given by the relation
W= -% dZ (4.12)
98 m CHAPTER FOUR WORK AND HEAT
Since the current, /, equals dZldt (where t — time), we can also write
BW= ~%idt (4.12)
X W 2 =- f%idt (4.13)
Equation 4. 1 3 may also be written as a rate equation for work (the power).
dt
Since the ampere (electric current) is one of the fundamental units in the International
System, and the watt has been defined previously, this relation serves as the definition of
the unit for electric potential, the volt (V), which is one watt divided by one ampere.
4.5 CONCLUDING REMARKS REGARDING WORK
The similarity of the expressions for work in the three processes discussed in Section 4,4
and in the processes in which work is done at a moving boundary should be noted. In each
of these quasi-equilibrium processes, the work is given by the integral of the product of an
intensive property and the change of an extensive property. The following is a summary
list of these processes and their work expressions
•2
PdV
•2
Stretched wire , JF, = - f £T dL
•2
Surface film JF 2 = - I <f dA
Simple compressible system iW 2 — j
/:
System in which the work is completely electrical j W 2 ~ — j % dZ (4.1 5)
Although we will deal primarily with systems in which there is only one mode of work,
it is quite possible to have more than one work mode in a given process. Thus, we could write
$W=PdV- % dL - VdA -%dZ+ - • (4.16)
where the dots represent other products of an intensive property and the derivative of a re-
lated extensive property. In each term the intensive property can be viewed as the driving
force that causes a change to occur in the related extensive property, which is often
termed the displacement. Just as we could derive the expression for power for the single
point force in Eq. 4.2, the rate form of Eq. 4. 1 6 expresses the power as
jy = iK = P y- g-v -&A -%z+ • • ■ (4.17)
dt
It should also be noted that many other forms of work can be identified in processes
that are not quasi-equilibrium processes. For example, there is the work done by shearing
forces in the friction in a viscous fluid or the work done by a rotating shaft that crosses the
system boundary.
CONCLUDING REMARKS REGARDING WORK
m 99
FIGURE 4.17
Example of process
involving a change of
volume for which the
work is zero.
System^
boundary
(b)
The identification of work is an important aspect of many thermodynamic prob-
lems. We have already noted that work can be identified only at the boundaries of the sys-
tem. For example, consider Fig. 4.17, which shows a gas separated from the vacuum by a
membrane. Let the membrane rupture and the gas fill the entire volume. Neglecting any
work associated with the rupturing of the membrane, we can ask whether work is done in
the process. If we take as our system the gas and the vacuum space, we readily conclude
that no work is done because no work can be identified at the system boundary. If we take
the gas as a system, we do have a change of volume, and we might be tempted to calculate
the work from the integral
JW
However, this is not a quasi-equilibrium process, and therefore the work cannot be
calculated from this relation. Because there is no resistance at the system boundary as the
volume increases, we conclude that for this system no work is done in this process of fill-
ing the vacuum.
Another example can be cited with the aid of Fig. 4.18. In Fig. 4.18a the system
consists of the container plus the gas. Work crosses the boundary of the system at the
point where the system boundary intersects the shaft, and this work can be associated with
the shearing forces in the rotating shaft. In Fig. 4.1S& the system includes the shaft and
weight as well as the gas and the container. Therefore, no work crosses the system bound-
ary as the weight moves downward. As we will see in the next chapter, we can identify a
change of potential energy within the system, but this should not be confused with work
crossing the system boundary.
FIGURE 4.18
Example showing how
selection of the system
determines whether work
is involved in a process.
L
Gas
(a)
Gas :
ib)
100 B CHAPTERFOUR WORK AND HEAT
4,6 DEFINITION OF HEAT
The thermodynamic definition of heat is somewhat different from the everyday under-
standing of the word. It is essential to understand clearly the definition of heat given here,
because it plays a part in so many thermodynamic problems.
If a block of hot copper is placed in a beaker of cold water, we know from experi-
ence that the block of copper cools down and the water warms up until the copper and
water reach the same temperature. What causes this decrease in the temperature of the
copper and the increase in the temperature of the water? We say that it is the result of the
transfer of energy from the copper block to the water. It is out of such a transfer of energy
that we arrive at a definition of heat.
Heat is defined as the form of energy that is transferred across the boundary of a
ystem at a given temperature to another system (or the surroundings) at a lower tem-
perature by virtue of the temperature difference between the two systems. That is, heat
is transferred from the system at the higher temperature to the system at the lower tem-
perature, and the heat transfer occurs solely because of the temperature difference be-
tween the two systems. Another aspect of this definition of heat is that a body never
contains heat. Rather, heat can be identified only as it crosses the boundary. Thus, heat
is a transient phenomenon. If we consider the hot block of copper as one system and the
cold water in the beaker as another system, we recognize that originally neither system
contains any heat (they do contain energy, of course). When the copper block is placed
in the water and the two are in thermal communication, heat is transferred from the cop-
per to the water until equilibrium of temperature is established. At this point we no
longer have heat transfer, because there is no temperature difference. Neither system
contains heat at the conclusion of the process. It also follows that heat is identified at
the boundary of the system, for heat is defined as energy being transferred across the
system boundary.
Heat, like work, is a form of energy transfer to or from a system. Therefore, the
units for heat, and to be more general, for any other form of energy as well, are the same
as the units for work, or are at least directly proportional to them. In the International Sys-
tem the unit for heat (energy) is the joule. Similarly, in the English System, the footpound
force is an appropriate unit for heat. However, another unit came to be used naturally over
the years, the result of an association with the process of heating water, such as that used
in connection with defining heat in the previous section. Consider as a system 1 lbm of
water at 59.5 F. Let a block of hot copper of appropriate mass and temperature be placed
in the water so that when thermal equilibrium is established the temperature of the water
is 60.5 F. This unit amount of heat transferred from the copper to the water in this process
is called the British thermal unit (Btu). More specifically, it is called the 60-degree Btu,
defined as the amount of heat required to raise 1 lbm of water from 59.5 F to 60.5 F. (The
Btu as used today is actually defined in terms of the standard SI units.) It is worth noting
here that a unit of heat in metric units, the calorie, originated naturally in a manner similar
to the origin of the Btu in the English system. The cal orie i s defined as the amount of heat
required to raise 1 g of water from 14.5°C to 15.5°C. K^)
Heat transferred to a system is considered posnr
a system is negative. Thus, positive heat represents energy transferred to a system, and
negative heat represents energy transferred from a system. The symbol Q represents heat.
A process in which there is no heat transfer [Q ~ 0) is called an adiabatic process.
From a mathematical perspective, heat, like work, is a path function and is recog-
nized as an inexact differential. That is, the amount of heat transferred when a system un-
and heat transferred from
Heat Transfer Modes H 101
dergoes a change from state 1 to state 2 depends on the path that the system follows dur-
ing the change of state. Since heat is an inexact differential, the differential is written SQ,
On integrating, we write
In words, X Q 2 is the heat transferred during the given process between states 1 and 2.
The rate at which heat is transferred to a system is designated by symbol Q.
It is also convenient to speak of the heat transfer per unit mass of the system, q,
often termed specific heat transfer, which is defined as
4.7 Heat Transfer Modes
Heat transfer is the transport of energy due to a temperature difference between different
amounts of matter. We know that an ice cube taken out of the freezer will melt as it is
placed in a warmer environment such as a glass of liquid water or on a plate with room air
around it. From the discussion about energy in Section 2.6 we realize that molecules of
matter have translational (kinetic), rotational, and vibrational energy. Energy in these
modes can be transmitted to the nearby molecules by interactions (collisions) or by ex-
change of molecules such that energy is given out by molecules that have more in the av-
erage (higher temperature) to those that have less in the average (lower temperature). This
energy exchange between molecules is heat transfer by conduction, and it increases with
the temperature difference and the ability of the substance to make the transfer. This is ex-
pressed in Fourier's law of conduction
Q=~ M % (W) (4.18)
giving the rate of heat transfer as proportional to the conductivity, k, the total area, A, and
the temperature gradient. The minus sign gives a direction of the heat transfer from a
higher temperature to a lower temperature region. Often the gradient is evaluated as a
temperature difference divided by a distance when an estimate has to be done if a mathe-
matical or numerical solution is not available.
Values of the conductivity, k, range from the order of 100 W/m K for metals, 1 to
10 for nonmetallic solids as glass, ice and rock, from 0.1 to 10 for liquids, around 0.1 for
insulation materials, and from 0.1 down to less than 0.01 for gases.
A different mode of heat transfer takes place when a medium is flowing, called
convective heat transfer. In this mode the bulk motion of a substance moves matter with
a certain energy level over or near a surface with a different temperature. Now the heat
transfer by conduction is dominated by the manner in which the bulk motion brings the
two substances in contact or close proximity. Examples of this are the wind blowing
102 B CHAPTER FOUR. WORK AND HEAT
over a building or flow through heat exchangers, which can be air flowing over/through
a radiator with water flowing inside the radiator piping. The overall heat transfer is typ-
ically correlated with Newton's law of cooling as
where the transfer properties are lumped into the heat transfer coefficient, h, which then
becomes a function of the media properties, the flow and geometry. A more detailed study
of fluid mechanics and heat transfer aspects of the overall process is necessary to evaluate
the heat transfer coefficient for a given situation.
Typical values for the convection coefficient (all in W/mK 2 ) are
Natural convection h = 5-25, gas h = 50-1000, liquid
Forced convection h = 25-250, gas h = 50-20 000, liquid
Boiling phase change A = 2500-1 00 000
The final mode of heat transfer is radiation, which transmits energy as electromag-
netic waves in space. The transfer can happen in empty space and does not require any
matter, but the emission (generation) of the radiation and the absorption does require a
substance to be present. Surface emission is usually written as a fraction, emissivity s, of
a perfect black body emission as
with the surface temperature, T„ and the Stefan-Boltzmann constant, or. Typical values of
the emissivity range from 0.92 for nonmetallic surfaces to 0.6 to 0.9 for nonpolished
metallic surfaces, to less than 0.1 for highly polished metal surfaces. Radiation is distri-
buted over a range of wavelengths and it is emitted and absorbed differently for different
surfaces, but such a description is beyond the scope of the present text.
EXAMPLE 4.7 Consider the constant transfer of energy from a warm room at 20°C inside a house to
the colder ambient at ™10°C through a single-pane window as shown in Fig. 4.19.
The temperature variation with distance from the outside glass surface is shown with
an outside convection heat transfer layer, b\it no such layer is inside the room (as a
simplification). The glass pane has a thickness of 5 mm (0.005 m) with a conductiv-
ity of 1.4 W/m K and a total surface area of 0.5 m 2 . The outside wind is blowing so
that the convective heat transfer coefficient is 100 W/m 2 K. With an outer glass sur-
Q = Ah AT
(4.19)
Q - s(rAT$ (W)
(4.20)
FIGURE 4.19
Conduction and
convection heat transfer
through a window pane.
Comparison of Heat and Work H 103
face temperature of 12.1°C we would like to know the rate of heat transfer in the
glass and the convective layer.
For the conduction through the glass we have
n- iA dT ~ irA^T i/i W 2 20 - 12.1 K
Q "- M Tx~~ kA ^~ lA ^^ 5m 0.005 m=-» 06W
and the negative sign shows that energy is leaving the room. For the outside qonvection
layer we have'
Q = hA Af= 100—^— X 0.5 m 2 [12.1 - (-10)]K= 1105 W
m K
with a direction from the higher to the lower temperature, i.e., toward the outside.
4.8 COMPARISON OF HEAT AND WORK
At this point it is evident that there are many similarities between heat and work.
1. Heat and work are both transient phenomena. Systems never possess heat or work,
but either or both cross the system boundary when a system undergoes a change of
state.
2. Both heat and work are boundary phenomena. Both are observed only at the bound-
aries of the system, and both represent energy crossing the boundary of the system.
3. Both heat and work are path functions and inexact differentials.
It should also be noted that in our sign convention, +Q represents heat transferred
to the system and thus is energy added to the system, and + W represents work done by
the system and thus represents energy leaving the system.
A final illustration may help explain the difference between heat and work. Figure
4.20 shows a gas contained in a rigid vessel. Resistance coils are wound around the out-
side of the vessel. When current flows through the resistance coils, the temperature of the
gas increases. Which crosses the boundary of the system, heat or work?
In Fig. 4.20a we consider only the gas as the system. The energy crosses the bound-
ary of the system because the temperature of the walls is higher than the temperature of
the gas. Therefore, we recognize that heat crosses the boundary of the system.
.System boundary
FIGURE 4.20 An
example showing the
difference between heat
and work.
104 M Chapter Four work and Heat
FIGURE 4.21 The
effects of heat addition to
a control volume that also
can give out work.
In Fig. 4.206 the system includes the vessel and the resistance heater. Electricity
crosses the boundary of the system and, as indicated earlier, this is work.
Consider a gas in a cylinder fitted with a movable piston, as shown in Fig. 4.21.
There is a positive heat transfer to the gas, which tends to make the temperature increase.
It also tends to increase the gas pressure. However, the pressure is dictated by the external
force acting on its movable boundary, as discussed in Section 2.8. If this remains con-
stant, then the volume increases instead. There are also the opposite tendencies for a nega-
tive heat transfer, that is, one out of the gas. Consider again the positive heat transfer,
except that in this case the external force simultaneously decreases. This causes the gas
pressure to decrease, such that the temperature tends to go down. In this case, there are si-
multaneous tendencies for temperature change in the opposite direction, which effectively
decouples directions of heat transfer and temperature change.
Often when we want to evaluate a finite amount of energy transferred as either work
or heat we must integrate the instantaneous rate over time.
In order to perform the integration we must know how the rate varies with time. For time
periods where the rate does not change significantly, a simple average may be of suffi-
cient accuracy to allow us to write
which is similar to the information given on your electric utility bill as kilowatt-hours.
Work and heat are energy transfers between a control volume and its surroundings.
Work is energy that can be transferred mechanically (or electrically, or chemically) from
one system to another and must cross the control surface either as a transient phenome-
non or as a steady rate of work, which is power , Work is a function of the process path as
well as the beginning state and end state. The displacement work is equal to the area
below the process curve drawn in a P-V diagram if we have an equilibrium process. A
number of ordinary processes can be expressed as polytropic processes having a particu-
lar simple mathematical form for the P-V relation. Work involved by the action of sur-
face tension, single-point forces, or electrical systems should be recognized and treated
separately. Any nonequilibrium processes (say, dynamic forces, which are important due
to accelerations) should be identified so that only equilibrium force or pressure is used to
evaluate the work term.
Heat transfer is energy transferred due to a temperature difference, and the conduc-
tion, convection, andradiation modes are discussed.
(4.21)
r
; concept-Study guide problems B 105
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Recognize force and displacement in a system.
• Know power as rate of work (force X velocity, torque X angular velocity)
• Know work is a function of the end states aud the path followed in process
• Calculate the work term knowing the P~V or the F-x relationship
• Evaluate the work involved in a polytropic process between two states
• Know work is the area under the process curve in a P-V diagram
• Apply a force balance on a mass and determine work in a process from it
• Distinguish between an equilibrium process and a nonequilibrium process
• Recognize the three modes of heat transfer: conduction, convection and radiation
• Be familiar with Fourier's law of conduction and its use in simple applications
• Know the simple models for convection and radiation heat transfer
• Understand the difference between the rates (W, Q) and the amounts (JV^ X Q 2 ).
Key concepts
and formulas
Work
Heat
Displacement work
Specific work
Power, rate of work
Polytropic process
Polytropic process work
Conduction heat transfer
Conductivity
Convection heat transfer
Convection coefficient
Radiation heat transfer
(net to ambient)
Rate integration
W
TdB
Energy in transfer— mechanical, electrical, and chemical
Energy in transfer, caused by a AT
= j^Fdx = j 2 p dV = JV dA = j'
w = Wim (work per unit mass)
W=FV = PV= Toy (^displacement rate)
Velocity V = no, torque T ~ Fr, angular velocity = w
PV" = constant or Pv" = constant
1
1^2 =
1 - n
(PiVi-PiVx) (if»#l)
,w 2 = p { y,\n~i (if„=i)
0= ~u
dT
dx
k (W/mK)
Q = hA&T
A(W/m 2 K)
Q - eaA(Tt ~ T^J {a = 5.67 X 1 CT 8 W/m 2 K 4 )
i& = /
Concept-Study Guide Problems
4.1 The electric company charges the customers per
kilowatt-hour. What is that in SI units?
4.2 A car engine is rated at 1 60 hp. What is the power
in SI units?
4.3 A 1200-hp dragster engine has a drive shaft rotating
at 2000 RPM. How much torque is on the shaft?
4.4 A 1200-hp dragster engine drives the car with a
speed of 100 km/h. How much force is between the
tires and the road?
4.5 Two hydraulic piston/cylinders are connected
through a hydraulic line so that they have roughly
the same pressure. If they have diameters of D x and
106 H Chapter Four Work and Heat
D 2 ~ 2Z>! respectively, what can you say about the
piston forces F { and F 2 ?
4.6 Normally pistons have a flat head, but in diesel en-
gines pistons can have bowls in them and protrud-
ing ridges. Does this geometry influence the work
term?
4.7 What is roughly the relative magnitude of the work
in the process 1— 2c versus the process \~~2a shown
in Fig. 4.8?
4.8 A hydraulic cylinder of area 0.01 m 2 must push a
1000-kg arm and shovel 0.5 m straight up. What
pressure is needed and how much work is done?
4.9 A work of 2.5 kJ must be delivered on a rod from
a pneumatic piston/cylinder where the air pressure
is limited to 500 kPa. What diameter cylinder
should I have to restrict the rod motion to maxi-
mum 0.5 m?
4.10 Helium gas expands from 125 kPa, 350 K } and
0.25 m 3 to 100 kPa in a polytropic process with
n = 1.667. Is the work positive, negative, or
zero?
4.11 An ideal gas goes through an expansion process
whereby the volume doubles. Which process will
lead to the larger work output, an isothermal
process or a polytropic process with n — 1.25?
Homework problems "
Force Displacement Work
4.20 A piston of mass 2 kg is lowered 0.5 m in the stan-
dard gravitational field. Find the required force and
the work involved in the process.
4.21 An escalator raises a 100-kg bucket of sand 10 m
in 1 min. Determine the total amount of work done
during the process.
4.22 A bulldozer pushes 500 kg of dirt 100 m with a
force of 1500 N. It then lifts the dirt 3 m up to put
it in a dump truck. How much work did it do in
each situation?
4.23 A hydraulic cylinder has a piston of cross-sectional
area 25 cm 2 and a fluid pressure of 2 MPa. If the
piston is moved 0.25 m, how much work is done?
4.24 Two hydraulic cylinders maintain a pressure of
1200 kPa. One has a cross-sectional area of 0.01
m 2 , the other one of 0.03 m 2 . To deliver 1 kJ of
4.12 Show how the polytropic exponent n can be evalu-
ated if you know the end state properties; (P u V\)
and {P 2i V 2 ).
4.13 A drag force on an object moving through a
medium (like a car through air or a submarine
through water) is F d = 0.225 ApV 2 . Verify that the
unit become newtons.
4.14 A force of 1 .2 kN moves a truck with 60 km/h up a
hill. What is the pow r er?
4.15 Electric power is volts times amperes (P = Vi).
When a car battery at 12 V is charged with 6 A for
3 h, how much energy is delivered?
4.16 Torque and energy and work have the same units
(N m). Explain the difference.
4.17 Find the rate of conduction heat transfer through a
1.5-cm-thick hardwood board, k ~ 0.16 W/m K,
with a temperature difference between the two
sides of 20°C.
4.18 A 2-m 2 window has a surface temperature of 15°C,
and the outside wind is blowing air at 2°C across it
with a convection heat transfer coefficient of h =
125 W/m 2 K. What is the total heat transfer toss?
4.19 A radiant heating lamp has a surface temperature of
1000 K with s ~ 0.8. How large a surface area is
needed to provide 250 W of radiation heat transfer?
work to the piston, how large a displacement (V)
and piston motion H are needed for each cylinder?
Neglect P ztm .
4.25 A linear spring, F = k s (x - x ) with spring con-
stant k s ~ 500 N/m is stretched until it is 100 mm
longer. Find the required force and the work input.
4.26 A nonlinear spring has a force versus the displace-
ment relation ofF = k s (x - x ) n . If the spring end
is moved to x t from the relaxed state, determine the
formula for the required work.
4.27 The rolling resistance of a car depends on its
weight as F — 0,006 mg. How long will a car of
1400 kg drive for a work input of 25 kJ?
4.28 A car drives for half an hour at constant speed and
uses 30 MJ over a distance of 40 km. What was the
traction force to the road and its speed? '
4.29 The air drag force on a car is 0.225 ApV 2 . Assume
air at 290 K, 100 kPa and a car frontal area of 4 m 2
HOMEWORK PROBLEMS H 107
driving at 90 km/h. How much energy is used to
overcome the air drag driving for 30 min?
4.30 Two hydraulic piston/cylinders are connected with
a line. The master cylinder has an area of 5 cm 2 ,
creating a pressure of 1000 kPa. The slave cylinder
has an area of 3 cm 2 . If 25 J is the work input to the
master cylinder, what is the force and displacement
of each piston and the work output of the slave
cylinder piston?
Boundary Work: Simple One-Step Process
4.31 A constant-pressure piston/cylinder assembly con-
tains 0.2 kg of water as saturated vapor at 400 kPa.
It is now cooled so that the water occupies half the
original volume. Find the work done in the
process.
4.32 A steam radiator in a room at 25°C has saturated
water vapor at 1 1 kPa flowing through it when the
inlet and exit valves are closed. What are the pres-
sure and the quality of the water when it has cooled
to 25°C? How much work is done?
4.33 A 400-L tank, A (see Fig. P4.33), contains argon
gas at 250 kPa and 30°C. Cylinder B, having a Mo-
tionless piston of such mass that a pressure of 150
kPa will float it, is initially empty. The valve is
opened and argon flows into B and eventually
reaches a uniform state of 150 kPa and 30°C
throughout. What is the work done by the argon?
B
1 (g, 1
FIGURE P4.33
4.34 A pi ston/cy Under contains air at 600 kPa, 290 K
and a volume of 0.01 m 3 . A constant-pressure
process gives 54 kJ of work out. Find the final vol-
ume and temperature of the air.
4.35 Saturated water vapor at 200 kPa is in a constant-
pressure piston/cylinder. In this state, the piston is
0.1 m from the cylinder bottom and the cylinder
area is 0.25 m 2 . The temperature is then changed to
200°C. Find the work in the process.
4.36 A cylinder fitted with a frictionless piston contains
5 kg of superheated refrigerant R-134a vapor at
1000 kPa and 140°C. The setup is cooled at con-
stant pressure until the R-134a reaches a quality of
25%. Calculate the work done in the process.
4.37 Find the specific work in Problem 3.54 for the case
where the volume is reduced.
4.38 A piston/cylinder has 5 m of liquid 20°C water on
top of the piston (m = 0) with cross-sectional area
of 0.1 m 2 , see Fig. P2.57. Air is let in under the pis-
ton that rises and pushes the water out over the top
edge. Find the necessary work to push all the water
out and plot the process in a P-Fdiagram.
4.39 Air in a spring-loaded piston/cylinder setup has a
pressure that is linear with volume, P - A + BV.
With an initial state of P = 150 kPa, V~ 1 L and a
final state of 800 kPa and volume 1.5 L, it is simi-
lar to the setup in Problem 3.113. Find the work
done by the air.
4.40 Find the specific work in Problem 3.43.
4.41 A piston/cylinder contains 1 kg of water at 20°C
with volume 0. 1 m 3 . By mistake someone locks the
piston, preventing it from moving while we heat
the water to saturated vapor. Find the final temper-
ature and volume, and the process work.
4.42 A piston/cylinder assembly contains 1 kg of liquid
water at 20°C and 300 kPa, as shown in Fig. P4.42.
There is a linear spring mounted on the piston such
that when the water is heated the pressure reaches
3 MPa with a volume of 0. 1 m 3 .
a. Find the final temperature.
b. Plot the process in a P-v diagram.
c. Find the work in the process.
L I
H 2
FIGURE P4.42
108 B CHAPTER FOUR WORK AND HEAT
4.43 A piston/cylinder assembly contains 3 kg of air at
20°C and 300 kPa. It is now heated in a constant
pressure process to 600 K.
a. Find the final volume.
b. Plot the process path in a P — v diagram.
c. Find the work in the process.
4.44 A piston/cylinder assembly contains 0.5 kg of air at
500 kPa and 500 K. The air expands in a process
such that P is linearly decreasing with volume to a
final state of 100 kPa, 300 K. Find the work in the
process.
4.45 Consider the nonequilibrium process described hi
Problem 3.109. Determine the work done by the
carbon dioxide in the cylinder during the process.
4.46 Consider the problem of inflating the helium bal-
loon, as described in Problem 3.79. For a control
volume that consists of the helium inside the balloon
detennine the work done during the filling process
when the diameter changes from 1 m to 4 m.
Polytropic Process
4.47 Consider a mass going through a polytropic
process where pressure is directly proportional to
volume (n — —1). The process starts with P — 0,
V = and ends with P = 600 kPa, V = 0.01 m 3 .
Find the boundary work done by the mass.
4.48 The piston/cylinder arrangement shown in Fig. P4.48
contains carbon dioxide at 300 KPa and 100°C with a
volume of 0.2 m 3 . Weights are added to the piston
such that the gas compresses according to the rela-
tion PV X2 — constant to a final temperature of 200°C.
Determine the work done during the process.
^0
1 I 1 1
I
ccy :
8
FIGURE P4.4S
4.49 A gas initially at 1 MPa and 500°C is contained in
a piston and cylinder arrangement with an initial
volume of 0.1 m 3 . The gas is then slowly expanded
according to the relation PV = constant until a
final pressure of 100 kPa is reached. Determine the
work for this process.
4.50 Helium gas expands from 125 kPa, 350 K, and
0.25 m 3 to 100 kPa in a polytropic process with
n — 1.667. How much work does it give out?
4.51 Air goes through a polytropic process from 125 kPa
and 325 K to 300 kPa and 500 K. Find the polytropic
exponent n and the specific work in the process.
4.52 A piston/cylinder device contains 0.1 kg of air at 100
kPa and 400 K that goes through a polytropic com-
pression process with n — 1.3 to a pressure of 300
kPa. How much work has the air done in the
process?
4.53 A balloon behaves so the pressure is P ~ C 2 V i/:i >
C 2 = 100 kPa/m. The balloon is blown up with air
from a starting volume of 1 m 3 to a volume of
3 m 3 . Find the final mass of the air, assuming it is
at 25°C, and the work done by the air.
4.54 A balloon behaves such that the pressure inside is
proportional to the diameter squared. It contains
2 kg of ammonia at 0°C, with 60% quality. The
balloon and ammonia are now heated so that a final
pressure of 600 kPa is reached. Considering the
ammonia as a control mass, find the amount of
work done in the process.
4.55 Consider a piston/cylinder setup with 0.5 kg of.
R-134a as saturated vapor at — 10°C. It is now
compressed to a pressure of 500 kPa in a poly-
tropic process with n — 1.5. Find the final volume
and temperature, and determine the work done dur-
ing the process.
4.56 Consider the process described in Problem 3.98.
With 1 kg of water as a control mass, determine the
boundary work during the process.
4.57 Find the work done in Problem 3.106.
4.58 A piston/cylinder contains water at 500°C, 3 MPa.
It is cooled in a polytropic process to 200°C,
1 MPa. Find the polytropic exponent and the spe-
cific work in the process.
Boundary Work: Multistep Process
4.59 Consider a two-part process with an expansion
from 0.1 to 0.2 m 3 at a constant pressure of 150
kPa followed by an expansion from 0.2 to 0.4 m 3
with a linearly rising pressure from 150 kPa ending
Homework Problems M 109
at 300 kPa. Show the process in a P-V diagram and
find the boundary work.
4.60 A cylinder containing 1 kg of ammonia has an ex-
ternally loaded piston. Initially the ammonia is at 2
MPa and 180°C, It is now cooled to saturated vapor
at 40°C and then further cooled to 20°C, at which
point the quality is 50%. Find the total work for the
process, assuming a piecewise linear variation of P
versus V.
4.61 A piston/cylinder arrangement shown in Fig. P4.61
__Jnitially contains air at 150 kPa and 400°C. The
setup is allowed to cool to the ambient temperature
of20°C.
a. Is the piston resting on the stops in the final
state? What is the final pressure in the cylinder?
b. What is the specific work done by the air during
the process?
1
1
I
Air
"1 l~
1m
FIGURE P4.61
4.62 A piston/cylinder has 1.5 kg of air at 300 K and 150
kPa. It is now heated up in a two step process. First
constant volume to 1000 K (state 2) then followed by
a constant pressure process to 1500 K, state 3. Find
the final volume and the work in the process.
4.63 A piston/cylinder assembly (Fig. P4.63) has 1 kg
of R- 134a at state 1 with 1 10°C, 600 kPa. It is then
R-134a
FIGURE P4.63
brought to saturated vapor, state 2, by cooling
■ while the piston is locked with a pin. Now the pis-
ton is balanced with an additional constant force
and the pin is removed. The cooling continues to a
state 3, where the R-134a is saturated liquid. Show
the processes in a P-V diagram and find the work
in each of the two steps, 1 to 2 and 2 to 3.
4.64 The refrigerant R-22 is contained in a piston/
cylinder as shown in Fig. P4.64, where the volume is
1 1 L when the piston hits the stops. The initial state
is -30°C, 150 kPa, with a volume of 10 L. This sys-
tem is brought indoors and warms up to 15°C.
a. Is the piston at the stops in the final state?
b. Find the work done by the R-22 during this
process.
ZD
S
R-22
' I FIGURE P4.64
4.65 A piston/cylinder assembly contains 50 kg of water
at 200 kPa with a volume of 0.1 m 3 . Stops in the
cylinder restrict the enclosed volume to 0.5 m 3 ,
similar to the setup in Problem 4.64. The water is
now heated to 200°C. Find the final pressure, vol-
ume, and work done by the water.
4.66 Find the work in Problem 3.108.
4.67 A piston/cylinder assembly contains 1 kg of liquid
water at 20°C and 300 kPa. Initially the piston
floats, similar to the setup in Problem 4.64, with a
maximum enclosed volume of 0.002 m 3 if the pis-
ton touches the stops. Now heat is added so that a
final pressure of 600 kPa is reached. Find the final
volume and the work in the process,
4.68 Ten kilograms of water in a piston/cylinder arrange-
ment exist as saturated liquid/vapor at 100 kPa, with
a quality of 50%. It is now heated so the volume
triples. The mass of the piston is such that a cylinder
pressure of 200 kPa will float it (see Fig. P4.68).
a. Find the final temperature and volume of the
water.
b. Find the work given out by the water.
110 H CHAPTER FOUR WORK AND HEAT
H O
FIGURE P4.68
4.69 Find the work in Problem 3,63.
4.70 A piston/cylinder setup similar to Problem 4.68
contains 0. 1 kg saturated liquid and vapor water at
100 kPa with quality 25%. The mass of the piston
is such that a pressure of 500 kPa will float it. The
water is heated to 300°C. Find the final pressure,
volume, and work, JV 2 .
Other Types of Work and General Concepts
4.71 A 0.5-m-long steel rod with a 1-cm diameter is
stretched in a tensile test. What is the work re-
quired to obtain a relative strain of 0.1%? The
modulus of elasticity of steel is 2 X 10 8 kPa.
4.72 A copper wire of diameter 2 mm is 10 m long and
stretched out between two posts. The nor-
mal stress (pressure) a = E(L ~ L^)!L , depends
on the length I versus the unstretched length L and
Young's modulus E = 1.1 X 10 6 kPa. The force is
F — Aa and measured to be 110 N. How much
longer is the wire and how much work was put in?
4.73 A film of ethanol at 20°C has a surface tension of
22.3 mN/m and is maintained on a wire frame as
shown in Fig. P4.73. Consider the film with two
surfaces as a control mass and find the work done
when the wire is moved 10 mm to make the film
20 X 40 mm.
4.74 Assume a balloon material with a constant surface
tension of if = 2N/m. What is the work required to
stretch a special balloon up to a radius of r = 0.5 m?
Neglect any effect from atmospheric pressure.
4.75 A soap bubble has a surface tension of if = 3 X 10 -4
N/cm as it sits flat on a rigid ring of diameter 5 cm.
You now blow on the film to create a half-sphere sur-
face of diameter 5 cm. How much work was done?
4.76 Assume we fill a spherical balloon from a bottle of
helium gas. The helium gas provides work / P dV
that stretches the balloon material / S dA and
pushes back the atmosphere / P dV, Write the in-
cremental balance for dW hi] i om = dW^^ + dW 2tm
to establish the connection between the helium
pressure, the surface tension S t and P as a function
of radius.
4.77 A sheet of rubber is stretched out over a ring of ra-
dius 0.25 m. I pour liquid water at 20°C on it, as in
Fig. P4.77, so that the rubber forms a half sphere
(cup). Neglect the rubber mass and find the surface
tension near the ring.
30 mm.
frame
/
Rubber sheet
FIGURE P4.77
FIGURE P4.73
4.78 Consider a window-mounted air-conditioning unit
used in the summer to cool incoming air. Examine
the system boundaries for rates of work and heat
transfer, including signs.
4.79 Consider a hot-air heating system for a home. Ex-
amine the following systems for heat transfer.
a. The combustion chamber and combustion gas
side of the heat transfer area
b. The furnace as a whole, including the hot- and
cold-air ducts and chimney
4.80 Consider a household refrigerator that has just been
filled up with room-temperature food. Define a
control volume (mass) and examine its boundaries
for rates of work and heat transfer, including sign,
a. Immediately after the food is placed in the re-
frigerator
b. After a long period of time has elapsed and the
food is cold
Homework Problems M 111
4.81 A room is heated with an electric space heater on a
winter day. Examine the following control volumes,
regarding heat transfer and work, including sign:
a. The space heater
b. Room
c. The space heater and the room together
Rates of Work
4.82 An escalator raises a 100-kg bucket 10 m in 1 min.
Determine the rate of work in the process.
4.83 A car uses 25 hp to drive at a horizontal level at a
constant speed of 100 km/h. What is the traction
force between the tires and the road?
4.84 A piston/cylinder of cross-sectional area 0.01 m 2
maintains constant pressure. It contains 1 kg of
water with a quality of 5% at 150°C. If we heat so
that 1 g/s of liquid turns into vapor, what is the rate
of work out?
4.85 A crane lifts a bucket of cement with a total mass of
450 kg vertically upward with a constant velocity of
2 m/s. Find the rate of work needed to do this.
4.86 Consider the car with the rolling resistance as in
Problem 4.27. How fast can it drive using 30 hp?
4.87 Consider the car with the air drag force as in Prob-
lem 4.29. How fast can it drive using 30 hp?
4.88 Consider a 1400-kg car having rolling resistance as
in Problem 4.27 and air resistance as in Problem
4.29. How fast can it drive using 30 hp?
4.89 A battery is well insulated while being charged by
12,3 V at a current of 6 A. Take the battery as a
control mass and find the instantaneous rate of
work and the total work done over 4 h.
4.90 A current of 10 A runs through a resistor with a re-
sistance of 1 5 XI. Find the rate of work that heats
the resistor up,
4.91 A pressure of 650 kPa pushes a piston of diameter
0.25 m with V = 5 m/s. What is the volume dis-
placement rate, the force, and the transmitted
power?
4.92 Assume the process in Problem 4.59 takes place
with a constant rate of change in volume over 2
minutes. Show the power (rate of work) as a func-
tion of time.
4.93 Air at a constant pressure in a piston/cylinder is at
300 kPa, 300 K and has a volume of 0.1 m 3 . It is
heated to 600 K over 30 s in a process with con-
stant piston velocity. Find the power delivered to
. the piston.
4.94 A torque of 650 N m rotates a shaft of diameter
0.25 m with &> = 50 rad/s. What is the shaft surface
speed and the transmitted power?
Heat Transfer Rates
4.95 The sun shines on a 150-m 2 road surface so that it
is at 45°C. Below the 5-cm-thick asphalt, with av-
erage conductivity of 0.06 W/m K, is a layer of
compacted rubble at a temperature of 15°C. Find
the rate of heat transfer to the rubble.
4.96 A steel pot, with conductivity of 50 W/m K and a
5-mm-thick bottom, is filled with 1 5°C liquid water.
The pot has a diameter of 20 cm and is now placed
on an electric stove that delivers 250 W as heat
transfer. Find the temperature on the outer pot bot-
tom surface assuming the inner surface is at 15°C.
4.97 A water heater is covered up with insulation
boards over a total surface area of 3 m 2 . The in-
side board surface is at 75°C, the outside surface
is at 20°C, and the board material has a conductiv-
ity of 0.08 W/m K. How thick should the board be
to limit the heat transfer loss to 200 W?
4.98 You drive a car on a winter day with the atmos-
pheric air at - 1 5°C, and you keep the outside front
windshield surface temperature at +2°C by blow-
ing hot air on the inside surface. If the windshield
is 0.5 m 2 and the outside convection coefficient is
250 W/m 2 K, find the rate of energy loss through
the front windshield. For that heat transfer rate and
a 5-mm-thick glass with k = 1.25 W/m K, what is
then the inside windshield surface temperature?
4.99 A large condenser (heat exchanger) in a power
plant must transfer a total of 100 MW from steam
running in a pipe to seawater being pumped
through the heat exchanger. Assume the wall sep-
arating the steam and'seawater is 4 mm of steel,
with conductivity of 15 W/m K, and that a maxi-
mum of 5°C difference between the two fluids is
allowed in the design. Find the required minimum
area for the heat transfer, neglecting any convec-
tive heat transfer in the flows.
4.100 The black grille on the back of a refrigerator has a
surface temperature of 35°C with a total surface
area of 1 m 2 . Heat transfer to the room air at 20°C
takes place with an average convective heat transfer
112 M CHAPTER FOUR WORK AND HEAT
coefficieni of 15 W/m 2 K. How much energy can
be removed during 15 rriiriutes of operation?
4.101 Owing to a faulty door contact, the small light
bulb (25 W) inside a refrigerator is kept on and
limited insulation lets 50 W of energy from the
outside seep into the refrigerated space. How
much of a temperature difference to the ambient
surroundings at 20°C must the refrigerator have in
its heat exchanger with an area of 1 m 2 and an av-
erage heat transfer coefficient of 15 W/m 2 K to re-
ject the leaks of energy?
4.102 The brake shoe and steel drum of a car continu-
ously absorb 25 W as the car slows down. As-
sume a total outside surface area of 0.1 m 2 with a
convective heat transfer coefficient of 10 W/m 2 K
to the air at 20°C. How hot does the outside brake
and drum surface become when steady conditions
are reached?
4.103 A wall surface on a house is 30°C with an emis-
sivity of e = 0.7. The surrounding ambient air is
at 15°C with an average emissfvity of 0.9. Find
the rate of radiation energy from each of those
surfaces per unit area.
4.104 A log of burning wood in the fireplace has a sur-
face temperature of 450°C. Assume the emissivity
is 1 (perfect black body) and find the radiant
emission of energy per unit surface area.
4.105 A radiant heat lamp is a rod, 0,5 m long and 0.5
cm in diameter, through which 400 W of electric
energy is deposited. Assume the surface has an
emissivity of 0.9 and neglect incoming radiation.
What will the rod surface temperature be?
Review Problems
4.106 A vertical cylinder (Fig. P4.106) has a 61.18 kg
piston locked with a pin, trapping 10 L of R-22 at
10°C with 90% quality inside. Atmospheric pres-
sure is 100 kPa, and the cylinder cross-sectional
area is 0.006 m 2 . The pin is removed, allowing the
piston to move and come to rest with a final tem-
perature of 10°C for the R-22. Find the final pres-
sure, final volume, and work done by the R-22.
4.107 A piston/cylinder assembly contains butane,
C 4 H 10) at 300°C and 100 kPa with a volume of
0.02 m\ The gas is now compressed slowly in an
isothermal process to 300 kPa.
a. Show that it is reasonable to assume that butane
behaves as an ideal gas during this process.
b. Determine the work done by the butane during
the process.
4.108 A cylinder fitted with a piston contains propane
gas at 100 kPa and 300 K with a volume of 0.2
m 3 . The gas is now slowly compressed according
to the relation PK 11 = constant to a final tempera-
ture of 340 K. Justify the use of the ideal-gas
model. Find the final pressure and the work done
during the process.
4.109 The gas space above the water in a closed storage
tank contains nitrogen at 25°C and 100 kPa. Total
tank volume is 4 m 3 , and there is 500 kg of water at
25°C. An additional 500 kg of water is now forced
into the tank. Assuming constant temperature
throughout, find the final pressure of the nitrogen
and the work done on the nitrogen in this process.
4.110 Two kilograms of water are contained in a pis-
ton/cylinder (Fig. P4.110) with a massless piston
loaded with a linear spring and the outside atmos-
phere. Initially the spring force is zero and P x =
P - 100 kPa with a volume of OJ^m 3 . If the pis-
ton just hits the upper stops, the volume is 0.8 m 3
and T = 600°C. Heat is now added until the pres-
sure reaches 1.2 MPa. Find the final temperature,
show the P-V diagram and find the work done
during the process.
Air
J
R-22
Pin
FIGURE P4.106
HoO
FIGURE P4.110
ENGLISH UNFT PROBLEMS
□ 113
4.111 A cylinder having an initial volume of 3 m 3 con-
tains 0.1 kg of water at 40°C. The water is then
compressed in an isothermal quasi-equilibrium
process until it has a quality of 50%. Calculate the
work done by splitting the process into two steps.
Assume the water vapor is an ideal gas during the
fist step of the process.
4.112 Air at 200 kPa, 30°C is contained in a
cylinder/piston arrangement with initial volume
0.1 m 3 . The inside pressure balances ambient
pressure of 100 kPa pins an externally imposed
force that is proportional to V 05 . Now heat is
transferred to the system to a final pressure of 225
kPa. Find the final temperature and the work done
in the process.
4.113 A spring-loaded piston/cylinder arrangement con-
tains R-134a at 20°C, 24% quality with a volume
50 L. The setup is heated and thus expands, mov-
ing the piston. It is noted that when the last drop
of liquid disappears the temperature is 40°C. The
heating is stopped when T = 130°C. Verify that
the final pressure is about 1200 kPa by iteration
and find the work done in the process.
4.114 A piston/cylinder setup (Fig. P4.68) contains 1 kg
of water at 20°C with a volume of 0.1 m 3 . Ini-
tially, the piston rests on some stops with the top
surface open to the atmosphere, P , and a mass
such that a water pressure of 400 kPa will lift it.
English Unit Problems
English Unit Concept Problems
4.1 17E The electric company charges the customers per
kilowatt-hour. What is that in English units?
4.1 18 E Work as F Ax has units of lbf/ft. What is that in
Btu?
4.119E Work of 2.5 Btu must be delivered on a rod
from a pneumatic piston/cylinder where the air
pressure is limited to 75 psia. What diameter
cylinder should I have to restrict the rod motion
to maximum 2 ft?
4.120E A force of 300 lbf moves a truck at 40 mi/h up a
hill. What is the power?
4.121E A 1200-hp dragster engine drives a car with a
speed of 65 mi/h. How much force is between
the tires and the road?
To what temperature should the water be heated
, to lift the piston? If it is heated to saturated vapor,
find the final temperature, volume, and work, x W 2 .
4.115 Two springs with the same spring constant are
installed in a massless piston/cylinder arrange-
ment with the outside air at 100 kPa. If the pis-
ton is at the bottom, both springs are relaxed,
and the second spring comes in contact with the
piston stV=2 m 3 . The cylinder (Fig. P4.115)
contains ammonia initially at — 2°C, x ~ 0.13,
V = 1 m 3 , which is then heated until the pres-
sure finally reaches 1200 kPa. At what pressure
will the piston touch the second spring? Find
the final temperature and the total work done by
the ammonia.
NH 3
FIGURE P4.115
4.116 Find the work in the process described in Problem
3.101.
4.122E A 1200-hp dragster engine has a drive shaft ro-
tating at 2000 RPM. How much torque is on the
shaft?
English Unit Problems
4.123E A bulldozer pushes 1000 ibm of dirt 300 ft with
a force of 400 lbf. It then lifts the dirt 10 ft up to
put it in a dump truck. How much work did it do
in each situation?
4.124E A steam radiator in a room at 75 F has saturated
water vapor at 16 lbf/in 2 flowing through it,
when the inlet and exit valves are closed. What
is the pressure and the quality of the water when
it has cooled to 75 F? How much work is done?
4.125E A linear spring, F = k£x - x ), with spring
constant k s = 35 lbf/ft, is stretched until it is
114 U Chapter four work and heat
2.5 in. longer. Find the required force and work
input.
4.126E Two hydraulic cylinders maintain a pressure
of 175 psia. One has a cross-sectional area of
0.1 ft 2 , the other one of 0.3 ft 2 . To deliver 1 Btu
of work to the piston, how large a displacement
(V) and piston motion H are needed for each
cylinder? Neglect P &im .
4.127E A piston/cylinder has 15 ft of liquid 70 F water
on top of the piston (m — 0) with cross-sectional
area of 1 ft 2 (see Fig. P2.57). Air is let in under
the piston, which rises and pushes the water out
over the top edge. Find the necessary work to
push all the water out, and plot the process in a
P-V diagram.
4.128E A cylinder fitted with a frictionless piston con-
tains 10 Ibm of superheated refrigerant R-134a
vapor at 100 lbf/in 2 , 300 F. The setup is cooled
at constant pressure until the R-134a reaches a
quality of 25%. Calculate the work done in the
process.
4.129E The gas space above the water in a closed stor-
age tank contains nitrogen at 80 F, 15 lbf/in 2 .
Total tank volume is 150 ft 3 , and there is 1000
lbm of water at 80 F. An additional 1000 Ibm of
water is now forced into the tank. Assuming
constant temperature throughout, find the final
pressure of the nitrogen and the work done on
the nitrogen in this process.
4.130E A cylinder having an initial volume of 100 ft 3
contains 0.2 lbm of water at 100 F. The water
is then compressed in an isothermal quasi-
equilibrium process until it has a quality of
50%. Calculate the work done in the process as-
suming water vapor is an ideal gas.
4.131E Helium gas expands from 20 psia, 600 R, and 9
ft 3 to 15 psia in a polytropic process with n —
1.667. How much work does it give out?
4.132E Consider a mass going through a polytropic
process where pressure is directly proportional
to volume (n = —1). The process starts with
P = 0, V = and ends with P = 90 lbf/in 2 ,
V = 0.4 ft 3 . The physical setup could be as in
Problem 2.83. Find the boundary work done by
the mass.
4.133E The piston/cylinder shown in Fig. P4.48 con-
tains carbon dioxide at 50 lbf/in 2 , 200 F with a
volume of 5 ft 3 . Mass is added at such a rate
that the gas compresses according to the rela-
tion PV xa = constant to a final temperature of
350 F. Determine the work done during the
process.
4.134E Find the specific work for Problem 3.156E.
4.135E Consider a two-part process with an expansion
from 3 to 6 ft 3 at a constant pressure of 20
lbf/in 2 followed by an expansion from 6 to 12 ft 3
with a linearly rising pressure from 20 lbf/in 2
ending at 40 lbf/in 2 . Show the process in a P-V
diagram and find the boundary work.
4.136E A piston/cylinder has 2 lbm of R- 134a at state
1 with 200 F, 90 lbf/in 2 , and is then brought to
saturated vapor, state 2, by cooling while the
piston is locked with a pin. Now the piston is
balanced with an additional constant force and
the pin is removed. The cooling continues to
state 3, where the R-134a is saturated liquid.
Show the processes in a P-V diagram and find
the work in each of the two steps, 1 to 2 and
2 to 3.
4.137E A cylinder containing 2 lbm of ammonia has an
externally loaded piston. Initially the ammonia
is at 280 lbf/in 2 , 360 F. It is now cooled to satu-
rated vapor at 105 F, and then further cooled to
65 F, at which point the quality is 50%. Find the
total work for the process, assuming a piecewise •
linear variation of P versus V.
4.138E A 1-ft-long steel rod with a 0.5-in. diameter is
stretched in a tensile test. What is the required
work to obtain a relative strain of 0.1%? The
modulus of elasticity of steel is 30 X 10 6 lbf/in 2 .
4.139E An escalator raises a 200-lbm bucket of sand 30
ft in 1 min. Determine the total amount of work
done and the instantaneous rate of work during
the process.
4.140E A ptston/cylinder of diameter 10 in. moves a
piston with a velocity of 18 ft/s. The instanta-
neous pressure if 100 psia. What is the volume
displacement rate, the force and the transmitted
power?
4.141E The sun shines on a 1500-ft 2 road surface so it is
at 1 1 5 F. Below the 2-in.-thick asphalt, average
conductivity of 0.035 Btu/h ft F, is a layer of
compacted rubble at a temperature of 60 F. Find
the rate of heat transfer to the rubble.
Computer, design, and open-ended Problems B 115
4.142E A water heater is covered up with insulation
boards over a total surface area of 30 ft 2 . The in-
side board surface is at 175 F, the outside sur-
face is at 70 F f and the board material has a
conductivity of 0.05 Btu/h ft F. How thick
should the board be to limit the heat transfer
loss to 720 Btu/h?
4.143E The black grille on the back of a refrigerator has
a surface temperature of 95 F with a total sur-
face area of 10 ft 2 . Heat transfer to the room air
at 70 F takes place with an average convective
heat transfer coefficient of 3 Btu/h ft 2 R. How
much energy can be removed during 15 min of
operation?
Computer, Design, and Open-Ended Problems
4.144 In Problem 4,48, determine the work done by the
carbon dioxide at any point during the process.
4.145 In Problem 4,112, determine the work done by
the air at any point during the process.
4.146 A piston/cylinder arrangement of initial volume
0.025 m 3 contains saturated water vapor at 200°C.
The steam now expands in a quasi-equilibrium
isothermal process to a final pressure of 200 kPa
while it does work against the piston. Determine
the work done in this process by a numerical inte-
gration (summation) of the area below the P-V
process curve. Compute about 10 points along the
curve by using the computerized software to get
the volume at 200°C and the various pressures.
How different is the work calculated if ideal gas
is assumed?
4.147 Reconsider the process in Problem 4.60 in which
three states were specified. Solve the problem by
fitting a single smooth curve (P versus v) through
the three points. Map out the path followed (includ-
ing temperature and quality) during the process.
4.148 Write a computer program to determine the
boundary movement work for a specified sub-
stance undergoing a process for a given set of
data (values of pressure and corresponding vol-
ume during the process).
4.149 Ammonia vapor is compressed inside a cylinder
by an external force acting on the piston. The am-
monia is initially at 30°C, 500 kPa, and the final
pressure is 1400 kPa. The following data have
been measured for the process:
Pressure, 500 653 802 945 1 100 1248 1400
kPa
Volume, L 1.25 1.08 0.96 0.84 0.72 0.60 0.50
Determine the work done by the ammonia by
summing the area below the P—V process curve.
As you plot it, P is the height and the change in
volume is the base of a number of rectangles.
4.150 A substance is brought from a state of P u v { to a
state of P 2 , u 2 in a piston/cylinder arrangement.
Assume that the process can be approximated as a
polytropic process. Write a program that will find
the polytropic exponent, n, and the boundary
work per unit mass. The four state properties are
input variables. Check the program with cases
that you can easily hand calculate.
4.151 Assume that you have a plate of A = 1 m 2 with
thickness L = 0.02 m over which there is a tem-
perature difference of 20°C. Find the conductiv-
ity, k, from the literature and compare the heat
transfer rates if the plate substance is a metal like
aluminum or steel, or wood, foam insulation, air,
argon, or liquid water. Assume the average sub-
stance temperature is 25°C.
4.152 Make a list of household appliances such as re-
frigerators, electric heaters, vacuum cleaners, hair
dryers, TVs, stereo sets, and any others you may
think of. For each, list its energy consumption and
explain where you have energy transfer as work
and where there is heat transfer.
THE FIRST LAW OF
THERMODYNAMICS
Having completed our consideration of basic definitions and concepts, we are ready to
proceed to a discussion of the first law of thermodynamics. This law is often called the
conservation of energy law and, as we will see later, this is essentially true. Our proce-
dure will be to state this law for a system (control mass) undergoing a cycle and then for
a change of state of a system.
After the energy equation is formulated we will use it to relate change of state
inside a control volume to the amount of energy that is transferred in a process as
work or heat transfer k When a car engine has transferred some work to the car, the
car's speed is increased, so we can relate the kinetic energy increase to the work, or if
a stove provides a certain amount of heat transfer to a pot with water we can relate the
water temperature increase to the heat transfer. More complicated processes can also
occur, such as the expansion of very hot gases in a piston cylinder, as in a car engine,
in which work is given out and at the same time heat is transferred to the colder walls.
In other applications we can also see a change in the state without any work or heat
transfer, such as a falling object that changes kinetic energy at the same time it is
changing elevation. The energy equation then relates the two forms of energy of the
object.
5.1 The First law of thermodynamics for
a Control Mass Undergoing a cycle
The first law of thermodynamics states that during any cycle a system (control mass)
undergoes, the cyclic integral of the heat is proportional to the cyclic integral of the
work.
To illustrate this law, consider as a control mass the gas in the container shown in
Fig. 5.1. Let this system go through a cycle that is made up of two processes. In the first
process work is done on the system by the paddle that turns as the weight is lowered. Let
the system then return to its initial state by transferring heat from the system until the
cycle has been completed.
Historically, work was measured in mechanical units of force times distance, such
as foot pounds force or joules, and heat was measured in thermal units, such as the British
thermal unit or the calorie. Measurements of work and heat were made during a cycle for
a wide variety of systems and for various amounts of work and heat. When the amounts
of work and heat were compared, it was found that they were always proportional. Such
116
The first law of thermodynamics for a Change in state of a Control mass
m 111
Gas
FIGURE 5.1 Example
of a control mass
undergoing a cycle.
1
observations led to the formulation of the first law of thermodynamics, which in equation
form is written
jj>8Q = j>5W (5.1)
The symbol $ SQ, which is called the cyclic integral of the heat transfer, represents the net
heat transfer during the cycle, and $ SW, the cyclic integral of the work, represents the net
work during the cycle. Here, J is a proportionality factor that depends on the units used
for work and heat.
The basis of every law of nature is experimental evidence, and this is true also of
the first law of thermodynamics. Many different experiments have been conducted on the
first law, and every one thus far has verified it either directly or indirectly. The first law
has never been disproved.
As was discussed in Chapter 4, the units for work and heat or for any other form of
energy either are the same or are directly proportional. In SI units, the joule is used as the
unit for both work and heat and for any other energy unit. In English units, the basic unit
for work is the foot pound force, and the basic unit for heat is the British thermal unit
(Btu). James P. Joule (1818-1889) did the first accurate work in the 1840s on measure-
ment of the proportionality factor J, which relates these units. Today, the Btu is defined in
terms of the basic SI metric units,
1 Btu = 778.17 ft lbf
This unit is termed the International British thermal unit. For much engineering work, the
accuracy of other data does not warrant more accuracy than the relation 1 Btu = 778 ft
lbf, which is the value used with English units in the problems in this text. Because these
units are equivalent, it is not necessary to include the factor J explicitly in Eq. 5.1, but
simply to recognize that for any system of units, each equation must have consistent units
throughout. Therefore, we may write Eq. 5.1 as
j>8Q = j>SW (5.2)
which can be considered the basic statement of the first law of thermodynamics.
5.2 THE FIRST law of thermodynamics for
a Change in state of a Control mass
Equation 5.2 states the first law of thermodynamics for a control mass during a cycle.
Many times, however, we are concerned with a process rather than a cycle. AVe now con-
sider the first law of thermodynamics for a control mass that undergoes a change of state.
118 M Chapter Five The First Law of Thermodynamics
THmwNET
We begin by introducing a new property, the energy, which is given the symbol E. Con-
sider a system that undergoes a cycle in which it changes from state 1 to state 2 by process
A and returns from state 2 to state 1 by process B. This cycle is shown in Fig. 5,2 on a
pressure (or other intensive property)-volume (or other extensive property) diagram.
From the first law of thermodynamics, Eq. 5.2, we have
j> 8Q = j> SW
Considering the two separate processes, we have
j* + f 2 SQb = | 2 sw A + f SW B
Now consider another cycle in which the control mass changes from state 1 to state 2 by
process C and returns to state 1 by process 5, as before. For this cycle we can write
/* SQ C + £ SQs = j 2 8W c + j 1 8W B
Subtracting the second of these equations from the first, we obtain
5W r
or, by rearranging,
j\8Q-8W) A = \\dQ-8W) c
(5.3)
Since A and C represent arbitrary processes between states 1 and 2, the quantity 8Q — 8 W
is the same for all processes between states 1 and 2. Therefore, 8Q - SW depends only on
the initial and final states and not on the path followed between the two states. We con-
clude that this is a point function, and therefore it is the differential of a property of the
mass. This property is the energy of the mass and is given the symbol E. Thus we can write
dE = 8Q- 8W
(5.4)
Because E is a property, its derivative is written dE. When Eq. 5.4 is integrated from an
initial state 1 to a final state 2, we have
^-^ = 102-102 (5-5)
where E v and£ 2 are the initial and final values of the energy E of the control mass, x Qi is
the heat transferred to the control mass during the process from state 1 to state 2, and Y W 2
is the work done by the control mass during the process.
FIGURE 5.2
Demonstration of the
existence of
thermodynamic property £.
the First Law of thermodynamics for a Change in State of a Control Mass
m 119
Note that a control mass may be made up of several different subsystems, as shown
in Fig. 5.3, In this case, each part must be analyzed and included separately in applying
the first law, Eq. 5.5. We further note that Eq. 5,5 is an expression of the general form
A Energy = 4- in — out
in terms of the standard sign conventions for heat and work.
The physical significance of the property E is that it represents all the energy of the
system in the given state. This energy might be present in a variety of forms, such as the
kinetic or potential energy of the system as a whole with respect to the chosen coordinate
frame, energy associated with the motion and position of the molecules, energy associated
with the structure of the atom, chemical energy present in a storage battery, energy pre-
sent in a charged condenser, or any of a number of other forms.
In the study of thermodynamics, it is convenient to consider the bulk kinetic and po-
tential energy separately and then to consider all the other energy of the control mass in a
single property that we call the internal energy and to which we give the symbol U. Thus,
we would write
E = Internal energy + kinetic energy + potential energy
or
E = U + KE + PE
The kinetic and potential energy of the control mass are associated with the coordi-
nate frame that we select and can be specified by the macroscopic parameters of mass, ve-
locity, and elevation. The internal energy U includes all other forms of energy of the
control mass and is associated with the thermodynamic state of the system.
Since the terms comprising E are point functions, we can write
dE = dU + 4KE) + rf(PE) (5.6)
The first law of thermodynamics for a change of state may therefore be written
dE=dU + rf(KE) + rf(PE) = 80 - 5 IF (5.7)
In words this equation states that as a control mass undergoes a change of state, en-
ergy may cross the boundary as either heat or work, and each may be positive or negative.
The net change in the energy of the system will be exactly equal to the net energy that
120 K Chapter five the first law of thermodynamics
crosses the boundary of the system. The energy of the system may change in any of three
ways— by a change in internal energy, in kinetic energy, or in potential energy.
This section concludes by deriving an expression for the kinetic and potential en-
ergy of a control mass. Consider a mass that is initially at rest relative to the earth, which
is taken as the coordinate frame. Let this system be acted on by an external horizontal
force F that moves the mass a distance dx in the direction of the force. Thus, there is no
change in potential energy. Let there be no heat transfer and no change in internal energy.
Then from the first law, Eq, 5.7, we have
8W= -Fdx = ~dKE
But
„ dV dxdV w^V
F — ma = m — r- = m —r — r~ = mv -7-
dt dt dx dx
Then
dKE = Fdx = mV dV
Integrating, we obtain
c KE f v
^KE = mVdV
Jke=o Jv=o
KE = ^mV 2 (5.8)
A similar expression for potential energy can be found. Consider a control mass that
is initially at rest and at the elevation of some reference level. Let this mass be acted on by
a vertical force F of such magnitude that it raises (in elevation) the mass with constant ve-
locity an amount dZ. Let the acceleration due to gravity at this point be g. From the first
law, Eq. 5.7, we have
SW = -FdZ= ~d~PE
F — ma = mg
Then
Integrating gives
dm =FdZ= mgdZ
rpE 2 rz,
dPE = m gdZ
J PE, ■> Z,
Assuming that g does not vary with-Z (which is a very reasonable assumption for moder-
ate changes in elevation), we obtain
PE 2 - PEi = mgfo - Z,) (5.9)
EXAMPLE 5.1 A car of mass 1 1 00 kg drives with a velocity such that it has a kinetic energy of 400 kJ (see
Fig. 5.4). Find the velocity. If the car is raised with a crane how high should it be lifted in
the standard gravitational field to have a potential energy that equals the kinetic energy?
The first law of thermodynamics for a change in state of a control mass E 121
FIGURE 5.4 Sketch
for Example 5.1.
Solution
The standard kinetic energy of the mass is
KE = i m V 2 = 400kJ
From this we can solve for the velocity
V -
2KE
2 X 400kJ
noo kg
S00Xl000Nm„ /8000kgms l m
1100 kg V 11kg
Standard potential energy is
PE = mgH
so when this is equal to the kinetic energy we get
j j KE _ 400 000Nm -371m
m S 1100kgX 9.807ms^ 2
Notice the necessity of converting the kJ to J in both calculations.
- 27m/s
EXAMPLE 5. IE A car of mass 2400 Ibm drives with a velocity such that it has a kinetic energy of 400
Btu. Find the velocity. If the car is raised with a crane, how high should it be lifted in the
standard gravitational field to have a potential energy that equals the kinetic energy?
Solution
The standard kinetic energy of the mass is
KE = I mV 2 = 400 Btu
From this we can solve for the velocity
V -
2 KE
\2 X 400 Btu X 778.17 4Jr x 32.174 ^|
Btu
2400 Ibm
= 91.4 ft/s
122 H Chapter Five The First Law of Thermodynamics
Standard potential energy is
PE = mgH
so when this is equal to the kinetic energy KE we get
400 Btu X 778.17 X 32.174 ^Mp
j j „ KE _ Btu lbfs ?
nig ft
2400 lbmX 32.174^
= 129.7 ft
Ibm ft
Note the necessity of using the conversion constant 32. 174 - — — in both calculations.
Ms 2
Now, substituting the expressions for kinetic and potential energy into Eq. 5.6, we have
dE = dU + mV dV + mg dZ
Integrating for a change of state from state 1 to state 2 with constant g, we get
E 2 - E x = U 2 - U x + ^ - ^ + mgTa ~ mgZ x
Similarly, substituting these expressions for kinetic and potential energy into Eq.
5.7, we have
dE = dU + + d(mgZ) = SQ-8W (5.10)
Assuming g is a constant, in the integrated form of this equation,
U 2 - U x + m{y2 \ V ' j + mg(Z 2 - Z x ) = X Q 2 - X W 2 (5.11)
Three observations should be made regarding this equation. The first observation is
that the property E, the energy of the control mass, was found to exist, and we were able
to write the first law for a change of state using Eq, 5.5. However, rather than deal with
this property E, we find it more convenient to consider the internal energy and the kinetic
and potential energies of the mass. In general, this procedure will be followed in the rest
of this book.
The second observation is that Eqs. 5.10 and 5.11 are in effect a statement of the
conservation of energy. The net change of the energy of the control mass is always equal
to the net transfer of energy across the boundary as heat and work. This is somewhat anal-
ogous to a joint checking account shared by a husband and wife. There are two ways in
which deposits and withdrawals can be made— either by the husband or by the wife — and
the balance will always reflect the net amount of the transaction. Similarly, there are two
ways in which energy can cross the boundary of a control mass — either as heat or as
work — and the energy of the mass will change by the exact amount of the net energy
crossing the boundary. The concept of energy and the law of the conservation of energy
are basic to thermodynamics.
The First law of thermodynamics for a Change in State of a Control mass
m 123
The third observation is that Eqs. 5.10 and 5.11 can give only changes in internal
energy, kinetic energy, and potential energy. We can leam nothing about absolute values
of these quantities from these equations. If we wish to assign values to internal energy, ki-
netic energy, and potential energy, we must assume reference states and assign a value to
the quantity in this reference state. The kinetic energy of a body with zero velocity rela-
tive to the earth is assumed to be zero. Similarly, the value of the potential energy is as-
sumed to be zero when the body is at some reference elevation. With internal energy,
therefore, we must also have a reference state if we wish to assign values of this property.
This matter is considered in the following section.
EXAMPLE 5,2 A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle
wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the
fluid inside a control surface and determine the change in internal energy of this control
mass.
The first law of thermodynamics is (Eq. 5.11)
U 2 -U,+^ m(V 2 2 - V?) + mgiZj - Z,) = ,g 2 " 1^2
Since there is no change in kinetic and potential energy, this reduces to
U 2 -U t =* -1500 - (-5090) - 3590 kJ
Consider a stone having a mass of 10 kg and a bucket containing 100 kg of liquid water.
Initially the stone is 10.2 m above the water, and the stone and the water are at the same
temperature, state 1. The stone then falls into the water.
Determine A 17, AKE, APE, Q, and fFfor the following changes of state, assuming
standard gravitational acceleration of 9.806 65 m/s 2 .
a. The stone is about to enter the water, state 2.
b. The stone has just come to rest in the bucket, state 3.
c. Heat has been transferred to the surroundings in such an amount that the stone and
water are at the same temperature, 7*!, state 4.
Analysis and Solution
The first law for any of the steps is
Q = AC/+ AKE + APE + W
and each term can be identified for each of the changes of state.
a. The stone has fallen from Z x to 2^, and we assume no heat transfer as it falls. The
water has not changed state; thus
A£/=0, i& = 0, ,JT 2 =
EXAMPLE 5.3
124 H CHAPTER FIVE the first law of thermodynamics
and the first law reduces to
AKE + APE
AKE
-APE = -mg(Z 2 - Z,)
- 10 kg X 9.806 65 m/s 2 X (-10.2 m)
1000 J = 1 kJ
That is, for the process from state 1 to state 2,
AKE = 1 kJ and APE = — 1 kJ
AU + AKE =
AU - -AKE = 1 kJ
c. In the final state, there is no kinetic, nor potential energy, and the internal energy is
the same as in state 1,
5,3 INTERNAL ENERGY — A THERMODYNAMIC
PROPERTY
Internal energy is an extensive property because it depends on the mass of the system.
Similarly, kinetic and potential energies are extensive properties.
The symbol U designates the internal energy of a given mass of a substance. Fol-
lowing the convention used with other extensive properties, the symbol u designates the
internal energy per unit mass. We could speak of it as the specific internal energy, as we
do with specific volume. However, because the context will usually make it clear whether
u or U is referred to, we will simply use the term internal energy to refer to both internal
energy per unit mass and the total internal energy.
In Chapter 3 we noted that in the absence of motion, gravity, surface effects, elec-
tricity, or other effects, the state of a pure substance is specified by two independent prop-
erties. It is very significant that, with these restrictions, the internal energy may be one of
the independent properties of a pure substance. This means, for example, that if we spec-
ify the pressure and internal energy (with reference to an arbitrary base) of superheated
steam, the temperature is also specified.
Thus, in a table of thermodynamic properties such as the steam tables, the' value of
internal energy can be tabulated along with other thermodynamic properties. Tables 1
and 2 of the steam tables (Tables B.l.l and B.1.2) list the internal energy for saturated
AE/-= -lkJ,
AKE = 0, APE = 0,
3 Q A = AU= -lkJ
Internal Energy— A thermodynamic Property H 125
states. Included are the internal energy of saturated liquid u fi the internal energy of satu-
rated vapor u g , and the difference between the internal energy of saturated liquid and sat-
urated vapor Uf r The values are given in relation to an arbitrarily assumed reference
state, which, for water in the steam tables, is taken as zero for saturated liquid at the
triple-point temperature, 0.0 TC. All values of internal energy in the steam tables are
then calculated relative to this reference (note that the reference state cancels out when
finding a difference in u between any two states). Values for internal energy are found in
the steam tables in the same manner as for specific volume. In the liquid-vapor satura-
tion region,
or
mu = m liq ti f + m^u s
Dividing by m and introducing the quality x gives
u = (1 — x)itf+ xu g
u=u f +xu fg
As an example, the specific internal energy of saturated seam having a pressure of
0.6 MPa and a quality of 95% can be calculated as
= u f + xu fg = 669.9 + 0.95(1897.5) = 2472.5 kJ/kg
Values for u in the superheated vapor region are tabulated in Table B.1.3, for compressed
liquid in Table B.1.4, and for solid-vapor in Table B. 1.5.
EXAMPLE 5.4 Determine the missing property (P, T or x) and also v for water at each of the following
states:
a. r= 300°C J i/ = 2780kJ/kg
b. P = 2000 kPa, it = 2000 kJ/kg
For each case, the two properties given are independent properties and therefore fix the
state. For each, we must first determine the phase by comparison of the given informa-
tion with phase boundary values.
a. At 300°C 3 from Table B.l.l, u g = 2563.0 kJ/kg. The given u > u g> so the state is in
the superheated vapor region at some P less than P gi which is 8581 kPa. Searching
through Table B.1.3 at 300°C, we find that the value u = 2780 is between given
values of u at 1600 kPa (2781.0) and 1800 kPa (2776.8). Interpolating linearly, we
obtain
P= 1648kPa.
Note that quality is undefined in the superheated vapor region. At this pressure, by
linear interpolation, we have v = 0.1542 mVkg.
126 H Chapter Five The First Law of Thermodynamics
b. At P = 2000 kPa, from Table B.1.2, the given u of 2000 kJ/kg is greater than a f
(906.4) but less than u g (2600,3). Therefore, this state is in the two-phase region with
T=T g = 212.4°C, and
u = 2000 = 906.4 + xl693.8, x = 0.6456
Then,
v = 0.001 177 + 0.6456 X 0.098 45 = 0.064 74 m 3 /kg. .
5 A PROBLEM ANALYSIS AND
SOLUTION TECHNIQUE
At this point in our study of thermodynamics, we have progressed sufficiently far (that is,
we have accumulated sufficient tools with which to work) that it is worthwhile to develop
a somewhat formal technique or procedure for analyzing and solving thermodynamic
problems. For the time being it may not seem entirely necessary to use such a rigorous
procedure for many of our problems, but we should keep in mind that as we acquire more
analytical tools the problems that we are capable of dealing with will become much more
complicated. Thus, it is appropriate that we begin to practice this technique now in antici-
pation of these future problems.
Our problem analysts and solution technique is contained within the framework of
the following set of questions that must be answered in the process of an orderly solution
of a thermodynamic problem.
1. What is the control mass or control volume? Is it useful, or necessary, to choose
more than one? It may be helpful to draw a sketch of the system at this point, illus-
trating all heat and work flows, and indicating forces such as external pressures and
gravitation.
2. What do we know about the initial state (i.e., which properties are known)?
3. What do we know about the final state?
4. What do we know about the process that takes place? Is anything constant or zero?
Is there some known functional relation between two properties?
5. Is it helpful to draw a diagram of the information in steps 2 to 4 (for example, a T-v
or P-u diagram)?
6. What is our thermodynamic model for the behavior of the substance (for example,
steam tables, ideal gas, and so' on)?
7. What is our analysis of the problem (i.e., do we examine control surfaces for vari-
ous work modes or use the first law or conservation of mass)?
8. What is our solution technique? In other words, from what we have done so far in
steps 1-7, how do we proceed to find whatever it is that is desired? Is a trial-and-
error solution necessary?
It is not always necessary to write out all these steps, and in the majority of the examples
throughout this text we will not do so. However, when faced with a new and unfamiliar
problem Analysis and solution technique 9 127
problem, the student should always at least think through this set of questions to develop
the ability to solve more challengingproblems. In solving the following example, we will
use this technique in detail.
EXAMPLE 5.5 A vessel having a volume of 5 m 3 contains 0.05 m 3 of saturated liquid water and 4.95 m 3
of saturated water vapor at 0. 1 MPa. Heat is transferred until the vessel is filled with sat-
urated vapor. Determine the heat transfer for this process.
Control mass: All the water inside the vessel.
Sketch: Fig. 5.5.
Initial state: Pressure, volume of liquid, volume of vapor; therefore, state 1 is fixed. :
Final state: Somewhere along the saturated- vapor curve; the water was heated,
so l\ > Pi.
Process: Constant volume and mass; therefore, constant specific volume.
Diagram: Fig. 5.6,
Model: Steam tables.
Analysis
From the first law we have
iQz=U 2 -U l + m-
V?- V?
From examining the control surface for various work modes, we conclude that the work
for this process is zero. Furthermore, the system is not moving, so there is no change in
kinetic energy. There is a small change in the center of mass of the system but we will
assume that the corresponding change in potential energy is negligible (in kilojoules).
Therefore,
iQi =U 2 -U l
Solution
The heat transfer will be found from the first law. State 1 is known, so U x can be calcu-
lated. The specific volume at state 2 is also known (from state 1 and the process). Since
FIGURE 5.5 Sketch
for Example 5.5.
128 H Chapter Five the First Law of Thermodynamics
FIGURE 5.6 Diagram
for Example 5.5.
Vo=V,
state 2 is saturated vapor, state 2 is fixed, as is seen from Fig. 5.6. Therefore, U 2 can also
be found.
The solution proceeds as follows:
0.05
lliq v f 0.001043
= 47.94 kg
'1 vap
4.95
' vap
~Vg~ 1.6940
= 2.92 kg
Then
&l ~ m lliq H Itiq + w *lvap H Ivap
= 47.94(417.36) + 2.92(2506.1) = 27 326 kJ
To deterrnrne u 2 we need to know two thermodynamic properties, since this determines
the final state. The properties we know are the quality, x = 100%, and v 2i the final spe-
cific volume, which can readily be determined.
m = mi li, + m l vap = 47.94 + 2.92 = 50.86 kg
y 2 = »: =
5.0
™ 50.86
= 0.098 31 m 3 /kg
In Table B.1.2 we find, by interpolation, that at a pressure of 2.03 MPa, v g = 0.098 31
m 3 /kg. The final pressure of the steam is therefore 2.03 MPa. Then
« 2 = 2600.5 kJ/kg
U 2 = mu 2 = 50.86(2600.5) = 132 261 kJ >
t Q 2 = U 2 ~ = 132 261 - 27 326 = 104 935 kJ
PROBLEM ANALYSIS AND SOLUTION TECHNIQUE H 129
EXAMPLE 5.SE A vessel having a volume of 100 ft 3 contains 1 ft 3 of saturated liquid water and 99 ft 3 of
saturated water vapor at 14,7 lbf/in 2 . Heat is transferred until the vessel is filled with sat-
urated vapor. Determine the heat transfer for this process.
Control mass:
All the water inside the vessel.
Sketch:
Fig. 5.5.
Initial state:
Pressure, volume of liquid, volume of vapor; therefore, state 1 is
fixed. '■
Final state:
Somewhere along the saturated- vapor curve; the water .-was
heated, so P 2 > Pi-
Process:
Constant volume and mass; therefore, constant specific volume. :
Diagram:
Fig. 5.6.
Model:
Steam tables.
Analysis
First law: X Q 2 - U 2 - £/, + m 2 2 + mgt& - Z x ) + X W 2
From examining the control surface for various work modes, we conclude that the work
for this process is zero. Furthermore, the system is not moving, so there is no change in
kinetic energy. There is a small change in center of mass of the system, but we will as-
sume that the corresponding change in potential energy is negligible (compared to other
terms). Therefore,
X Q 2 =U 2 ~U X
Solution
The heat transfer will be found from the first law. State 1 is known, so U x can be calcu-
lated. Also, the specific volume at state 2 is known (from state 1 and the process). Since
state 2 is saturated vapor, state 2 is fixed, as is seen from Fig. 5.6. Therefore, U 2 can also
be found.
The solution proceeds as follows:
'ilk - — ~ =59.81 Ibm
1
0.016 72
v
' vap
99
26.80
< lvap - ^ - - 3.69 Ibm
Then,
U\ — ^lUq u lliq ' M lvap W lvap
= 59.81(180.1) + 3.69(1077.6) = 14 748 Btu
130 Chapter Five The First Law of Thermodynamics
To determine u 2 we need to know two thermodynamic properties, since this determines
the final state. The properties we know are the quality, x = 100%, and v 2 , the final spe- ■
cific volume, which can readily be determined.
m •-- m m + m lvsp = 59.81 I 3.69 - 63.50 lbm
y? ~^"63fo : " L575 ftV,bra
In Table F7.1 of the steam tables we find, by interpolation, that at a pressure of 294
lbf/in 2 , u g = 1.575 ftMbm. The final pressure of the steam is therefore 294 lbf/in 2 . Then,
u ? = 1117.0 Btu/lbm
U 2 -■ mu 2 - 63.50(1 1 17.0) - 70 930 Htu
,0 2 = U 2 - U x = 70 930 - 14 748 - 56 182 Btu
-£> Gas
FIGURE 5.7 The
constant-pressure quasi-
equilibrium process.
5.5 the thermodynamic
Property Enthalpy
In analyzing specific types of processes, we frequently encounter certain combinations of
thermodynamic properties, which are therefore also properties of the substance undergo-
ing the change of state. To demonstrate one such situation, let us consider a control mass
undergoing a quasi-equilibrium constant-pressure process, as shown in Fig. 5,7. Assume
that there are no changes in kinetic or potential energy and that the only work done during
the process is that associated with the boundary movement. Taking the gas as our control
mass and applying the first law, Eq. 5.11, we have, in terms of Q,
xQx=V 2 -V x + l W 2
The work done can be calculated from the relation
,W 2
-j:
PdV
Since the pressure is constant,
= P j 2 dV^P(V 2 - V x )
Therefore,
iQ 2 =U 2 ~U l + P 2 V 2 -P l V l
= (u 2 + p 2 r 3 )-(v l +p l v l )
We find that, in this very restricted case, the heat transfer during the process is
given in terms of the change in the quantity U + PV between the initial and final states.
Because all these quantities are thermodynamic properties, that is, functions only of the
state of the system, their combination must also have these same characteristics. There-
fore, we find it convenient to define a new extensive property, the enthalpy,
H^U + PV (5.12)
The thermodynamic Property enthalpy fl 131
or, per unit mass,
h = u + p v (5.13)
As for internal energy, we could speak of specific enthalpy, h, and total enthalpy, H.
However, we will refer to both as enthalpy, since the context will make it clear which is
being discussed.
The heat transfer in a constant-pressure quasi-equilibrium process is equal to the
change in enthalpy, which includes both the change in internal energy and the work for
this particular process. This is by no means a general result. It is valid for this special case
only because the work done during the process is equal to the difference m the PV product
for the final and initial states. This would not be true if the pressure had not remained con-
stant during the process.
The significance and use of enthalpy is not restricted to the special process just de-
scribed. Other cases in which this same combination of properties u + Pv appear will be
developed later, notably in Chapter 6 in which we discuss control volume analyses. Our
reason for introducing enthalpy at this time is that although the tables in Appendix B list
values for internal energy, many other tables and charts of thermodynamic properties give
values for enthalpy but not for the internal energy. Therefore, it is necessary to calculate
the internal energy at a state using the tabulated values and Eq. 5.13:
u = h - Pv
Students often become confused about the validity of this calculation when analyz-
ing system processes that do not occur at constant pressure, for which enthalpy has no
physical significance. We must keep in mind that enthalpy, being a property, is a state or
point function, and its use in calculating internal energy at the same state is not related to,
or dependent on, any process that may be taking place.
Tabular values of internal energy and enthalpy, such as those included in Table's B.l
through B.7, are all relative to some arbitrarily selected base. In the steam tables, the in-
ternal energy of saturated liquid at 0.0 TC is the reference state and is given a value of
zero. For refrigerants, such as ammonia and chlorofmorocarbons R-12 and R-22, the ref-
erence state is arbitrarily taken as saturated liquid at — 40°C. The enthalpy in this refer-
ence state is assigned the value of zero. Cryogenic fluids, such as nitrogen, have other
arbitrary reference states chosen for enthalpy values listed in their tables. Because each of
these reference states is arbitrarily selected, it is always possible to have negative values
for enthalpy, as for saturated-solid water in Table B.1.5. When enthalpy and internal en-
ergy are given values relative to the same reference state, as they are in essentially all
thermodynamic tables, the difference between internal energy and enthalpy at the refer-
ence state is equal to Pv. Since the specific volume of the liquid is very small, this product
is negligible as far as the significant figures of the tables are concerned, but the principle
should be kept in mind, for in certain cases it is significant.
In many thermodynamic tables, values of the specific internal energy w are not given.
As mentioned earlier, these values can be readily calculated from the relation u = h — Pv,
though it is important to keep the units in mind. As an example, let us calculate the internal
energy u of superheated R-134a at 0.4 MPa, 70°C.
u = h-Pv
= 460.545 - 400 X 0.066 484
= 433.951 kJ/kg
132 H CHAPTER FIVE the first law of thermodynamics
The enthalpy of a substance in a saturation state and with a given quality is found
in the same way as the specific volume and internal energy. The enthalpy of saturated
liquid has the symbol hp saturated vapor h g , and the increase in enthalpy during vapor-
ization hf g . For a saturation state, the enthalpy can be calculated by one of the following
relations:
The enthalpy of compressed liquid water may be found from Table B.1.4. For sub-
stances for which compressed-liquid tables are not available, the enthalpy is taken as that
of saturated liquid at the same temperature.
EXAMPLE 5.6 A cylinder fitted with a piston has a volume of 0.1 m 3 and contains 0,5 kg of steam at 0.4
MPa. Heat is transferred to the steam until the temperature is 300°C, while the pressure
remains constant.
Determine the heat transfer and the work for this process.
Control mass: Water inside cylinder.
Initial state: P h V u m\ therefore u u is known, state 1 is fixed (at v u check
steam tables — two-phase region).
Final state: P 2 , T 2 ; therefore state 2 is fixed (superheated). '
Process: Constant pressure.
Diagram: Fig. 5.8.
Model: Steam tables.
There is no change in kinetic energy or change in potential energy. Work is done by
movement at the boundary. Assume the process to be quasi-equilibrium. Since the pres-
sure is constant, we have
h - (1 - x)h f + xh t
h = hj- + xhf g
Analysis
X W 2 = \ PdV=P dV^P(V 2 ~V 1 ) = 7n(P 2 v 2 ~P 1 v 1 )
T
P
V
FIGURE 5.8 The constant-pressure quasi-equifibrium process.
y
THE CONSTANT-VOLUME AND CONSTANT-PRESSURE SPECIFIC HEATS M 133
Therefore, the first law is, in terms of Q,
= m(u 2 - «j) + /n(F 2 y2 - P\Vi)'= m(h 2 - hi)
Solution
There is a choice of procedures to follow. State 1 is known, so v x and h x (or ui) can be
found. State 2 is also known, so v 2 and h 2 (or n 2 ) can be found. Using the first law and
the work equation, we can calculate the heat transfer and work. Using the enthalpies, we
have
Vl = h = M = o.2 = 0.001 084 + ^0.4614
1 0.4614
//, ^ h f + x x h /g
= 604.74 + 0.4311 X 2133.8 = 1524.7kJ/kg
/j 2 •- 3066.8 kJ/kg
i0z = 0.5(3066.8 - 1524.7) = 771.1 kJ
J¥ 2 - mP(v 2 ~ i>i) = 0.5 X 400(0.6548 - 0.2) = 91.0 kJ
Therefore,
U 2 ~U X = X Q 2 - X W 2 - 771.1 - 91.0 = 680.1 kJ ■
The heat transfer could also have been found from and u 2 :
11 x — UfT X\llj g
- 604.31 -f 0.4311 X 1949.3 = 1444.7 kJ/kg
n 2 = 2804.8 kJ/kg
and
xQ 2 -U 2 ■ <7, -t- X W 2
= 0.5(2804.8 - 1444.7) + 91.0 = 771.1 kJ
5.6 The Constant-Volume and
Constant-Pressure Specific Heats
In this section we will consider a homogeneous phase of a substance of constant composi-
tion. This phase may be a solid, a liquid, or a gas, but no change of phase will occur. We
will then define a variable termed the specific heat, the amount of heat required per unit
mass to raise the temperature by one degree. Since it would be of interest to examine the
relation between the specific heat and other thermodynamic variables, we note first that
134 H Chapter Five The First Law of Thermodynamics
the heat transfer is given by Eq. 5.10. Neglecting changes in kinetic and potential ener-
gies, and assuming a simple compressible substance and quasi-equilibrium process, for
which the work in Eq. 5. 10 is given by Eq. 4.2, we have
8Q = dU + 6W=dU + PdV
We find that this expression can be evaluated for two separate special cases.
(^Constant volume, for which the work term (P dV) is zero, so that the specific heat
(at constant volume) is
c » = k (S\ = ^ (jt) v = (5,14)
( 2. Constant pressure, for which the work term can be integrated and the resulting PV
^-^terms at the initial and final states can be associated with the internal energy terms,
as in Section 5.5, thereby leading to the conclusion that the heat transfer can be ex-
pressed tn terms of the enthalpy change. The corresponding specific heat (at con-
stant pressure) is
C =1(^1 = (&) (5 15)
^ ™\8Tj p ™{dT) p \dTj p P ' 1DJ
Note that in each of these special cases, the resulting expression, Eq. 5.14 or 5.15,
contains only thermodynamic properties, from which we conclude that the constant-
volume and constant-pressure specific heats must themselves be thermodynamic proper-
ties. This means that, although we began this discussion by considering the amount of
heat transfer required to cause a unit temperature change and then proceeded through a
very specific development leading to Eq. 5.14 (or 5.15), the result ultimately expresses a
relation among a set of thermodynamic properties and therefore constitutes a definition
that is independent of the particular process leading to it (in the same sense that the defini-
tion of enthalpy in the previous section is independent of the process used to illustrate one
situation in which the property is useful in a thermodynamic analysis). As an example,
consider the two identical fluid masses shown in Fig. 5.9. In the first system 100 kJ of
heat is transferred to it, and in the second system 100 kj of work is done on it. Thus, the
change of internal energy is the same for each, and therefore the final state and the final
temperature are the same in each. In accordance with Eq. 5.14, therefore, exactly the same
value for the average constant-volume specific heat would be found for this substance for
the two processes, even though the two processes are very different as far as heat transfer
is concerned.
FIGURE 5.9 Sketch
showing two ways in
which a given AC may be
achieved.
-iy= too kJ
Rufd
£=100 kJ
ft ft ft
1
Ml
The Internal Energy, Enthalpy, and Specific Heat of Ideal Gases H 135
Solids and Liquids
' As a special case, consider either a solid or a liquid. Since both of these phases are nearly
incompressible,
dh = du + d{Pv) ~du + v dP (5.16)
Also, for both of these phases, the specific volume is very small, such that in many cases
dh^du^CdT (5.17)
where C is either the constant- volume or the constant-pressure specific heat, as the two
would be nearly the same. In many processes involving a solid or a liquid, we might
further assume that the specific heat in Eq. 5.17 is constant (unless the process occurs
at low temperature or over a wide range of temperatures). Equation 5.17 can then be
integrated to
h 2 - A, - u 2 - h, - C(T 7 - TO (5.18)
Specific heats for various solids and liquids are listed in Tables A.3 and A.4.
In other processes for which it is not possible to assume constant specific heat, there
may be a known relation for C as a function of temperature. Equation 5.17 could then also
be integrated.
5.7 the internal energy, enthalpy,
and Specific Heat of Ideal gases
In general, for any substance the internal energy u depends on the two independent prop-
erties specifying the state. For a low-density gas, however, a depends primarily on T and
much less on the second property, P or v. For example, consider several values for super-
heated vapor steam from Table B.1.3, shown in Table 5.1. From these values, it is evident
that u depends strongly on T, but not much on P. Also, we note that the dependence of u
on P is less at low pressure and is much less at high temperature; that is, as the density de-
creases, so does dependence of u on P (or v). It is therefore reasonable to extrapolate this
behavior to very low density and to assume that as gas density becomes so low that the
ideal-gas model is appropriate, internal energy does not depend on pressure at all but is a
function only of temperature. That is, for an ideal gas,
Pv=RT and u =f{T) only (5.19)
Table 5.1
Internal Energy for Superheated Vapor Steam
r,°c
P, kPa
10
100
500
1000
200
2661.3
2658.1
2642.9
2621.9
700
3479.6
3479.2
3477.5
3475.4
1200
4467.9
4467.7
4466.S
4465.6
136 M Chapter Five The first Law of thermodynamics
The relation between the internal energy u and the temperature can be established
by using the definition of constant- volume specific heat given by Eq. 5,14:
C = (^i
Because the internal energy of an ideal gas is not a function of specific volume, for an
ideal gas we can write
r = ^"
C " dT
du = C e0 dT (5.20)
where the subscript denotes the specific heat of an ideal gas. For a given mass m,
dU=mC v0 dT (5.21)
From the definition of enthalpy and the equation of state of an ideal gas, it follows
that
h = u + Pv = u + RT (5.22)
Since R is a constant and u is a function of temperature only, it follows that the en-
thalpy, h } of an ideal gas is also a function of temperature only. That is,
h -AT) (5.23)
The relation between enthalpy and temperature is found from the constant-pressure spe-
cific heat as defined by Eq. 5.15:
Since the enthalpy of an ideal gas is a function of the temperature only and is independent
of the pressure, it follows that
r =4L
dT
dh = C^dT (5.24)
For a given mass m,
dH = mC pa dT (5.25)
The consequences ofEqs. 5.20 and 5.24 are demonstrated in Fig. 5.10, which shows
two lines of constant temperature. Since internal energy and enthalpy are functions of
temperature only, these lines of constant temperature are also lines of constant internal en-
ergy and constant enthalpy. From state 1 the high temperature can be reached by a variety
of paths, and in each case the final state is different. However, regardless of the path, the
change in internal energy is the same, as is the change in enthalpy, for lines of constant
temperature are also lines of constant u and constant h.
Because the internal energy and enthalpy of an ideal gas are functions of tempera-
ture only, it also follows that the constant-volume and constant-pressure specific heats are
also functions of temperature only. That is,
C v0 =AT), C p0 =AT) (5.26)
The Internal Energy, Enthalpy, and Specific Heat of Ideal Gases H 137
Because all gases approach ideal-gas behavior as the pressure approaches zero, the ideal-
gas specific heat for a given substance is often called the zero-pressure specific heat, and
the zero-pressure, constant-pressure specific heat is given the symbol C fQ . The zero-pres-
sure, constant-volume specific heat is given the symbol C„ . Figure 5.11 shows as a
function of temperature for a number of different substances. These values are determined
by the techniques of statistical thermodynamics and will not be discussed here. A brief
summary presentation of this subject is given in Appendix C. It is noted there that the
principal factor causing specific heat to vary with temperature is molecular vibration.
More complex molecules have multiple vibrational modes and therefore show a greater
temperature dependency, as is seen in Fig. 5.1 1. This is an important consideration when
deciding whether or not to account for specific heat variation with temperature in any par-
ticular application.
8
7
6
T 5
4
3
FIGURE 5.11 Heat 2
capacity for some gases as 500 1000 1500 2000 2500 3000 3500
function of temperature. T [K]
138 H chapter Five the first law of thermodynamics
A very important relation between the constant-pressure and constant-volume spe-
cific heats of an ideal gas may be developed from the definition of enthalpy:
A = a + Pu ~ u + RT
Differentiating and substituting Eqs. 5.20 and 5.24, we have
dh = du + R dT
Cp dT = C vQ dT+RdT
Therefore,
C p0 ~C v0 = R (5.27)
On a mote basis this equation is written
Cpo ~C v0 = R (5.27)
This tells us that the difference between the constant-pressure and constant-volume spe-
cific heats of an ideal gas is always constant, though both are functions of temperature.
Thus, we need examine only the temperature dependency of one, and the other is given by
Eq. 5.27.
Let us consider the specific heat C^. There are three possibilities to examine. The
situation is simplest if we assume constant specific heat, that is, no temperature depen-
dence. Then it is possible to integrate Eq. 5.24 directly to
h ~ h x = C^(r 2 - TO (5.29)
We note from Fig. 5.11 the circumstances under which this will be an accurate
model. It should be added, however, that it may be a reasonable approximation under
other conditions, especially if an average specific heat in the particular temperature range
is used in Eq. 5.29. Values of specific heat at room temperature and gas constants for vari-
ous gases are given in Table A.5.
The second possibility for the specific heat is to use an analytical equation for as
a function of temperature. Because the results of specific-heat calculations from statistical
thermodynamics do not lend themselves to convenient mathematical forms, these results
have been approximated empirically. The equations for as a function of temperature
are listed in Table A.6 for a number of gases.
The third possibility is to integrate the results of the calculations of statistical ther-
modynamics from an arbitrary reference temperature to any other temperature T and to
define a function
This function can then be tabulated in a Tingle-entry (temperature) table. Then, between
any two states 1 and 2,
h 2 - ft, = \ Tl dT- P dT = h Ti - h 2
J T<> •>T,
(5.30)
and it is seen that the reference temperature cancels out. This function h T (and a similar
function i/ r = h T — RT) is listed for air in Table A. 7. These functions are listed for other
gases in Table A. 8.
The Internal Energy, Enthalpy, and Specihc Heat of Ideal Gases Ml 139
To summarize the three possibilities, we note that using the ideal-gas tables, Tables
A. 7 and A. 8, gives us the most accurate answer, but that the equations in Table A.6 would
give a close empirical approximation. Constant specific heat would be less accurate, ex-
cept for monatomic gases and gases below room temperature. It should be remembered
that all these results are a part of the ideal-gas model, which in many of our problems is
not a valid assumption for the behavior of the substance.
EXAMPLE 5.7 Calculate the change of enthalpy as 1 kg of oxygen is heated from 300 to 1500 K. As-
sume ideal-gas behavior.
Solution
For an ideal gas, the enthalpy change is given by Eq. 5.24. However, we also need to
make an assumption about the dependence of specific heat on temperature. Let us solve
this problem in several ways and compare the answers.
Our most accurate answer for the ideal-gas enthalpy change for oxygen between 300
and 1500 K would be from the ideal-gas tables, Table A.8. This result is, using Eq. 5.30,
h 2 -h } = 1540.2 - 273.2 = 1267.0 kJ/kg
The empirical equation from Table A.6 should give a good approximation to this
result. Integrating Eq. 5.24, we have
h 2 -h x = Oo dT = f 92 CJff) X 1000 dB
= 100 o lasso - M501 02 + 0|4 fl 3 _ 033 J'*""
L 2 3 4 Je,=o.3
= 1241.5 kJ/kg
which is lower than the first result by 2.0%.
If we assume constant specific heat, we must be concerned about what value
we are going to use. If we use the value at 300 K from Table A. 5, we find, from Eq.
5.29, that
h 2 ~h l = CpaiTz ~ T x ) = 0.922 X 1200 = 1106.4 kJ/kg
which is low by 12.7%. However, suppose we assume that the specific heat is constant
at its value at 900 K, the average temperature. Substituting 900 K into the equation for
specific heat from Table A.6, we have
= 0.88 - 0.0001(0.9) + 0'.54(0.9) 2 - 0.33(0.9) 3
= 1.0767 kJ/kgK
Substituting this value into Eq. 5.29 gives the result
h 2 -h x = 1.0767 X 1200 = 1292.1 kJ/kg
which is high by about 2.0%, a much closer result than the one using the room tempera-
ture specific heat. It should be kept in mind that part of the model involving ideal gas
with constant specific heat also involves a choice of what value is to be used.
140 H Chapter Five The First Law of Thermodynamics
Example 5.8
A cylinder fitted with a piston has an initial volume of 0.1 m 3 and contains nitrogen at 150
kPa, 25°C. The piston is moved, compressing the nitrogen until the pressure is 1 MPa and
the temperature is 1 50°C. During this compression process heat is transferred from the nitro-
gen, and the work done on the nitrogen is 20 kJ. Determine the amount of this heat transfer.
Control mass:
Initial state:
Final state:
Process:
Model:
Nitrogen.
P u T u Fi; state 1 fixed.
P 2 , T 2 ; state 2 fixed.
Work input known.
Ideal gas, constant specific heat with value at 300 K, Table A.5.
Analysis
From the first law we have
102 = m(u 2 ~ «j) + X W 2
Solution
The mass of nitrogen is found from the equation of state with the value of R from Table A.5:
■ PV 150 kPaX 0.1m 3
m —
RT
0.2968
kJ
k g K
0.1695 kg
X 298.15 K
Assuming constant specific heat as given in Table A.5, we have
] Q 2 = mC v0 (T 2 -T l ) + 1 W 2
= 0.1695 kg X 0.745 ™^ X (150 - 25) K- 20.0
= 15.8 -20.0- -4.2 kJ
It would, of course, be somewhat more accurate to use Table A.8 than to assume con-
stant specific heat (room temperature value), but often the slight increase in accuracy
does not warrant the added difficulties of manually interpolating the tables.
EXAMPLE 5.8E A cylinder fitted with a piston has an initial volume of 2 ft 3 and contains nitrogen at 20
lbf/in 2 , 80 F. The piston is moved, compressing the nitrogen until the pressure is 160
lbf/in 2 and the temperature is 300 F. During this compression process heat is transferred
from the nitrogen, and the work done on the nitrogen is 9.15 Btu, Determine the amount
of this heat transfer.
Control mass: Nitrogen.
Initial state: P u T u V t ; state 1 fixed.
Final state: P 2l T 2 ; state 2 fixed. '
Process: Work input known.
Model: Ideal gas, constant specific heat with value at 540 R, Table P.4.
The First Law as a Rate Equation H 141
Analysis
First law: jg 2 - m(ti 2 - u t ) + JV 2
Solution
The mass of nitrogen is found from the equation of state with the value ofR from Table FA
20^| X 144 X % 2 ft 3
PV m ft 2 '
»i = = — - — — i± = 0.19341bm
RT 55.15,^X 540/?
lbm R
Assuming constant specific heat as given in Table F.4,
= 0.1934 lbm X 0.177 X (300 - 80) R - 9.15
lbm R J
= 7.53 - 9.15 = -1.62 Btu
It would, of course, be somewhat more accurate to use Table F.6 than to assume con-
stant specific heat (room temperature value), but often the slight increase in accuracy
does not warrant the added difficulties of manually interpolating the tables.
5.8 The First Law as a rate Equation
We frequently find it desirable to use the first law as a rate equation that expresses either
the instantaneous or average rate at which energy crosses the control surface as heat and
work and the rate at which the energy of the control mass changes. In so doing we are de-
parting from a strictly classical point of view, because basically classical thermodynamics
deals with systems that are in equilibrium, and time is not a relevant parameter for sys-
tems that are in equilibrium. However, since these rate equations are developed from the
concepts of classical thermodynamics and are used in many applications of thermodynam-
ics, they are included in this book. This rate form of the first law will be used in the devel-
opment of the first law for the control volume in Section 6.2, and in this form the first law
finds extensive applications in thermodynamics, fluid mechanics, and heat transfer.
Consider a time interval St during which an amount of heat SQ crosses the control
surface, an amount of work S W is done by the control mass, the internal energy change is
A 17, the kinetic energy change is AKE, and the potential energy change is APE. From the
first law we can -write
AC/ + AKE + APE = SQ- 8W
Dividing by St we have the average rate of energy transfer as heat work and increase of
the energy of the control mass:
142 ffl CHAPTER FIVE THE FIRST LAW OF THERMODYNAMICS
Taking the limit for each of these quantities as St approaches zero, we have
AC/ _ du A(KE) _ rf(KE) A(PE) _ dm
ln~sJ-lu' dT> JSS 5/ dt
lim -~ = Q (the heat transfer rate)
$i~*Q St
lim ^ = JF (the power)
Therefore, the rate equation form of the first law is
dU + dm + m = Q-w (5.31)
dt dt dt *
We could also write this in the form
^l = Q~JV (5.32)
at
EXAMPLE 5.9 During the charging of a storage battery, the current i is 20 A and the voltage % is 12.8
V. The rate of heat transfer from the battery is 10 W. At what rate is the internal energy
increasing?
Solution
Since changes in kinetic and potential energy are insignificant, the first law can be writ-
ten as a rate equation in the form of Eq. 5.31:
dt *
iy=%i=~ 12.8 X 20 = -256 W = -256 J/s
Therefore,
^=Q - W= -10 -(-256) = 246 J/s
EXAMPLE 5.10 A 25-kg cast-iron wood-burning stove, shown in Fig. 5.12, contains 5 kg of soft pine
wood and 1 kg of air. All the masses are at room temperature, 20°C, and pressure, 101
kPa. The wood now bums and heats all the mass uniformly, releasing 1500 watts. Ne-
glect any air flow and changes in mass and heat losses. Find the rate of change of the
temperature (dT/dt)' mi estimate the time it will take to reach a temperature of 75°C.
Solution
C. V,: The iron, wood and air.
This is a control mass.
Energy equation rate form:
E=Q-W
Conservation of Mass B 143
FIGURE 5.12 Sketch
for Example 5.10.
We have no changes in kinetic or potential energy and no change in mass, so
U ~~ f aifWaif m wood'Arood ^ ^tron^iron
E — U = WairKafc + »* WO od"wood + m iron H ir(ra
Now the energy equation has zero work, an energy release of Q, and becomes
* .
dT '
( m airQ r air + m woodC*-ood ^iiraCiroa) ^ = £? —
rfr = . 2
1500
- 0.0828 K/s
1 X 0.717 + 5 X 1.38 + 25 X 0.42 kg (kJ/kg)
Assuming the rate of temperature rise is constant, we can find the elapsed time as
J dt dt
Ar 75 - 20
41
dt
0.0828
664 s = 1 1 min
5,9 CONSERVATION OF MASS
In the previous sections we considered the first law of thermodynamics for a control mass
undergoing a change of state. A control mass is defined as a fixed quantity of mass. The
question now is whether the mass of such a system changes when its energy changes. If it
does, our definition of a control mass as a fixed quantity of mass is no longer valid when
the energy changes.
We know from relativistic considerations that mass and energy are related by the
well-known equation
E = m<? (5.33)
where c = velocity of light and E = energy. We conclude from this equation that the
mass of a control mass does change when its energy changes. Let us calculate the magni-
tude of this change of mass for a typical problem and determine whether this change in
mass is significant.
Consider a rigid vessel that contains a 1-kg stoichiometric mixture of a hydrocarbon
fuel (such as gasoline) and air. From our knowledge of combustion, we know that after
combustion takes place it wilt be necessary to transfer about 2900 kj from the system to
restore it to its initial temperature. From the first law
ifi^Oi-t/i + i^
144 H Chapter Five The First Law of Thermodynamics
we conclude that since X W 2 = and X Q 2 = -2900 kJ, the internal energy of this system
decreases by 2900 kj during the heat transfer process. Let us now calculate the decrease
in mass during this process using Eq. 5.33.
The velocity of light, c, is 2.9979 X 10 8 m/s. Therefore,
2900 kJ = 2 900 000 J = m (kg) X (2.9979 X 10 8 m/s) 2
and so
m -3.23 X 10" u kg
Thus, when the energy of the control mass decreases by 2900 kJ, the decrease in mass is
3.23 X 10~ 15 kg.
A change in mass of this magnitude cannot be detected by even our most accurate
chemical balance. Certainly, a fractional change in mass of this magnitude is beyond the
accuracy required in essentially all engineering calculations. Therefore, if we use the laws
of conservation of mass and conservation of energy as separate laws, we will not intro-
duce significant error into most thermodynamic problems and our definition of a control
mass as having a fixed mass can be used even though the energy changes.
UMMARY Conservation of energy is expressed for a cycle, and changes of total energy are then writ-
ten for a control mass. Kinetic and potential energy can be changed through the work of a
force acting on the control mass, and they are part of the total energy.
The internal energy and the enthalpy are introduced as substance properties with the
specific heats (heat capacity) as derivatives of these with temperature. Property variations
for limited cases are presented for incompressible states of a substance such as liquids and
solids, and for a highly compressible state as an ideal gas. The specific heat for solids and
liquids changes little with temperature, whereas the specific heat for a gas can change
substantially with temperature.
The energy equation is also shown in a rate form to cover transient processes.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Recognize the components of total energy stored in a control mass
• Write the energy equation for a single uniform control mass
• Find the properties u and h for a given state in the Appendix B tables
• Locate a state in the tables with an entry such as (P, h)
• Find changes in u and h for liquid or solid states using Tables A. 3 and A.4 or F.2
and F.3
• Find changes in u and h for ideal-gas states using Table A.5 or F.4
• Find changes in u and h for ideal-gas states using Tables A.7 and A. 8 or F. 5 and
F.6
• Recognize that forms for C p in Table A.6 are approximations to what is shown in
Fig. 5.11 and the more accurate tabulations in Tables A.7, A,8 3 F.5, and F.6
• Formulate the conservation of mass and energy for a control mass that goes through
a process involving work and heat transfers and different states
• Formulate the conservation of mass and energy for a more complex control mass
where there are different masses with different states
Concept-Study guide problems ffl 145
• Use the energy equation in a rate form
• Know the difference between the general laws as the conservation of mass (continu-
ity equation), conservation of energy (first law) and the specific laws that describes
a device behavior or process
Hey Concepts
and formulas
Total energy
Kinetic energy
Potential energy
Specific energy
Enthalpy
Two-phase mass average
Specific heat, heat capacity
Solids and liquids
Ideal gas
Energy equation rate form
Energy equation integrated
Multiple masses, states
E= U+KE + KE = mil + ^ mV 2 + mgZ
KE = | mV 2
KE = mgZ
h^u + Pv
U — Uj + XUjg — (1 — x)Uf+XU g
h = hf+ xh fg = (1 — x)hf+ xh g
Incompressible, sou = constant = tyand v very small
C=C v =C p [Tables A.3 and A.4 (F.2 and F.3)]
k 2 - »i = C(T 2 - T x )
h 2 ~ hi = u 2 — «i + v(P 2 — Pi) (Often the second term
is small.)
h = hf+ Vf{P - P s n); u = iif (saturated at same T)
h = u+Pv-u + RT (only functions of T)
C v \ gT )',c p
dT' p
dT
= C a + R
w 2 -«i = \c v dT=C,{Tj-T{)
h 2 -h x = \c p dT^C p {T 2 ~T x )
Left-hand side from Table A.7 or A. 8, middle from Table
A.6 and right-hand side from Table A ,6 at a T avg or from
Table A.5 at25°C
Left-hand side from Table F.5 or F.6, right-hand side from
Table F.4 at 77 F
E = Q- W (rate = + in - out)
E 2 - #i = \Q 2 " \W 2 (change = + in - out)
1
ci) = m(«2 " «i) + \ >«(Vi - V?) + mg{Z 2 ~ Z,)
m{e 2
E - m A e A + m B e B + m c e c + • •
Loncept-Study Guide problems
5.1 What is 1 cal in SI units and what is the name
given to 1 N m?
5.2 In a complete cycle what is the net change in en-
ergy and in volume?
5.3 Why do we write Aii or E 2 - E % whereas we write
i0 2 and
5.4 When you wind a spring up in a toy or stretch a
rubber band, what happens in terms of work, en-
146 ^ Chapter Five The Hrst Law of thermodynamics
ergy, and heat transfer? Later, when they are re-
leased, what happens then?
5.5 Explain in words what happens with the energy
terms for the stone in Example 5.2. What would
happen if it were a bouncing ball falling to a hard
surface?
5.6 Make a list of at least five systems that store en-
ergy, explaining which form of energy.
5.7 A 1200-kg car is accelerated from 30 to 50 km/h in
5 s. How much work is that? If you continue from
50 to 70 km/h in 5 s, is that the same?
5.8 A crane used 2 kW to raise a 100-kg box 20 m.
How much time did it take?
5.9 Saturated water vapor has a maximum for u and h
at around 235°C. Is this similar to other substances?
5.10 A pot of water is boiiing on a stove supplying 325
W to the water. What is the rate of mass (kg/s) va-
porizing, assuming a constant pressure process?
5.11 A constant mass goes through a process whereby
100 W of heat transfer comes in and 100 W of
work leaves. Does the mass change state?
5.12 I have 2 kg of liquid water at 20°C, 100 kPa. I now
add 20 kJ of energy at a constant pressure. How hot
does the water get if it is heated? How fast does it
move if it is pushed by a constant horizontal force?
How high does it go if it is raised straight up?
5.13 Water is heated from 100 kPa, 20°C to 1000 kPa,
200°C. In one case pressure is raised at T = C, then
T is raised at P = C. In a second case the opposite
order is followed. Does that make a difference for
i&and.JF;?
5.14 Two kilograms of water at 120°C with a quality of
25% has its temperature raised 20°C in a constant-
volume process. What are the new quality and spe-
cific internal energy?
5.15 Two kilograms of water at 200 kPa with a quality
25% has its temperature raised 20°C in a constant-
pressure process. What is the change in enthalpy?
5.16 You heat a gas 10 K at P = C. Which one in Table
A. 5 requires the most energy? Why?
5.17 Air is heated from 300 to 350 K at V = C. Find x q % .
What if the air is heated from 1300 to 1350 K?
5.18 A mass of 3 kg of nitrogen gas at 2000 K, V = C,
cools with 500 W. What is dTldfl
5.19 A drag force on a car, with frontal area i=2m 2 ,
driving at 80 km/h in air at 20°C, is F d = 0.225
Apz k V 2 . How much power is needed and what is
the traction force?
Homework problems
Kinetic and Potential Energy
5.20 A hydraulic hoist raises a 1750-kg car 1.8 m in an
auto repair shop. The hydraulic pump has a con-
stant pressure of 800 kPa on its piston. What is the
increase in potential energy of the car and how
much volume should the pump displace to deliver
that amount of work?
5.21 A piston motion moves a 25-kg hammerhead verti-
cally down 1 m from rest to a velocity of 50 m/s in
a stamping machine. What is the change in total
energy of the hammerhead?
5.22 Airplane takeoff from an aircraft carrier is assisted
by a steam-driven piston/cylinder with an average
pressure of 1250 kPa. A 17 500-kg airplane should
be accelerated from zero to a speed of 30 m/s with
30% of the energy coming from the steam piston.
Find the needed piston displacement volume.
5.23 Solve Problem 5.22, but assume the steam pressure
in the cylinder starts at 1000 kPa, dropping linearly
with volume to reach 100 kPa at the end of the
process.
5.24 A 1200-kg car accelerates from zero to 100 km/h
over a distance of 400 m. The road at the end of the
400 m is at 10 m higher elevation. What is the total
increase in the car kinetic and potential energy?
5.25 A 25 kg piston is above a gas in a long vertical
cylinder. Now the piston is released from rest and
accelerates up in the cylinder reaching the end 5 m
higher at a velocity of 25 m/s. The gas pressure
drops during the process, so the average is 600 kPa
with an outside atmosphere at 100 kPa. Neglect the
change in gas kinetic and potential energy, and find
the needed change in the gas volume.
5.26 The rolling resistance of a car depends on its
weight as F= 0.006 nig. How far will a car of 1200
kg roll if the gear is put in neutral when it drives at
90 km/h on a level road without air resistance?
5.27 A mass of 5 kg is tied to an elastic cord 5 m long
and dropped from a tall bridge. Assume the cord,
homework problems B 147
once straight, acts as a spring with k = 100 N/m.
Find the velocity of the mass when the cord is
straight (5 m down). At what level does the mass
come to rest after bouncing up and down?
Properties («, h) from General Tables
5.28 Find the missing properties.
a. HA T = 250°C, /> = ?« = ?
v = 0.02 mVkg,
b. N 2 , T= 120 K, x = ?h = 1
P = 0.8 MPa,
c. H 2 0, T=-2°C i u=?u=?
P - 100 kPa,
d. R-134a, P = 200 kPa, U = 7T=?
v - 0.12m 3 /kg,
5.29 Find the missing properties of T, P, v, u, h, and x if
applicable and plot the location of the three states
as points in the T-v and the P-v diagrams.
a. Water at 5000 kPa, u = 800 kJ/kg
b. Water at 5000 kPa, v = 0.06 m 3 /kg
c. R-134aat35°C, u = 0.01 mVkg
5.30 Find the missing properties and give the phase of
the substance.
a. NH 3 , T = 65°C, « = ? u = ?
P = 600 kPa,
b. NH 3 , T = 20°C, u = ? v = ?
P= 100 kPa, x = ?
c. Ammonia, r = 50°C } i* = ? « = ?
y = 0.1I85mVkg, * = ?
5.31 i. Find the phase and missing properties of P, T,
v, u, and x.
a. H 2 at P = 5000 kPa, w = 1000 kJ/kg
(steam table reference)
b. R-134a at T = 20°C, k = 300 kJ/kg
c. N 2 at250K,P = 200 kPa
ii. Show the three states as labeled dots in a T-v
diagram with correct position relative to the
two-phase region.
5.32 Find the missing properties and give the phase of
the substance.
a. H 2 0, T = 120°C, u = ? P = ?x = ?
v = 0.5 m 3 /kg,
b. F£ 2 0, T = 100°C, u = ?x=?u=?
P= 10 MPa,
c. N 2 , r= 200K, u = ?» = ?
P - 200 kPa,
d. nh 3 , r=ioo°c, p = ?.x = ?
v = 0.1 nrVkg,
e. N 2 , 7 1 = 100 K, u = ? K = ?
.r - 0.75,
5.33 Find the missing properties among T, P, f, u, h,
and (if applicable), give the phase of the sub-
stance, and indicate the states relative' to the two-
phase region in both a T-v and a P-v diagram.
a. R-12, P = 500 kPa, h = 230 kJ/kg
b. R-22, T = 10°C, u = 200 kJ/kg
c. R-134a, T = 40°C, h = 400 kJ/kg
5.34 Saturated liquid water at 20°C is compressed to a
higher pressure with constant temperature. Find the
changes in u and h from the initial state when the
final pressure is
a. 500 kPa
b. 2000 kPa
c. 20 000kPa
Energy Equation: Simple Process
5.35 A 100-L rigid tank contains nitrogen (N 2 ) at 900 K
and 3 MPa. The tank is now cooled to 100 K. What
are the work and heat transfer for the process? '
5.36 A rigid container has 0.75 kg of water at 300°C,
1200 kPa. The water is now cooled to a final pres-
sure of 300 kPa. Find the final temperature, the
work, and the heat transfer in the process.
5.37 A cylinder fitted with a fricfionless piston contains
2 kg of superheated refrigerant R-134a vapor at
350 kPa, 100°C. The cylinder is now cooled so that
the R-134a remains at constant pressure until it
reaches a quality of 75%. Calculate the heat trans-
fer in the process.
5.38 Ammonia at 0°C with a quality of 60% is con-
tained in a rigid 200-L tank. The tank and ammonia
are now heated to a final pressure of 1 MPa. Deter-
mine the heat transfer for the process.
5.39 Water in a 150-L closed, rigid tank is at 100°C and
90% quality. The tank is then cooled to -10°C.
Calculate the heat transfer for the process.
5.40 A piston/cyltnder contains 1 kg of water at 20°C
with volume 0.1 m 3 . By mistake someone locks
the piston, preventing it from moving while we
heat the water to saturated vapor. Find the final
148 m CHAPTER Five the first law of thermodynamics
temperature and the amount of heat transfer in
the process.
5.41 A test cylinder with constant volume of 0. 1 L con-
tains water at the critical point. It now cools down
to room temperature of 20°C. Calculate the heat
transfer from the water.
5.42 A 10-L rigid tank contains R-22 at -10°C with a
quality of 80%. A 10-A electric current (from a 6-
V battery) is passed through a resistor inside the
tank for 10 min, after which the R-22 temperature
is 40°C. What was the heat transfer to or from the
tank during this process?
5.43 A piston/cylinder device contains 50 kg of water at
200 kPa with a volume of 0.1 m 3 . Stops in the
cylinder are placed to restrict the enclosed volume
to a maximum of 0.5 m 3 . The water is now heated
until the piston reaches the stops. Find the neces-
sary heat transfer.
5.44 A constant-pressure piston/cylinder assembly con-
tains 0.2 kg of water as saturated vapor at 400 kPa. It
is now cooled so that the water occupies half the
original volume. Find the heat transfer hi the process.
5.45 Two kilograms of water at 120°C with a quality of
25% has its temperature raised 20°C in a constant-
volume process as in Fig. P5.45. What are the heat
transfer and work in the process?
5.47 An insulated cylinder fitted with a piston contains
R-12 at 25°C with a quality of 90% and a volume
of 45 L. The piston is allowed to move, and the R-
12 expands until it exists as saturated vapor. Dur-
ing this process the R-12 does 7.0 kJ of work
against the piston. Determine the final temperature,
assuming the process is adiabatic.
5.48 A water-filled reactor with volume of 1 m 3 is at 20
MPa and 360°C and placed inside a containment
room, as shown in Fig. P5.48. The room is well in-
sulated and initially evacuated. Due to a failure, the
reactor ruptures and the water fills the containment
,fl fl ft,
=i - — 1
FIGURE P5.45
5.46 A 25-kg mass moves at 25 m/s. Now a brake sys-
tem brings the mass to a complete stop with a con-
stant deceleration over a period of 5 s. The brake
energy is absorbed by 0.5 kg of water initially at
20°C and 100 kPa. Assume the mass is at constant
P and T. Find the energy the brake removes from
the mass and the temperature increase of the water,
assuming its pressure is constant.
% FIGURE P5.48
room. Find the minimum room volume so that the
final pressure does not exceed 200 kPa.
5.49 A piston/cylinder arrangement contains water of
quality x = 0.7 in the initial volume of 0. 1 m 3 ,
where the piston applies a constant pressure of 200
kPa. The system is now heated to a final tempera-
ture of 200°C. Determine the work and the heat
transfer in the process.
A piston/cylinder arrangement has the piston
loaded with outside atmospheric pressure and the
piston mass to a pressure of 150 kPa, as shown in
Fig. P5.50. It contains water at -2°C, which is
then heated until the water becomes saturated
vapor. Find the final temperature and specific work
and heat transfer for the process.
5.50
Po
_
1
H 2 6 . .."
FIGURE P5.50
5.51 A piston/cylinder assembly contains 1 kg of liquid
water at 20°C and 300 kPa. There is a linear spring
mounted on the piston such that when the water is
heated the pressure reaches 1 MPa with a volume
Homework Problems M 149
of 0.1 m 3 . Find the final temperature and the heat
transfer in the process.
5.52 A closed steel bottle contains ammonia at -20°C,
x = 20% and the volume is 0.05 m 3 . It has a safety-
valve that opens at a pressure of 1.4 MPa. By acci-
dent, the bottle is heated until the safety valve
opens. Find the temperature and heat transfer when
the valve first opens.
FIGURE P5.52
FIGURE P5.57
5.58 A rigid tank is divided into two rooms, both con-
taining water, by a membrane, as shown in Fig.
P5.58. Room^ is at 200 kPa, v = 0.5 m 3 /kg, V A =
1 m 3 , and room B contains 3.5 kg at 0.5 MPa,
400°C. The membrane now ruptures and heat
transfer takes place so the water comes to a uni-
form state at 100°C. Find the heat transfer during
the process.
5.53 Two kilograms of water at 200 kPa with a quality
of 25% has its temperature raised 20°C in a con-
stant-pressure process. What are the heat transfer
and work in the process?
5.54 Two kilograms of nitrogen at 100 K, x = 0.5 are
heated in a constant pressure process to 300 K in a
piston/cylinder arrangement. Find the initial and
final volumes and the total heat transfer required.
5.55 A 1-L capsule of water at 700 kPa and 150°C is
placed in a larger insulated and otherwise evacu-
ated vessel. The capsule breaks and its contents fill
the entire volume. If the final pressure should not
exceed 125 kPa, what should the vessel volume be?
5.56 Superheated refrigerant R-134a at 20°C and 0.5
MPa is cooled in a piston/cylinder arrangement at
constant temperature to a final two-phase state with
quality of 50%. The refrigerant mass is 5 kg, and
during this process 500 kJ of heat is removed. Find
the initial and final volumes and the necessary
work.
5.57 A cylinder having a piston restrained by a linear
spring (of spring constant 15 kN/m) contains 0,5
kg of saturated vapor water at 120°C, as shown in
Fig. P5.57. Heat is transferred to the water, causing
the piston to rise. If the piston cross-sectional area
is 0.05 m 2 and the pressure varies linearly with vol-
ume until a final pressure of 500 kPa is reached,
find the final temperature in the cylinder and the
heat transfer for the process.
A
FIGURE P5.58
5.59 A 10-m-high open cylinder, with A cyl = 0.1 m 2 ,
contains 20°C water above and 2 kg of 20°C water
below a 198.5-kg thin insulated floating piston, as
shown in Fig. P5.59. Assume standard g t P . Now
heat is added to the water below the piston so that
it expands, pushing the piston up, causing the
water on top to spill over the edge. This process
continues until the piston reaches the top of the
cylinder. Find the final state of the water below
the piston (7*, P, v) and the heat added during the
process.
HoO
FIGURE P5.59
150 M Chapter Five the First Law of Thermodynamics
5.60 Assume the same setup as in Problem 5.48, but the
room has a volume of 100 m 3 . Show that the final
state is two phase and find the final pressure by
trial and error.
Energy Equation: Multistep Solution
5.61 Ten kilograms of water in a piston/cylinder
arrangement exist as saturated liquid/vapor at 100
kPa, with a quality of 50%. The system is now
heated so that the volume triples. The mass of the
piston is such that a cylinder pressure of 200 kPa
will float it, as in Fig. P4.68. Find the final temper-
ature and the heat transfer in the process.
5.62 Two tanks, each with a volume of 1 m 3 , are con-
nected by a valve and line, as shown in Fig. P5.62.
Tank A is filled with R- 134a at 20°C with a quality
of 15%. Tank B is evacuated. The valve is opened
and saturated vapor flows from A into B until the
pressures become equal. The process occurs slowly
enough that all temperatures stay at 20°C during
the process. Find the total heat transfer to the
R-134a during the process.
R-22
FIGURE P5.64
5.65 Find the heat transfer in Problem 4.67.
5.66 Refrigerant- 12 is contained in a piston/cylinder
arrangement at 2 MPa and 150°C with a massless
piston against the stops, at which point V = 0.5 m 3 .
The side above the piston is connected by an open
valve to an air line at 10°C and 450 kPa, as shown
in Fig. P5.66. The whole setup now cools to the
surrounding temperature of 10°C. Find the heat
transfer and show the process in a P-v diagram.
T— " , , -i
Air line
FIGURE PS.62
5.63 Consider the same system as in the previous prob-
lem. Let the valve be opened and transfer enough
heat to both tanks so that all the liquid disappears.
Find the necessary heat transfer.
5.64 A vertical cylinder fitted with a piston contains 5
kg of R-22 at 10°C, as shown in Fig. P5.64. Heat is
transferred to the system, causing the piston to rise
until it reaches a set of stops, at which point the
volume has doubled. Additional heat is transferred
until the temperature inside reaches 50°C, at which
point the pressure inside the cylinder is 1.3 MPa.
a. What is the quality at the initial state?
b. Calculate the heat transfer for the overall process.
R-12
FIGURE P5.66
5.67 Find the heat transfer in Problem 4. 1 14.
5.68 A rigid container has two rooms filled with water,
each of 1 m 3 , separated by a wall (see Fig. P5.58).
Room A has P - 200 kPa with a quality of x =
0.80. Room B has P = 2 MPa and T = 400°C. The
partition wall is removed, and because of heat
transfer the water comes to a uniform state with a
temperature of 200°C Find the final pressure and
the heat transfer in the process.
5.69 The cylinder volume below the constant loaded
piston has two compartments A and B filled with
water, as shown in Fig. P5.69. A has 0.5 kg at 200
kPa and 150°C and B has 400 kPa with a quality of
50% and a volume of 0.1 m 3 . The valve is opened
and heat is transferred so that the water comes to a
homework problems H 151
uniform state with a total volume of 1 .006 m 3 . Find
the total mass of water and the total initial volume.
Find the work and the heat transfer in the process.
A
FIGURE P5.69
5.70 A rigid tank A of volume 0.6 m 3 contains 3 kg of
water at 120°C, and rigid tank B is 0.4 m 3 with
water at 600 kPa, 200°C. They are connected to a
piston/cylinder initially empty with closed valves
as shown in Fig. P5.70. The pressure in the cylin-
der should be 800 kPa to float the piston. Now the
valves are slowly opened and heat is transferred so
the water reaches a uniform state at 250°C with the
valves open. Find the final volume and pressure,
and the work and heat transfer in the process.
V////////////////////////A
' /////////////////////////A
W////////////////////M
FIGURE P5.70
5.71 Calculate the heat transfer for the process de-
scribed in problem 4.60.
5.72 Calculate the heat transfer for the process de-
scribed in Problem 4.70.
5.73 A cylinder/piston arrangement contains 5 kg of
water at 100°C with x = 20% and the piston, of
m p = 75 kg, resting on some stops, similar to Fig.
P5.73. The outside pressure is 100 kPa, and the
cylinder area is A ~ 24.5 cm 2 . Heat is now added
until the water reaches a saturated vapor state. Find
the initial volume, final pressure, work, and heat
transfer terms and show the P-v diagram.
FIGURE P5.73
Energy Equation: Solids and Liquids
5.74 Because a hot water supply must also heat some
pipe mass as it is turned on, the water does not
come out hot right away. Assume 80°C liquid
water at 1 00 kPa is cooled to 45°C as it heats 1 5 kg
of copper pipe from 20 to 45°C. How much mass
(kg) of water is needed?
5.75 A house is being designed to use a thick concrete
floor mass as thermal storage material for solar en-
ergy heating. The concrete is 30 cm thick and the
area exposed to the sun during the daytime is
4 m X 6 m. It is expected that this mass will un-
dergo an average temperature rise of about 3°C
during the day. How much energy will be available
for heating during the nighttime hours?
5.76 A copper block of volume 1 L is heat treated at
500°C and now cooled in a 200-L oil bath initially
at 20°C, as shown in Fig. P5.76. Assuming no heat
transfer with the surroundings, what is the final
temperature?
Copper
y/, FIGURE P5.76
5.77 A 1-kg steel pot contains 1 kg of liquid water, both
at 15°C. It is now put on the stove, where it is
heated to the boiling point of the water. Neglect
152 m Chapter five the First law of thermodynamics
any air being heated and find the total amount of
energy needed,
5.78 A car with mass 1275 kg is driven at 60 km/h
when the brakes are applied quickly to decrease its
speed to 20 km/h. Assume that the brake pads have
a 0.5-kg mass with a heat capacity of 1.1 kJ/kg K
and that the brake disks/drums are 4.0 kg of steel.
Further assume that both masses are heated uni-
formly. Find the temperature increase in the brake
assembly.
5.79 Saturated, x = 1%, water at 25°C is contained in a
hollow spherical aluminum vessel with inside di-
ameter of 0,5 m and a 1 -cm- thick wall. The vessel
is heated until the water inside becomes saturated
vapor. Considering the vessel and water together as
a control mass, calculate the heat transfer for the
process.
5.80 A 25-kg steel tank initially at -10°C is filled up
with 100 kg of milk (assumed to have the same
properties as water) at 30X. The milk and the steel
come to a uniform temperature of +5°C in a stor-
age room. How much heat transfer is needed for
this process?
5.81 An engine, shown in Fig. P5.81, consists of a 100-
kg cast iron block with a 20-kg aluminum head, 20
kg of steel parts, 5 kg of engine oil, and 6 kg of
glycerine (antifreeze). Everything begins at 5"C,
and as the engine starts we want to know how hot
it becomes if it absorbs a net of 7000 kJ before it
reaches a steady uniform temperature.
Automobile engine FIGURE P5.81
Properties («, /i } C Uj and C p ), Ideal Gas
5,82 Use the ideal-gas air Table A.7 to evaluate the heat
capacity C p at 300 K as a slope of the curve h(T) by
A/t/AT 1 . How much larger is it at 1000 K and at
1500 K?
5.83 We want to find the change in u for carbon dioxide
between 600 K and 1200 K.
a. Find it from a constant C vQ from Table A.5.
b. Find it from a evaluated from the equation
in Table A.6 at the average T,
c. Find it from the values of u listed in Table A. 8.
5.84 Do Problem 5.83 for oxygen gas.
5.85 Water at 20°C and 100 kPa is brought to 100 kPa
and 1500°C. Find the change in the specific internal
energy, using the water tables and ideal gas tables.
5.86 We want to find the increase in temperature of ni-
trogen gas at 1200 K when the specific internal en-
ergy is increased with 40 kJ/kg.
a. Find it from a constant C v0 from Table A.5.
b. Find it from a evaluated from the equation
in Table A.6 at 1200 K.
c. Find it from the values of u listed in Table A,8.
5.87 For a particular application the change in enthalpy
of carbon dioxide from 30 to I SOOT at 100 kPa is
needed. Consider the following methods and indi-
cate the most accurate one.
a. Using a constant specific heat and reading the
value from Table A.5.
b. Using a constant specific heat and obtaining the
value at average temperature from the equation,
in Table A.6,
c. Using a variable specific heat and integrating
the equation in Table A.6.
d. Reading the enthalpy from ideal gas tables in
Table A.8.
5.88 An ideal gas is heated from 500 to 1500 K. Find
the change in enthalpy using constant specific heat
from Table A.5 (room temperature value) and dis-
cuss the accuracy of the result if the gas is
a. Argon
b. Oxygen
c. Carbon dioxide
Energy Equation: Ideal Gas
5.89 A 250-L rigid tank contains methane' at 500 K,
1500 kPa. It is now cooled down to 300 K. Find
the mass of methane and the heat transfer using
(a) the ideal-gas and (b) the methane tables.
HOMEWORK PROBLEMS M 153
5.90 A rigid insulated tank is separated into two
rooms by a stiff plate. Room A of 0.5 m 3 con-
tains air at 250 kPa and 300 K and room B of 1
m 3 has air at 150 kPa and 1000 K. The plate is
removed and the air comes to a uniform state
without any heat transfer. Find the final pres-
sure and temperature.
5.91 A rigid container has 2 kg of carbon dioxide gas
at 100 kPa and 1200 K that is heated to 1400 K.
Solve for the heat transfer using (a) the heat ca-
pacity from Table A.5 and (b) properties from
Table A.8.
5.92 Do the previous problem for nitrogen, N 2) gas.
5.93 A 10-m-high cylinder, with a cross-sectional area
of 0.1 m 2 , has a massless piston at the bottom
with water at 20°C on top of it, as shown in Fig.
P5.93. Air at 300 K, with a volume of 0.3 m 3 ,
under the piston is heated so that the piston moves
up, spilling the water out over the side. Find the
total heat transfer to the air when all the water has
been pushed out.
Air
FIGURE P5.93
5.94 Find the heat transfer in Problem 4.43.
5.95 An insulated cylinder is divided into two parts of
1 m 3 each by an initially locked piston, as shown
in Fig. P5.95. Side A has air at 200 kPa, 300 K,
and side B has air at 1.0 MPa, 1000 K. The piston
is now unlocked so that it is free to move, and it
A
Air
B
Air
conducts heat so that the air comes to a uniform
temperature T A = T B . Find the mass in both A and
5 and the final 7 1 and P.
5.96 A piston/cylinder contains air at 600 kPa, 290 K
and a volume of 0.01 m 3 . A constant-pressure
process gives 54 kJ of work out. Find the final
temperature of the air and the heat transfer
input.
5.97 A cylinder with a piston restrained by a linear
spring contains 2 kg of carbon dioxide at 500 kPa
and 400°C. It is cooled to 40°C, at which point the
pressure is 300 kPa. Calculate the heat transfer for
the process.
5.98 Water at 100 kPa and 400 K is heated electrically
adding 700 kJ/kg in a constant pressure process.
Find the final temperature using
a. The water Table B.l
b. The ideal-gas Table A.8
c. Constant specific heat from Table A.5
5.99 A piston/cylinder has 0.5 kg of air at 2000 kPa,
1000 K as shown. The cylinder has stops, so
Pmin = 0.03 m 3 . The air now cools to 400 K by
heat transfer to the ambient. Find the final volume
and pressure of the air (does it hit the stops?) and
the work and heat transfer in the process.
II I Mil
I I I I
FIGURE P5.99
FIGURE P5.95
5.100 A spring-loaded piston/cylinder contains 1,5 kg
of air at 27°C and 1 60 kPa. It is now heated to 900
K in a process where the pressure is linear in vol-
ume to a final volume of twice the initial volume.
Plot the process in a P-u diagram and find the
work and heat transfer.
5.101 Air in a piston/cylinder assembly at 200 kPa and
600 K is expanded in a constant-pressure pro-
cess to twice the initial volume, state 2, as
shown in Fig. P5.101. The piston is then locked
with a pin and heat is transferred to a final tem-
perature of 600 K. Find P, T, and h for states 2
154 M Chapter Five the First Law of Thermodynamics
and 3, and find the work and heat transfer in
both processes.
Air
FIGURE P5.101
5.102 A vertical piston/cylinder setup has a linear spring
mounted on the piston so that at zero cylinder vol-
ume a balancing pressure inside is zero. The
cylinder contains 0.25 kg of air at 500 kPa and
2TC. Heat is now added so that the volume dou-
bles.
a. Show the process path in a P—V diagram.
b. Find the final pressure and temperature.
c. Find the work and heat transfer.
Energy Equation: Polytropic Process
5.103 A piston/cylinder device contains 0. 1 kg of air at
300 K. and 100 kPa. The air is now slowly com-
pressed in an isothermal (T = constant) process to
a final pressure of 250 kPa. Show the process in a
P-V diagram and find both the work and heat
transfer in the process.
5.104 Oxygen at 300 kPa and 100°C is in a piston/cylin-
der arrangement with a volume of 0.1 m\ It is
now compressed in a polytropic process with ex-
ponent n = 1.2 to a final temperature of 200°C.
Calculate the heat transfer for the process.
5.105 A piston/cylinder setup contains 0.001 m 3 air at
300 K and 150 kPa. The air is now compressed in
a process in which PV l 2S = C to a final pressure
of 600 kPa. Find the work performed by the air
and the heat transfer.
5.106 Helium gas expands from 125 kPa, 350 K and
0.25 m 3 to 100 kPa in a polytropic process with
n = 1,667. How much heat transfer is involved?
5.107 A piston/cylinder assembly in a car contains 0.2 L
of air at 90 kPa and 20°C, as shown in Fig. P5. 107.
The air is compressed in a quasi-equilibrium poly-
tropic process with polytropic exponent n - 1.25
to a final volume six times smaller. Determine the
final pressure and temperature, and the heat trans-
fer for the process.
] FIGURE P5.107
5.108 A piston/cylinder has nitrogen gas at 750 K. and
1500 kPa shown in Fig. P5.108. Now it is ex-
panded in a polytropic process with n — 1.2 to
P = 750 kPa. Find the final temperature, the spe-
cific work and specific heat transfer in the
process.
FIGURE P5.108
5.109 A piston/cylinder arrangement of initial volume
0.025 m 3 contains saturated water vapor at 180°C.
The steam now expands in a polytropic process
with exponent n = 1 to a final pressure of 200
kPa while it does work against the piston. Deter-
mine the heat transfer for this process.
5.110 Air is expanded from 400 kPa and 600 K in a
polytropic process to 150 kPa and 400 K in
a piston/cy Under arrangement. Find the poly-
tropic exponent n and the work and heat transfer
per kg of air using constant heat capacity from
Table A.5.
5.111 A piston/cylinder assembly has 1 kg of propane
gas at 700 kPa and 40°C. The piston cross-
sectional area is 0.5 m 2 , and the total external
force restraining the piston is directly propor-
tional to the cylinder volume squared. Heat is
transferred to the propane until its temperature
reaches 700°C. Determine the final pressure in-
side the cylinder, the work done by the propane,
and the heat transfer during the process.
[
homework Problems m 155
5.112 An air pistol contains compressed air in a small
cylinder, as shown in Fig. P5.112. Assume that
the volume is 1 cm 3 , the pressure is 1 MPa, and
the temperature is 27°C when armed. A bullet,
with m = 1 5 g, acts as a piston initially held by a
pin (trigger); when released, the air expands in an
isothermal process (T = constant). If the air pres-
sure is 0.1 MPa in the cylinder as the bullet leaves
the gun, find
a. the final volume and the mass of air
b. the work done by the air and work done on the
atmosphere
c. the work done to the bullet and the bullet exit
velocity
FIGURE P5.112
5.113 A spherical balloon contains 2 kg of R-22 at 0°C
with a quality of 30%. This system is heated until
the pressure in the balloon reaches 600 kPa. For
this process, it can be assumed that the pressure in
the balloon is directly proportional to the balloon
diameter. How does pressure vary with volume
and what is the heat transfer for the process?
5.114 Calculate the heat transfer for the process in Prob-
lem 4.55.
5.115 A piston/cylinder setup contains argon gas at 140
kPa and lO'C, and the volume is 100 L. The gas
is compressed in a polytropic process to 700 kPa
and 280°C. Calculate the polytropic exponent and
the heat transfer during the process.
Energy Equation in Rate Form
5.116 A crane lifts a load of 450 kg vertically upward
with a power input of 1 kW. How fast can the
crane lift the load?
5.117 A computer in a closed room of volume 200 m 3
dissipates energy at a rate of 10 kW. The room
has 50 kg of wood, 25 kg of steel, and air, with all
material at 300 K and 100 kPa. Assuming all the
mass heats up uniformly, how long will it take to
increase the temperature 10°C?
5.118 The rate of heat transfer to the surroundings from
a person at rest is about 400 kJ/h. Suppose that
the ventilation system fails in an auditorium con-
taining 100 people. Assume the energy goes into
the air of volume 1500 m 3 initially at 300 K and
101 kPa. Find the rate (degrees per minute) of the
air temperature change.
5.119 A piston/cylinder of cross-sectional area 0.01 m 2
maintains constant pressure. It contains 1 kg of
water with a quality of 5% at 150°C. If we heat so
that 1 g/s liquid rums into vapor, what is the rate
of heat transfer needed?
5.120 The heaters in a spacecraft suddenly fail. Heat is
lost by radiation at the rate of 100 kJ/h, and the
electric instruments generate 75 kJ/h. Initially, the
air is at 100 kPa and 25°C with a volume of
10m 3 . How long will it take to reach an air tem-
perature of — 20°C?
5.121 A steam-generating unit heats saturated liquid
water at constant pressure of 200 kPa in a pis-
ton/cylinder device. If 1.5 kW of power is added
by heat transfer, find the rate (kg/s) at which satu-
rated vapor is made.
5.122 A small elevator is being designed for a construc-
tion site. It is expected to carry four 75-kg work-
ers to the top of a 100-m-taIl building in less than
2 min. The elevator cage will have a counter-
weight to balance its mass. What is the smallest
size (power) electric motor that can drive this
unit?
5.123 As fresh poured concrete hardens, the chemical
transformation releases energy at a rate of 2 W/kg.
Assume the center of a poured layer does not
have any heat loss and that it has an average heat
capacity of 0.9 kJ/kg K. Find the temperature rise
during 1 h of the hardening (curing) process.
5.124 A 100-W heater is used to melt 2 kg of solid ice at
- 10°C to liquid at +5°C at a constant pressure of
150 kPa.
a. Find the change in the total volume of the
water.
b. Find the energy the heater must provide to the
water.
c. Find the time the process will take assuming
uniform T in the water,
5.125 Water is in a piston/cylinder maintaining constant
P at 700 kPa, quality 90% with a volume of 0.1
156 @ Chapter five the First law of thermodynamics
m 3 , A heater is turned on, heating the water with
2.5 kW. How long does it take to vaporize all the
liquid?
Review Problems
5.126 Ten kilograms of water in a piston/cylinder setup
with constant pressure are at 450°C and occupy a
volume of 0.633 m 3 . The system is now cooled to
20°C. Show the P-v diagram and find the work
and heat transfer for the process.
5.127 Consider the system shown in Fig. P5.127. Tank
A has a volume of 100 L and contains saturated
vapor R-134a at 30°C. When the valve is cracked
open, R-134a flows slowly into cylinder B. The
piston requires a pressure of 200 kPa in cylinder
B to raise it. The process ends when the pressure
in tank A has fallen to 200 kPa. During this
process heat is exchanged with the surroundings
such that the R-134a always remains at 30°C. Cal-
culate the heat transfer for the process.
Tank
A
Cylinder
B
ft"
-Piston
<g>
Valve
FIGURE P5.127
5.128 Ammonia, NH 3) is contained in a sealed rigid tank
at 0°C, x = 50% and is then heated to 100°C. Find
the final state P 2i u 2 and the specific work and
heat transfer.
5.129 A piston/cylinder setup contains 1 kg of ammonia
at 20°C with a volume of 0. 1 m 3 , as shown in Fig.
P5.129. Initially the piston rests on some stops
with the top surface open to the atmosphere, P ,
so that a pressure of 1400 kPa is required to lift it.
To what temperature should the ammonia be
heated to lift the piston? If it is heated to saturated
vapor find the final temperature, volume, and heat
transfer, X Q 2 .
NH,
FIGURE P5.129
5,130 A piston held by a pin in an insulated cylinder,
shown in Fig. P5.130, contains 2 kg of water at
100°C, with a quality of 98%. The piston has a
mass of 102 kg, with cross-sectional area of 100
cm 2 , and the ambient pressure is 100 kPa. The pin
is released, which allows the piston to move. De-
termine the final state of the water, assuming the
process to be adiabatic.
Fir b
H-,0
wmm^mmm. figure ps.i30
5.131 A piston/cylinder arrangement has a linear spring
and the outside atmosphere acting on the piston
shown in Fig. P5.131. It contains water at 3 MPa
and 400°C with a volume of 0.1 m 3 . If the piston
is at the bottom, the spring exerts a force such that
a pressure of 200 kPa inside is required to balance
the forces. The system now cools until the pres-
sure reaches 1 MPa. Find the heat transfer for the
process.
FIGURE P5.131
5,132 Consider the piston/cylinder arrangement shown
in Fig. P5.I32. A frictionless piston is free to
Homework problems h 157
move between two sets of stops. When the piston
rests on the lower stops, the enclosed volume is
400 L. When the piston reaches the upper stops,
the volume is 600 L. The cylinder initially con-
tains water at 100 kPa, with 20% quality. It is
heated until the water eventually exists as satu-
rated vapor. The mass of the piston requires 300
kPa pressure to move it against the outside ambi-
ent pressure. Determine the final pressure in the
cylinder, the heat transfer, and the work for the
overall process.
H,0
FIGURE P5.132
5.133 A piston/cylinder setup, shown in Fig. P5.133,
contains R-12 at -30°C, x = 20%. The volume is
0.2 m 3 . It is known that K stop = 0.4 m 3 , and if the
piston sits at the bottom, the spring force balances
the other loads on the piston. The system is now
heated up to 20°C. Find the mass of the fluid and
show the P~v diagram. Find the work and heat
transfer.
FIGURE PS. 133
5,134 A piston/cylinder arrangement B is connected to
a 1-m 3 tank A by a line and valve, shown in
Fig. P5.134. Initially both contain water, with A at
100 kPa, saturated vapor and B at 400°C, 300 kPa,
1 m 3 . The valve is now opened, and the water in
both A and B comes to a uniform state.
a. Find the initial mass in^ and B.
b. If the process results in T 2 = 200°C, find the
heat transfer and the work.
FIGURE P5.134
5.135 A small flexible bag contains 0.1 kg of ammonia
at - 10°C and 300 kPa. The bag material is such
that the pressure inside varies linear with volume.
The bag is left in the sun with an incident radia-
tion of 75 W, losing energy with an average 25 W
to the ambient ground and air. After a while the
bag is heated to 30°C at which time the pressure is
1000 kPa. Find the work and heat transfer in the
process and the elapsed time.
5.136 Water at 150°C, 50% quality is contained in a
cylinder/piston arrangement with initial volume
0.05 m 3 . The loading of the piston is such that the
inside pressure is linear with the square root of
volume as P - 100 + CF° S kPa. Now heat is
transferred to the cylinder to a final pressure of
600 kPa. Find the heat transfer in the process.
5.137 A 1-m 3 tank containing air at 25°C and 500 kPa is
connected through a valve to another tank con-
taining 4 kg of air at 60°C and 200 kPa. Now the
valve is opened and the entire system reaches
thermal equilibrium with the surroundings at
20°C. Assume constant specific heat at 25°C and
determine the final pressure and the heat transfer.
FIGURE P5.137
5.138 A closed cylinder is divided into two rooms by a
frictionless piston held in place by a pin, as shown
158 H CHAPTERFIVE THE FIRST LAW OF THERMODYNAMICS
A
f
R
Air
FIGURE P5.138
in Fig. P5.138. Room A has 10 L of air at 100
kPa, 30°C, and room B has 300 L of saturated
water vapor at 30°C. The pin is pulled, releasing
the piston, and both rooms come to equilibrium at
30°C, and as the water is compressed it becomes
two-phase. Considering a control mass of the air
and water, determine the work done by the system
and the heat transfer to the cylinder.
English Unit Problems
English Unit Concept Problems
5.139E What is 1 cal in English units? What is 1 Btu in
ftlbf?
5.140E Work as F Ax has units of lbf ft. What is that in
Btu?
5.141E A 2500-lbm car is accelerated from 25 mi/h to
40 mi/h. How much work is that?
5.142E A crane uses 7000 Btu/h to raise a 200-lbm box
60 ft. How much time does it take?
5.143E I have 4 Ibm of liquid water at 70 F, 15 psia. I
now add 20 Btu of energy at a constant pressure.
How hot does it get if it is heated? How fast
does it move if it is pushed by a constant hori-
zontal force? How high does it go if it is raised
straight up?
5.144E Air is heated from 540 R to 640 R at V - C.
Find x q % . What if the air is heated from 2400 to
2500 R?
English Unit Problems
5.145E Airplane takeoff from an aircraft carrier is as-
sisted by a steam-driven piston/cylinder with an
average pressure of 200 psia. A 38 500-lbm air-
plane should be accelerated from zero to a speed
of 100 ft/s with 30% of the energy coming from
the steam piston. Find the needed piston dis-
placement volume.
5.146E A hydraulic hoist raises a 3650-lbm car 6 ft in
an auto repair shop. The hydraulic pump has a
constant pressure of 100 lbf/in 2 on its piston.
What is the increase in potential energy of the
car and how much volume should the pump dis-
place to deliver that amount of work?
5.147E A piston motion moves a 50-lbm hammerhead
vertically down 3 ft from rest to a velocity of
150 ft/s in a stamping machine. What is the
change in total energy of the hammerhead?
5.148E Find the missing properties and give the phase
of the substance.
a. HA
u ~
1000 Btu/lbm,
h =
?y
r =
270 F,
x =
?
b. H 2 0,
u =
450 Btu/lbm,
r=
?x
p =
1500 lbf/in 2 ,
?
c. R-22,
T =
P =
30 F,
75 lbf/in 2 ,
h =
?x
5.149E Find the missing properties among (P, T, v, u, h)
together with x, if applicable, and give the phase
of the substance.
a. R-22, r = 50 F, u = 85 Btu/lbm
b. HA T = 600 F, h- 1322 Btu/lbm
c. R-22, P = 150 lbf/in 2 , h = 115.5 Btu/lbm
5.150E Find the missing properties among (P, T, v, u, h)
together with x, if applicable, and give the phase
of the substance.
a. R-I34a, T = 140 F, h = 185 Btu/lbm
b. NH 3> T = 170 F, P = 60 lbf/in 2
c. R-134a, r = 100 F, w = 175 Btu/lbm
5.151E A cylinder fitted with a frictionless piston con-
tains 4 lbm of superheated refrigerant R-134a
vapor at 400 lbf/in 2 , 200 F. The cylinder is now
cooled so that the R-134a remains at constant
pressure until it reaches a quality of 75%. Cal-
culate the heat transfer in the process.
5.152E Ammonia at 30 F, quality 60% is contained in a
rigid 8-ft 3 tank. The tank and ammonia are now
heated to a final pressure of 150 lbf/in 2 . Deter-
mine the heat transfer for the process.
ENGLISH UNIT PROBLEMS H 159
5.153E Water in a 6-ft 3 closed, rigid tank is at 200 .F,
90% quality. The tank is then cooled to 20 F.
Calculate the heat transfer during the process,
5.154E A constant-pressure piston/cylinder has 2 lbm of
water at 1100 F and 2.26 ft 3 . It is now cooled to
occupy 1/10 of the original volume. Find the
heat transfer in the process.
5.155E A piston/cylinder arrangement has the piston
loaded with outside atmospheric pressure and
the piston mass to a pressure of 20 lbf/in 2 ,
shown in Fig. P5.50. It contains water at 25 F,
which is then heated until the water becomes
saturated vapor. Find the final temperature and
specific work and heat transfer for the process.
5.156E A water-filled reactor with volume of 50 ft 3 is at
2000 lbf/in 2 , 560 F and placed inside a contain-
ment room, as shown in Fig. P5.48. The room is
well insulated and initially evacuated. Due to a
failure, the reactor ruptures and the water fills
the containment room. Find the minimum room
volume so the final pressure does not exceed
30 lbf/in 2 .
5.157E A piston/cylinder contains 2 lbm of liquid water
at 70 F, and 30 lbf/in 2 . There is a linear spring
mounted on the piston such that when the
water is heated the pressure reaches 300 lbf/in 2
with a volume of 4 ft 3 . Find the final tempera-
ture and plot the P-v diagram for the process.
Calculate the work and the heat transfer for
the process.
5.158E A twenty-pound-mass of water in a piston/cylin-
der with constant pressure is at 1100 F and a
volume of 22.6 ft 3 . It is now cooled to 100 F.
Show the P-v diagram and find the work and
heat transfer for the process.
5.159E A vertical cylinder fitted with a piston contains
10 lbm of R-22 at 50 F, shown in Fig. P5.64*.
Heat is transferred to the system, causing the
piston to rise until it reaches a set of stops at
which point the volume has doubled. Additional
heat is transferred until the temperature inside
reaches 1 20 F, at which point the pressure inside
the cylinder is 200 lbf/in 2 .
a. What is the quality at the initial state?
b. Calculate the heat transfer for the overall
process.
5.160E A piston/cylinder contains 2 lbm of water at 70 F
with a volume of 0.1 ft 3 , shown in Fig. P5.129.
Initially the piston rests on some stops with the
top surface open to the atmosphere, Pq, so a pres-
sure of 40 lbf/in 2 is required to lift it. To what
temperature should the water be heated to lift the
piston? If it is heated to saturated vapor, find the
final temperature, volume, and the heat transfer.
5.161E Two tanks are connected by a valve and line, as
shown in Fig. P5.62. The volumes are both 35
ft 3 with R-134a at 70 F, quality 25% in A, and
tank B is evacuated. The valve is opened, and
saturated vapor flows from A into B until the
pressures become equal. The process occurs
slowly enough that all temperatures stay at 70 F
during the process. Find the total heat transfer to
the R-134a during the process.
5.162E Ammonia, NH 3 , is contained in a sealed rigid
tank at 30 F, x = 50% and is then heated to 200
F, Find the final state P 2 , u 2 and the specific
work and heat transfer.
5.163E Water at 70 F, 15 lbf/in 2 , is brought to 30 lbf/in 2 ,
2700 F. Find the change in the specific internal
energy, using the water tables and ideal-gas table.
5.164E A car with mass 3250 lbm is driven at 60 mi/h
when the brakes are applied to quickly decrease
its speed to 20 mi/h. Assume the brake pads are
1 lbm/in with a heat capacity of 0.2 Btu/lbm R,
the brake disks/drums are 8 lbm of steel, and
both masses are heated uniformly. Find the tem-
perature increase in the brake assembly.
5.165E A 2-lbm steel pot contains 2 lbm of liquid water
at 60 F. It is now put on the stove, where it is
heated to the boiling point of the water. Neglect
any air being heated and find the total amount of
energy needed.
5.166E A copper block of volume 60 in 3 is heat treated
at 900 F and now cooled in a 3 -ft 3 oil bath ini-
tially at 70 F. Assuming no heat transfer with
the surroundings, what is the final temperature?
5.167E An engine, shown in Fig. P5.81, consists of a
200-lbm cast iron block with a 40-lbm alu-
minum head, 40 lbm of steel parts, 10 lbm of
engine oil and 12 lbm of glycerine (antifreeze).
Everything begins at 40 F, and as the engine
starts it absorbs a net of 7000 Btu before it
160 H Chapter Five The First Law of Thermodynamics
reaches a steady uniform temperature. We want
to know how hot it becomes.
5.168E A cylinder with a piston restrained by a linear
spring contains 4 Ibm of carbon dioxide at 70
lbf/in 2 , 750 F. It is cooled to 75 F, at which
point the pressure is 45 lbf/in 2 . Calculate the
heat transfer for the process.
5.169E An insulated cylinder is divided into two parts
of 10 ft 3 each by an initially locked piston. Side
A has air at 2 atm, 600 R, and side B has air at
10 atm, 2000 R, as shown in Fig. P5.95. The
piston is now unlocked so it is free to move, and
it conducts heat so the air comes to a uniform
temperature T A = T B , Find the mass in both A
and B and also the final T and P.
5.170E A 65-gal rigid tank contains methane gas at
900 R, 200 psia. It is now cooled down to
540 R. Assume an ideal gas and find the needed
heat transfer.
5.171E Air in a piston/cylinder at 30 lbf/in 2 , 1080 R is
shown in Fig. P5.101. It is expanded in a con-
stant-pressure process to twice the initial volume
(state 2). The piston is then locked with a pin,
and heat is transferred to a final temperature of
1080 R. Find P, T, and h for states 2 and 3, and
find the work and heat transfer in both processes.
5.172E A 30-ft-high cylinder, cross-sectional area 1 ft 2 ,
has a massless piston at the bottom with water at
70 F on top of it, as shown in Fig. P5.93. Air at
540 R, volume 10 ft 3 under the piston is heated
so that the piston moves up, spilling the water
out over the side. Find the total heat transfer to
the air when all the water has been pushed out.
5.173 E An air pistol contains compressed air in a small
cylinder, as shown in Fig. P5.112. Assume that
the volume is 1 in 3 , pressure is 10 atm, and the
temperature is 80 F when armed. A bullet, m =
0.04 Ibm, acts as a piston initially held by a pin
(trigger); when released, the air expands in an
isothermal process (T = constant). If the air
pressure is 1 atm in the cylinder as the bullet
leaves the gun, find
a. the final volume and the mass of air.
b. the work done by the air and work done on
the atmosphere.
c. the work to the bullet and the bullet exit
velocity.
5.174 E A piston/cylinder in a car contains 12 in 3 of air
at 13 lbf/in 2 , 68 F, shown in Fig. P5.107. The air
is compressed in a quasi-equilibrium polytropic
process with polytropic exponent n — 1.25 to a
final volume six times smaller. Determine the
final pressure, temperature, and heat transfer for
the process.
5.175E Oxygen at 50 lbf/in 2 , 200 F is in a piston/cylin-
der arrangement with a volume of 4 ft 3 . It is now
compressed in a polytropic process with expo-
nent, n = 1.2, to a final temperature of 400 F.
Calculate the heat transfer for the process.
5.176E Helium gas expands from 20 psia, 600 R and
9 ft 3 to 15 psia in a polytropic process with
n — 1.667. How much heat transfer is involved?
5.177E A cylinder fitted with a frictionless piston con-
tains R-134a at 100 F, 80% quality, at which
point the volume is 3 gal. The external force on
the piston is now varied in such a manner that
the R-134a slowly expands in a polytropic
process to 50 lbf/in 2 , 80 F. Calculate the work
and the heat transfer for this process.
5.178E A piston/cylinder contains argon at 20 lbf/in 2 ,
60 F, and the volume is 4 ft 3 . The gas is com-
pressed in a polytropic process to 100 lbf/in 2 ,
550 F. Calculate the heat transfer during the
process.
5.179 E A small elevator is being designed for a con-
struction site. It is expected to carry four 150-
Ibm workers to the top of a 300-ft-taII building
in less than 2 min. The elevator cage will have a
counterweight to balance its mass. What is the
smallest size (power) electric motor that can
drive this unit?
5,180 E Water is in a piston/cylinder maintaining con-
stant P at 330 F, quality 90%, with a volume of
4 ft 3 . A heater is turned on heating the water
with 10 000 Btu/h. What is the elapsed time to
vaporize all the liquid?
5.181 E A computer in a closed room of volume 5000 ft 3
dissipates energy at a rate of 10 hp. The room
has 100 Ibm of wood, 50 Ibm of steel, and air,
with all material at 540 R, 1 atm. Assuming all
the mass heats up uniformly, how much time
will it take to increase the temperature by 20 F?
5.182 E A closed cylinder is divided into two rooms by
a frictionless piston held in place by a pin, as
Computer, design, and Open-Ended problems 9 161
shown in Fig. P5.138. Room A has 0.3 ft 3 air
at 14.7 Ibf/in 2 , 90 F, and room B has 10 ft 3 sat-
urated water vapor at 90 F. The pin is pulled,
releasing the piston, and both rooms come to
equilibrium at 90 F. Considering a control
mass of the air and water, determine the work
done by the system and the heat transfer to the
cylinder.
Computer, Design, and Open-Ended Problems
5.183 Use the supplied software to track the process in
Problem 5.37 in steps of 10°C until the two-phase
region is reached, after that step with jumps of 5%
in the quality. At each step write out T t x $ and the
heat transfer to reach that state from the initial state.
5.184 For one of the substances in Table A.6, compare
the enthalpy change between any two tempera-
tures, 7*1 and T 2 , as calculated by integrating the
specific heat equation; by assuming constant spe-
cific heat at the average temperature; and by as-
suming constant specific heat at temperature 7\.
5.185 Track the process described in Problem 5.54 so
that you can sketch the amount of heat transfer
added and the work given out as a function of the
volume.
5.186 Using states with given (P, v) and properties from
the supplied software, track the process in Prob-
lem 5,57. Select five pressures away from the ini-
tial toward the final pressure so that you can plot
the temperature, the heat added, and the work
given out as a function of the volume.
5.187 Examine the sensitivity of the final pressure to the
containment room volume in Problem 5.48. Solve
for the volume for a range of final pressures,
100-250 kPa, and sketch the pressure versus vol-
ume curve.
5.188 Write a program to solve Problem 5.78 for a
range of initial velocities. Let the car mass and
final velocity be input variables.
5.189 Write a program for Problem 5.107, where the
initial state, the volume ratio, and the polytropic
exponent are input variables. To ^simplify the
formulation, use constant specific heat.
5.190 Consider a general version of Problem 5.89 with a
substance listed in Table A.6. Write a program
where the initial temperature and pressure, and
the final temperature are program inputs.
5.191 Examine a process where air at 300 K, 100 kPa is
compressed in a piston/cylinder arrangement to
600 kPa. Assume the process is polytropic with
exponents in the 1 .2-1.6 range. Find the work and
heat transfer per unit mass of air. Discuss the dif-
ferent cases and how they may be accomplished
by insulating the cylinder or by providing heating
or cooling.
5.192 A cylindrical tank of height 2 m with a cross-
sectional area of 0.5 m 2 contains hot water at
80°C, 125 kPa. It is in a room with temperature
T = 20°C, so it slowly loses energy to the room
air proportional to the temperature difference as
Q ob = CA( l T-T q )
with the tank surface area, A, and C is a constant.
For different values of the constant C, estimate
the time it takes to bring the water to 50°C. Make
enough simplifying assumptions so that you can
solve the problem mathematically, that is find a
formula for T(t).
6
First-Law Analysis for
In the preceding chapter we developed the first-law analysis (energy balance) for a con-
trol mass going through a process. Many applications in thermodynamics do not readily
lend themselves to a control mass approach but are conveniently handled by the more
general control volume technique, as discussed in Chapter 2. The present chapter is con-
cerned with development of the control volume forms of the conservation of mass and
energy in situations where there are flows of substance present.
6.1 CONSERVATION OF MASS
AND THE CONTROL VOLUME
r\ control volume is a volume in space in which one has interest for a particular study or
Jnalysis. The surface of this control volume is referred to as a control surface and always
consists of a closed surface. The size and shape of the control volume are completely ar-
bitrary and are so denned as to best suit the analysis to be made. The surface may be
fixed, or it may move so that it expands or contracts. However, the surface must be de-
fined relative to some coordinate system. In some analyses it may be desirable to consider
a rotating or moving coordinate system and to describe the position of the control surface
relative to such a coordinate system.
T^llass as welt as heat and work can cross the control surface, and the mass in the
crJ^J volume, as well as the properties of this mass, can change with time. Figure 6.1
shows a schematic diagram of a control volume that includes heat transfer, shaft work,
moving boundary work, accumulation of mass within the control volume, and several
mass flows. It is important to identify and label each flow of mass and energy and the
parts of the control volume that can store (accumulate) mass.
Let us consider the conservation of mass law as it relates to the control volume.
The physical law concerning mass, recalling Section 5.9, says that we cannot create or
destroy mass. We will express this law in a mathematical statement about the mass in the
control volume. To do this we must consider all the mass flows into and out of the control
volume and the net increase of mass within the control volume. As a somewhat simpler
control volume we consider a tank with a cylinder and piston and two pipes attached as
shown in Fig. 6.2. The rate of change of mass inside the control volume can be different
from zero if we add or take a flow of mass out as
Rate of change - Hn - out
162
CONSERVATION OF MASS AND THE CONTROL VOLUME H 163
High-pressure steam
Mass rate of flow
/
FIGURE 6.1
Schematic diagram of a
control volume showing
mass and energy transfers
and accumulation.
Accumulator
initially
evacuated
Steam expanding
against a piston
Control surface
Steam /
radiator /
1 — <\f\f\j — + — >
V
(2 C v = heat transfer rate
Shaft connecting the
turbine to generator
Low-pressure steam
Mass rate of flow
= ('"e)low pressure steam
Condensate
Mass rate of flow
= ( nl e)condensa!e
With several possible flows this is written as
^ = 2>,-2>. C6.D
stating that if the mass inside the control volume changes with time it is because we add
some mass or take some mass out. There are no other means by which the mass inside the
control volume could change. Equation 6.1 expressing the conservation of mass is com-
monly termed the continuity equation. While this form of the equation is sufficient for the
majority of applications in thermodynamics, it is frequently rewritten in terms of the local
fluid properties in the study of fluid mechanics and heat transfer. In this text we are
Flow
FIGURE 6.2
Schematic diagram of a
control volume for the
analysis of the continuity
equation.
Control
surface
Flow
164 H chapter Six first-Law analysis for a Control volume
FIGURE 6.3 The flow across a control volume surface
with a flow cross-sectional area of A. Left of valve shown
as an average velocity and to the right of valve shown as a
distributed flow across area.
mainly concerned with the overall mass balance and thus consider Eq. 6.1 as the general
expression for the continuity equation.
Since Eq. 6.1 is written for the total mass (lumped form) inside the control volume
we may have to consider several contributions to the mass as
m c . v . = \ pdV^ j{Vv)dV = m A + m B + m c + • « ■
Such a summation is needed when the control volume has several accumulation units with
different states of the mass.
Let us now consider the mass flow rates across the control volume surface in a little
more detail. For simplicity we assume the fluid is flowing in a pipe or duct as illustrated in
Fig. 6.3. We wish to relate the total flow rate that appears in Eq. 6.1 to the local properties
of the fluid state. The flow across the control volume surface can be indicated with an av-
erage velocity shown to the left of the valve or with a distributed velocity over the cross
section as shown to the right of the valve.
The volume flow rate is
V=VA = jVtetdA (6.2)
so the mass flow rate becomes
m = p avg V=Vfv^j QJ^Jv)dA = VA/v (6.3) |q|
where often the average velocity is used. It should be noted that this result, Eq. 6.3, has
been developed for a stationary control surface and we tacitly assumed the flow was
normal to the surface. This expression for the mass flow rate applies to any of the vari-
ous flow streams entering or leaving the control volume, subject to the assumptions
mentioned.
EXAMPLE 6.1 Air is flowing in a 0.2-m-diameter pipe at a uniform velocity of 0.1 m/s. The tempera-
ture is 25°C and the pressure 150 kPa. Determine the mass flow rate.
Solution
From Eq. 6.3 the mass flow rate is
m = VA!v
For air, using R from Table A.5, we have ■■ '/
g _ 0.287 U/kgKX 298.2 K _ 5 ^
v P ■ 150 kPa
The first law of thermodynamics for a Control volume
b 165
fumoMET
The cross-sectional area is
A = ~(0.2) 2 = 0.0314 m 2
Therefore,
m = VAfv = 0.1 m/s X 0.0314 m 2 /0.5705 m 3 /kg = 0.0055 kg/s
6.2 The first Law of Thermodynamics
for a Control volume
We have already considered the first law of thermodynamics for a control mass, which
consists of a fixed quantity of mass, and noted, Eq. 5.5, that it may be written
We have also noted that this may be written as an instantaneous rate equation as
dEr " Q-iv (6.4)
dt
To write the first law as a rate equation for a control volume, we proceed in a man-
ner analogous to that used in developing a rate equation for the law of conservation of
mass. For this purpose a control volume is shown in Fig. 6.4 that involves rate of heat
transfer, rates of work, and mass flows. The fundamental physical law states that we can-
not create or destroy energy such that any rate of change of energy must be caused by
rates of energy into or out of the control volume. We have already included rates of heat
transfer and work in Eq. 6.4, so the additional explanations we need are associated with
the mass flow rates.
P,T:
FIGURE 6.4
Schematic diagram to
illustrate terms in the
energy equation for a
general control volume.
11 1 .1 1 I
~:y~~
^boundary
dE c
dt
a a ft
—J P.T.
166 B CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
The fluid flowing across the control surface enters or leaves with an amount of en-
ergy per unit mass as
relating to the state and position of the fluid. Whenever a fluid mass enters a control volume
at state / or exits at state e, there is a boundary movement work associated with that process
To explain this in more detail consider an amount of mass flowing into the control
volume As this mass flows in there is a pressure at its back surface, so as this mass moves
into the control volume it is being pushed by the mass behind it, which is the surround-
ings The net effect is that after the mass has entered the control volume the surroundings
have pushed it in against the local pressure with a velocity giving it a rate of work m the
process Similarly a fluid exiting the control volume at state e must push the surrounding
fluid ahead of it, doing work on it, which is work leaving the control volume. The velocity
and the area correspond to a certain volume per unit time entering the control volume en-
abling us to relate that to the mass flow rate and the specific volume at the state of the
mass going in. Now we are able to express the rate of flow work as
WW = FV = j PVdA=PV= Pvm (6-5)
For the flow that leaves the control volume work is being done by the control volume,
P vJn and for the mass that enters, the surroundings do the rate of work, Pp^. The flow
work per unit mass is then Pv, and the total energy associated with the flow of mass is
e + Pv = u + Pv+\v 2 +gZ=h+±V 2 + gZ (6.6)
In this equation we have used the definition of the thermodynamic property en-
thalpy and it is the appearance of the combination (a + Pv) for the energy in connection
with a mass flow that is the primary reason for the definition of the property enthalpy. Its
introduction earlier in conjunction with the constant-pressure process was to facilitate use-
of the tables of thermodynamic properties at that time.
EXAMPLE 6.2 Assume we are standing next to the local city's mam water line The liquid watery mside
flows at a pressure of say 600 kPa (6 arm) with a temperature of about 10 C. We want to
add a smaller amount, 1 kg, of liquid to the line through a side pipe and valve mounted
on the main line. How much work wilt be involved in this process?
If the 1 kg of liquid water is in a bucket and we open the valve to the water main
trying to pour it down into the pipe opening, we will realize that the water flows the
other way. The water will flow from a higher to a lower pressure, that is, from inside the
main line to the atmosphere (from 600 kPa to 101 fcPa). .
We must take the 1 kg of liquid water and put it into a piston cylmder (like a hand-
held pump) and attach the cylinder to the water pipe. Now we can press on the piston
until the water pressure inside is 600 kPa and then open the valve to the mam lme and
slowly squeeze the 1 kg of water in. The work done at the piston surface to the water is
W= j PdV= F^tet mv = 600 kPa X 1 kg x 0.001 mVkg = 0.6 kJ
and this is the necessary flow work for adding the 1 kg of liquid.
The Steady-State Process H 167
The extension of the first law of thermodynamics from Eq. 6.4 becomes
dE cv • •
and the substitution of Eq. 6.5 gives
dt
Qcv. ~ w cv. + m£e t + P t v,) - mfa + P e v e )
Qcv. ~ Wcv. + '»/ h t + ±Vf + gZ l \- m e \h e + ±V 2 e + gZ (
In this form of the energy equation the rate of work term is the sum of all shaft work
terms and boundary work terms and any other types of work given out by the control volume;
however, the flow work is now listed separately and included with the mass flow rate terms.
For the general control volume we may have several entering or leaving mass flow
rates, so a summation over those terms is often needed. The final form of the first law of
thermodynamics then becomes
d -^~ = Qcv. ~ Wcs. + £ >"/ (hi + \ V? + gz)j - S (a. + \ V5 + g^j (6.7)
expressing that the rate of change of energy inside the control volume is due to a net rate
of heat transfer, a net rate of work (measured positive out), and the summation of energy
fluxes due to mass flows into and out of the control volume. As with the conservation of
mass, this equation can be written for the total control volume and can therefore be put in
the lumped or integral form where
Eq.v. = j pedV= me = m A e A + m B e B + m& c + • * •
As the kinetic and potential energy terms per unit mass appear together with the en-
thalpy in all the flow terms, a shorter notation is often used
h l0X = h + ±V 2 + gZ
defining the total enthalpy and the stagnation enthalpy (used in fluid mechanics). The
shorter equation then becomes
dt
~ QCV. - Wcv. + 2 ™AoUi - 2) ™eKt,e ( 6 - 8 )
giving the general energy equation on a rate form. All applications of the energy equation
start with the form in Eq. 6.8, and for special cases this will result in a slightly simpler
form as shown in the subsequent sections.
6,3 The Steady-State Process
Our first application of the control volume equations will be to develop a suitable analyti-
cal model for the long-term steady operation of devices such as turbines, compressors,
nozzles, boilers, condensers— a very large class of problems of interest in thermodynamic
168 ■ CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
analysis. This model will not include the short-term transient start-up or shutdown of such
devices, but only the steady operating period of time.
Let us consider a certain set of assumptions (beyond those leading to Eqs. 6.1 and
6.7) that lead to a reasonable model for this type of process, which we refer to as the
steady-state process .
1. The control volume does not move relative to the coordinate frame.
2. The state of the mass at each point in the control volume does not vary with time.
3. As for the mass that flows across the control surface, the mass flux and the state of
this mass at each discrete area of flow on the control surface do not vary with time.
The rates at which heat and work cross the control surface remain constant.
As an example of a steady-state process consider a centrifugal air compressor that op-
erates with constant mass rate of flow into and out of the compressor, constant properties at
each point across the inlet and exit ducts, a constant rate of heat transfer to the surroundings,
and a constant power input. At each point in the compressor the properties are constant with
time, even though the properties of a given elemental mass of air vary as it flows through the
compressor. Often, such a process is referred to as a steady-flow process, since we are con-
cerned primarily with the properties of the fluid entering and leaving the control volume.
However, in the analysis of certain heat transfer problems in which the same assumptions
apply, we are primarily interested in the spatial distribution of properties, particularly temper-
ature/and such a process is referred to as a steady-state process. Since this is an introductory
book we will use the term steady-state process for both. The student should realize that the
terms steady-state process and steady-flow process are both used extensively in the literature.
Let us now consider the significance of each of these assumptions for the steady-
state process.
1. The assumption that the control volume does not move relative to the coordinate
frame means that all velocities measured relative to the coordinate frame are also
velocities relative to the control surface, and there is no work associated with the
acceleration of the control volume.
2. The assumption that the state of the mass at each point in the control volume does
not vary with time requires that
dm cy _
dt
and also
dE C y.
=
-0
dt
Therefore, we conclude that for the steady-state process we can write, from Eqs. 6.1
and 6.7,
Continuity equation; ^ m t — X ™ e
(6.9)
First law: fi cv . + 2 «h (*/ + y + 8 Z ) = 2 < [k + y + + ^c.v ; (6.10)
3. The assumption that the various mass flows, states, and rates at which heat and
work cross the control surface remain constant requires that every quantity in Eqs.
Examples of Steady-State Processes H 169
6.9 and 6,10 be steady with time. This means that application of Eqs. 6.9 and 6.10
to the operation of some device is independent of time.
Many of the applications of the steady-state model are such that there is only one flow
stream entering and one leaving the control volume. For this type of process, we can write
Continuity equation: »i ( = m e ~ in (6.11)
First law : 2c.v. + m + y + gz)j = m (ji e + ^ + gz)j + IF C;V . (6. 1 2)
Rearranging this equation, we have
V? V 2
q t n i -i- —
where, by definition,
q + h l + -£ + gZ i = h, + 'f + gZ t + w (6.13)
Qc.V. , **C.V. r£\A\
q = — : — and w = ■ ■ ■■ . (6.14)
m m
Note that the units for q and w are kj/kg. From their definition, q and w can be
thought of as the heat transfer and work (other than flow work) per unit mass flowing into
and out of the control volume for this particular steady-state process.
The symbols q and w are also used for the heat transfer and work per unit mass of a
control mass. However, since it is always evident from the context whether it is a control
mass (fixed mass) or control volume (involving a flow of mass) with which we are con-
cerned, the significance of the symbols q and w will also be readily evident in each situation.
The steady-state process is often used in the analysis of reciprocating machines,
such as reciprocating compressors or engines. In this case the rate of flow, which may ac-
tually be pulsating, is considered to be the average rate of flow for an integral number of
cycles. A similar assumption is made regarding the properties of the fluid flowing across
the control surface and the heat transfer and work crossing the control surface. It is also
assumed that for an integral number of cycles the reciprocating device undergoes, the en-
ergy and mass within the control volume do not change.
A number of examples are now given to illustrate the analysis of steady-state
processes.
6.4 EXAMPLES OF STEADY-STATE PROCESSES
In this section, we consider a number of examples of steady-state processes in which there
is one fluid stream entering and one leaving the control volume, such that the first law can
be written in the form of Eq. 6,13. Some may instead utilize control volumes that include
more than one fluid stream, such that it is necessary to write the first law in the more gen-
eral form of Eq. 6.10.
Heat Exchanger
A steady-state heat exchanger is a simple fluid flow through a pipe or system of pipes,
where heat is transferred to or from the fluid. The fluid may be heated or cooled, and may
or may not boil, liquid to vapor, or condense, vapor to liquid. One such example is the
170 M CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
FIGURE 6.5 A
refrigeration system
condenser.
R-134a
Vapor In
-2c.v. Cold water pipes
o
; C
J Q C
condenser in an R-I34a refrigeration system, as shown in Fig. 6.5. Superheated vapor en-
ters the condenser, and liquid exits. The process tends to occur at constant pressure, since
a fluid flowing in a pipe usually undergoes only a small pressure drop, because of fluid
friction at the walls. The pressure drop may or may not be taken into account in a particu-
lar analysis. There is no means for doing any work (shaft work, electrical work, etc.), and
changes in kinetic and potential energies are commonly negligibly small. (One exception
may be a boiler tube in which liquid enters and vapor exits at a much larger specific vol-
ume. In such a case, it may be necessary to check the exit velocity using Eq. 6.3.) The
heat transfer in most heat exchangers is then found from Eq. 6.13 as the change in en-
thalpy of the fluid. In the condenser shown in Fig. 6.5, the heat transfer out of the con-
denser then goes to whatever is receiving it, perhaps a stream of air or of cooling water. It
is often simpler to write the first law around the entire heat exchanger, including both flow
streams, in which case there is little or no heat transfer with the surroundings. Such a situ-
ation is the subject of the following example.
EXAMPLE 6.3 Consider a water-cooled condenser in a large refrigeration system in which R-134a is
the refrigerant fluid. The refrigerant enters the condenser at 1.0 MPa and 60°C, at the
rate of 0.2 kg/s, and exits as a liquid at 0.95 MPa and 35°C. Cooling water enters the
condenser at 10°C and exits at 20°C. Determine the rate at which cooling water flows
through the condenser.
Control volume:
Sketch:
Inlet states:
Exit states:
Process:
Model:
Condenser.
Fig. 6.6
R-134a— fixed; water — fixed.
R134a — fixed; water— fixed.
Steady-state.
R-134a tables; steam tables.
Analysis
With this control volume we have two fluid streams, the R-134a and the water, entering
and leaving the control volume. It is reasonable to assume that both kinetic and potential
energy changes are negligible. We note that the work is zero, and we make the other rea-
sonable assumption that there is no heat transfer across the control surface. Therefore,
the first law, Eq. 6.10, reduces to
Examples of Steady-State Processes B 171
R-134a Vapor In
Cooling
water in
FIGURE 6.6
Schematic diagram of an
R-134a condenser.
R-134a Liquid out
Cooling
water out
• Control
surface
Using the subscript r for refrigerant and w for water, we write
inXhiX + m»(hX = m/hX + mJJiX
Solution
From the R-134a and steam tables, we have
(hi) f: = 441.89 kJ/kg, (h t ) w = 42.00 kJ/kg
(h e ) r = 249.10 kJ/kg, (h e ) w = 83.95 kJ/kg
Solving the above equation for m m the rate of flow of water, we obtain
■ Vu-hX r (441.89 - 249.10) kJ/kg .„.„. ,
m " = mr Jh~^W, = °' 2 ^ (83.95 - 42.00) kJ/kg = °" 919 kg/s
This problem can also be solved by considering two separate control volumes, one
having the flow of R-134a across its control surface and the other having the flow of
water across its control surface. Further, there is heat transfer from one control volume
to the other.
The heat transfer for the control volume involving R-134a is calculated first. In
this case the steady-state energy equation, Eq. 6.10, reduces to
Qc.V. = ™£K -
= 0.2 kg/s X (249.10 - 441.89) kJ/kg = -38.558 kW
This is also the heat transfer to the other control volume, for which Q c v = +38.558 kW.
Qcv. = - hX
38.558 kW
w (83.95 - 42.00) kJ/kg
0.919 kg/s
172 B CHAPTER SIX FIRST-LAW ANALYSIS FOR. A COTJTROL VOLUME
Nozzle
A nozzle is a steady-state device whose purpose is to create a high-velocity fluid stream at
the expense of the fluid's pressure. It is contoured in an appropriate manner to expand a
flowing fluid smoothly to a lower pressure, thereby increasing its velocity. There is no
means to do any work— there are no moving parts. There is little or no change in potential
energy and usually little or no heat transfer. An exception is the large nozzle on a liquid-
propellant rocket, such as was described in Section 1.7, in which the cold propellant is
commonly circulated around the outside of the nozzle walls before going to the combus-
tion chamber, in order to keep the nozzle from melting. This case, a nozzle with signifi-
cant heat transfer, is the exception and would be noted in such an application. In addition,
the kinetic energy of the fluid at the nozzle inlet is usually small and would be neglected if
its value is not known.
EXAMPLE 6.4 Steam at 0.6 MPa and 200°C enters an insulated nozzle with a velocity of 50 m/s. It
leaves at a pressure of 0.15 MPa and a velocity of 600 m/s. Determine the final tempera-
ture if the steam is superheated in the final state and the quality if it is saturated.
Control volume: Nozzle.
Inlet state: Fixed (see Fig. 6.7).
Exit state: P e known.
Process: Steady-state.
Model: Steam tables.
Analysis
We have
j2 c v = (nozzle insulated)
W c .v. =
PE,~FE (
The first law (Eq. 6.13) yields
V- V 2
A, + y = A e + y
Examples of Steady-State Processes 1 173
Solution
Solving for h e we obtain
K - 2850.1 +
(50) a
(600) 2
2 X 1000 2 X 1000
m 2 /s 2
J/kJ
= 2671.4 kJ/kg
The two properties of the fluid leaving that we now know are pressure and en-
thalpy, and therefore the state of this fluid is determined. Since k e is less than h at 0.15
VI Pa, the quality is calculated. 8
h-h f +xh fg
2671.4 = 467.1 t-.r, 2226.5
x=0.99
Example 6.4E
Steam at 100 lbf/in 2 , 400 F, enters an insulated nozzle with a velocity of 200 ft/s. It
leaves at a pressure of 20.8 lbf/in 2 and a velocity of 2000 ft/s. Determine the final tem-
perature if the steam is superheated in the final state, and the quality if it is saturated.
Control volume:
Inlet state:
Exit state:
Process:
Model:
Nozzle.
Fixed (see Fig. 6.7E).
P e known.
Steady-state.
Steam tables.
Analysis
First law(Eq. 6.13):
Qc.v. ~ (nozzle insulated)
W c .m. = 0, PE f = PE e
V? V 2
1 2 e 2
174 M Chapter Six First-Law Analysis for. a control Volume
Solution
„ = 1227 5 + goof f^f im3Btanbm
The two properties of the fluid leaving that we now know are pressure and en-
thalpy, and therefore the state of this fluid is determined. Since h e is less than h s at 20.8
lbf/in 2 , the quality is calculated.
h = hf+xhf g
1148.3 - 198.31 + .v e 958.81
.v. - 0.99
Diffuser
A steady-state diffuser is a device constructed to decelerate a high-velocity fluid in a man-
ner that results in an increase in pressure of the fluid. In essence, it is the exact opposite of a
nozzle, and it may be thought of as a fluid flowing in the opposite direction through a noz-
zle, with the opposite effects. The assumptions are similar to those for a nozzle, with a large
kinetic energy at the diffuser inlet and a small, but usually not negligible, kinetic energy at
the exit being the only terms besides the enthalpies remaining in the first law, Eq. 6. 13.
Throttle
A throttling process occurs when a fluid flowing in a line suddenly encounters a restric-
tion in the flow passage. This may be a plate with a small hole in it, as shown in Fig. 6.8,
it may be a partially closed valve protruding into the flow passage, or it may be a change
to a much smaller diameter tube, called a capillary tube, which is normally found on a re-
frigerator. The result of this restriction is an abrupt pressure drop in the fluid, as it is
forced to find its way through a suddenly smaller passageway. This process is drastically
unlike the smoothly contoured nozzle expansion and area change, which results in a sig-
nificant velocity increase. There is typically some increase in velocity in a throttle, but
both inlet and exit kinetic energies are usually small enough to be neglected. There is no
means for doing work and little or no change in potential energy. Usually, there is neither
time nor opportunity for an appreciable heat transfer, such that the only terms left in the
first law, Eq. 6.13, are the inlet and exit enthalpies. We conclude that a steady-state throt-
tling process is approximately a pressure drop at constant enthalpy, and we will assume
this to be the case unless otherwise noted.
Frequently, a throttling process involves a change in the phase of the fluid. A typical
example is the flow through the expansion valve of a vapor-compression refrigeration
system. The following example deals with this problem.
Control surface
FIGURE 6.8 The > j I
throttling process. ' — — •
Examples of Steady-State processes B 175
EXAMPLE 6.5 Consider the throttling process across the expansion valve or through the capillary tube
in a vapor-compression refrigeration cycle. In this process the pressure of the refriger-
ant drops from the high pressure in the condenser to the low pressure in the evaporator,
and during this process some of the liquid flashes into vapor. If we consider this
process to be adiabatic, the quality of the refrigerant entering the evaporator can be
calculated.
Consider the following process, in which ammonia is the refrigerant. The ammo-
nia enters the expansion valve at a pressure of 1.50 MPa and a temperature of 35°C. Its
pressure on leaving the expansion valve is 291 kPa. Calculate the quality of 'the ammo-
nia leaving the expansion valve.
Control volume: Expansion valve or capillary tube.
Inlet state: P {> T t known; state fixed.
Exit state: P e known.
Process: Steady-state.
Model: Ammonia tables.
Analysis
We can use standard throttling process analysis and assumptions. The first law reduces
to
h, = h e
Solution
From the ammonia tables we get
h t = 346.8 kJ/kg
(The enthalpy of a slightly compressed liquid is essentially equal to the enthalpy of satu-
rated liquid at the same temperature.)
h e = k t = 346.8 = 134.4 + ,r e (1296.4)
x e = 0.1638 = 16.38%
Turbine
A turbine is a rotary steady-state machine whose purpose is to produce shaft work
(power, on a rate basis) at the expense of the pressure of the working fluid. Two general
classes of turbines are steam (or other working fluid) turbines, in which the steam exiting
the turbine passes to a condenser, where it is condensed to liquid, and gas turbines, in
which the gas usually exhausts to the atmosphere from the turbine. In either type, the tur-
bine exit pressure is fixed by the environment into which the working fluid exhausts, and
the turbine inlet pressure has been reached by previously pumping or compressing the
working fluid in another process. Inside the turbine, there are two distinct processes. In
the first, the working fluid passes through a set of nozzles, or the equivalent — fixed blade
passages contoured to expand the fluid to a lower pressure and to a high velocity. In the
second process inside the turbine, this high-velocity fluid stream is directed onto a set of
176 H CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
moving (rotating) blades, in which the velocity is reduced before being discharged from
the passage. This directed velocity decrease produces a torque on the rotating shaft, re-
sulting in a shaft work output. The low-velocity, low-pressure fluid then exhausts from
the turbine.
The first law for this process is either Eq. 6.10 or 6.13. Usually, changes in poten-
tial energy are negligible, as is the inlet kinetic energy. Often, the exit kinetic energy is
neglected, and any heat rejection from the turbine is undesirable and is commonly small.
We therefore normally assume that a turbine process is adiabatic, and the work output in
this case reduces to the decrease in enthalpy form the inlet to exit states. In the follow-
ing example, however, we include alt the terms in the first law and study their relative
importance.
EXAMPLE 6.6 The mass rate of flow into a steam turbine is 1.5 kg/s, and the heat transfer from the
turbine is 8.5 k\V. The following data are known for the steam entering and leaving the
turbine.
Inlet
Conditions
Exit
Conditions
Pressure
2.0 MPa
0.1 MPa
Temperature
350°C
Quality
100%
Velocity
50 m/s
100 m/s
Elevation above reference plane
6 m
3 m
g = 9.8066 m/s 2
Determine the power output of the turbine.
Control volume: Turbine (Fig. 6.9).
Inlet state: Fixed (above).
Exit state: Fixed (above).
Process: Steady-state.
Model: Steam tables.
FIGURE 6.9
Illustration for Example
6.6.
= 1.5 kg/s
= 2 MPa
= 350°C
= 50 m/s
= 6 m
Control
surface
5>W
m e = 1 .5 kg/s
P = 0.1 MPa
x e = 100%
v e = i0Om/s
Z =3 m
Examples of Steady-State Processes H 177
Analysis
From the first law (Eq. 6. 12) we have
Qcv. + m (h t + ^ + gz}j = m (k e + ^- + gZ^j + fF cv .
with
£ cv =-8.5kW
Solution
From the steam tables, ft, = 3137,0 kJ/kg. Substituting inlet conditions gives
V?_ 50X50
2 X 1000
= 1.25kJ/kg
g^= 6X 1Q 9 8 066 = 0.059kJ/kg
Similarly, for the exit h e = 2675.5 kJ/kg and
Ve _ 100 X 100 , ni Tfl
T~TYToob~ = 5 - 0kJ/kg
gZ ^ 3X iooo 066 ^ - 029kJ/kg
Therefore, substituting into Eq. 6. 1 2, we obtain
-8.5 4- 1.5(3137 + 1.25 + 0.059) - 1.5(2675.5 + 5.0 -f 0.029) + W^.
#c.v. = -8.5 + 4707.5 - 4020.8 = 678.2 kW
If Eq. 6.13 is used, the work per kilogram of fluid flowing is found first.
V? V 2
q + h ( + -£ + g Z l = h e + ^- + gZ, + w
q = -5.667 kJ/kg
Therefore, substituting into Eq. 6. 13, we get
-5.667 + 3137 + 1.25 + 0.059 = 2675.5 + 5.0 + 0.029 + w
w = 452.11 kJ/kg
W cy . = 1.5 kg/s X 452.1 1 kJ/kg = 678.2 kW
Two further observations can be made by referring to this example. First, in many
engineering problems, potential energy changes are insignificant when compared with the
other energy quantities. In the above example the potential energy change did not affect
any of the significant figures. In most problems where the change in elevation is small the
potential energy terms may be neglected.
Second, if velocities are small — say, under 20 m/s — in many cases the kinetic
energy is insignificant compared with other energy quantities. Furthermore, when the
178 H
Chapter Six First-Law analysis for a Control Volume
velocities entering and leaving the system are essentially the same, the change in ki-
netic energy is small. Since it is the change in kinetic energy that is important in the
steady-state energy equation, the kinetic energy terms can usually be neglected when
there is no significant difference between the velocity of the fluid entering and that
leaving the control volume. Thus, in many thermodynamic problems, one must make
judgments as to which quantities may be negligible for a given analysis.
The preceding discussion and example concerned the turbine, which is a rotary
work-producing device. There are other nonrotary devices that produce work, which can
be called expanders as a general name. In such devices, the first-law analysis and assump-
tions are generally the same as for turbines, except that in a piston/cylinder type expander,
there would in most cases be a larger heat loss or rejection during the process.
Compressor and Pump
The purpose of a steady-state compressor (gas) or pump (liquid) is the same: to increase
the pressure of a fluid by putting in shaft work (power, on a rate basis). There are two fun-
damentally different classes of compressors. The most common is a rotary-type compres-
sor (either axial flow or radial/centrifugal flow), in which the internal processes are
essentially the opposite of the two processes occurring inside a turbine. The working fluid
enters the compressor at low pressure, moving into a set of rotation blades, from which it
exits at high velocity, a result of the shaft work input to the fluid. The fluid then passes
through a diffuser section, in which it is decelerated in a manner that results in a pressure
increase. The fluid then exits the compressor at high pressure.
The first law for the compressor is either Eq. 6.10 or 6.13. Usually, changes in po-
tential energy are negligible, as is the inlet kinetic energy. Often the exit kinetic energy is
neglected, as well. Heat rejection from the working fluid during compression would be
desirable, but it is usually small in a rotary compressor, which is a high-volume flow-rate
machine, and there is not sufficient time to transfer much heat from the working fluid. We
therefore normally assume that a rotary compressor process is adiabatic, and the work
input in this case reduces to the change in enthalpy from the inlet to exit states.
In a piston/cylinder-type compressor, the cylinder usually contains fins to promote heat
rejection during compression (or the cylinder may be water-jacketed in a large compressor
for even greater cooling rates). In this type of compressor, the heat transfer from the working
fluid is significant and is not neglected in the first law. As a general rule, in any example or
problem in this text, we will assume that a compressor is adiabatic unless otherwise noted.
EXAMPLE 6.7 The compressor in a plant (see Fig. 6.10) receives carbon dioxide at 100 kPa, 280 K,
with a low velocity. At the compressor discharge, the carbon dioxide exits at 1 100 kPa,
500 K, with velocity of 25 m/s and then flows into a constant-pressure aftercooler (heat
exchanger) where it is cooled down to 350 K. The power input to the compressor is 50
kW. Determine the heat transfer rate in the aftercooler.
Solution
C.V. compressor, steady state, single inlet and exit flow.
Energy Eq. 6.13: q + h, + | V? = h 2 + | V\ + w
EXAMPLES OF STEADY-STATE PROCESSES H 179
for Example 6.7. Compressor section Cooler section
Here we assume q = and V l = 0, so, getting h from Table A. 8,
1 (25) 2
-w = A 2 - h x + ± V 2 2 = 401.52 - 198 + 2 ^ ^ = 203.5 + 0.3 - 203.8 kJ/kg
Remember here to convert kinetic energy J/kg to kJ/kg by division by 1000,
W —
7 " = ^ = ^203l = 0,245 kg/s
C.V. aftercooler, steady state, single inlet and exit flow, and no work.
Energy Eq. 6.13: q + h 2 + ± V\ = £ 3 + \
Here we assume no significant change in kinetic energy (notice how unimportant it was)
and again we look for h in Table A.8
q = hi~h 2 = 257.9 - 401.5 = - 143.6 kJ/kg
Q™i = -Qcv. = ~W = 0.245 kg/s X 143.6 kJ/kg = 35.2 kW
EXAMPLE 6.8 A small liquid water pump is located 15 m down in a well (see Fig. 6.11), taking water
in at 10°C, 90 kPa at a rate of 1.5 kg/s. The exit line is a pipe of diameter 0.04 m that
goes up to a receiver tank mamtaining a gauge pressure of 400 kPa. Assume the process
is adiabatic with the same inlet and exit velocities and that the water stays at 10°C. Find
the required pump work.
FIGURE 6.11 Sketch
for Example 6.8.
C.V. pump + pipe. Steady
state, 1 inlet, 1 exit flow.
Assume same velocity in and ■ j p »
out and no heat transfer.
<5
^4
180 H CHAPTERSIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
Solution
Continuity equation: m m = w ct — in
Energy Eq. 6.12: m {h^ + i V? a + gzj) = m ^ + ± V* + + ^ V^;
States: h eyL --- /i^ + (P ex — (u is constant and it is constant.) '
From the energy equation
W=m(h kl + gZ in -h ex -gZ^ = m[g{Z a - Z ex ) - (P^-PM :
= 1.5 ^ X j^9.807 ~ X "Iqqq ^- (400 + 101.3 - 90) kPa 0.001 001
= 1.5 X (-0.147 - 0.412) = 0.84 kW)
Tnat is, the pump requires a power input of 840 W,
Power Plant and Refrigerator
The following examples illustrate the incorporation of several of the devices and ma-
chines already discussed in this section into a complete thermodynamic system, which is
built for a specific purpose.
EXAMPLE 6.9 Consider the simple steam power plant, as shown in Fig. 6.12. The following data are
for such a power plant.
Location
Pressure
Temperature
or Quality
Leaving boiler
2.0 MPa
300°C
Entering turbine
1.9 MPa
290°C
Leaving turbine,
entering condenser
15kPa
90%
Leaving condenser,
entering pump
14kPa
45°C
Pump work = 4 kj/kg
Determine the following quantities per kilogram flowing through the unit:
a. Heat transfer in line between boiler and turbine.
b. Turbine work.
c. Heat transfer in condenser.
d. Heat transfer in boiler.
f
There is a certain advantage in assigning a number to various points in the cycle.
For this reason the subscripts i and e in the steady-state energy equation are often re-
placed by appropriate numbers.
Since there are several control volumes to be considered in the solution to this
problem, let us consolidate our solution procedure somewhat in this example. Using the
notation of Fig. 6.12, we have:
All processes: Steady-state.
Model: Steam tables.
From the steam tables:
h t - 3023.5 kJ/kg
h 2 = 3002.5 kJ/kg
h 3 = 226.0 + 0.9(2373.1) = 2361.8 kJ/kg
h 4 = 188.5 kJ/kg
All analyses: No changes in kinetic or potential energy will be considered in
the'solution. In each case, the first law is given by Eq. 6.13,
Now, we proceed to answer the specific questions raised in the problem statement.
a. For the control volume for the pipe line between the boiler and the turbine, the first
law and solution are
i?i + h = h 2
i?2 = h 2 -h x = 3002.5 - 3023.5 = -21.0 kJ/kg
182 H chapter six first-Law analysis for a Control volume
b. A turbine is essentially an adiabatic machine. Therefore, it is reasonable to neglect
heat transfer in the first law, so that
h 2 /i 3 4- 2 u 'j
2 u' } - 3002.5 2361.8 - 640.7 kJ/kg
c. There is no work for the control volume enclosing the condenser. Therefore, the first
law and solution are
//1 + lh r - h \
3 q 4 = 188.5 - 2361.8 = -2173.3 kJ/kg . . , :
d. If we consider a control volume enclosing the boiler, the work is equal to zero, so that
the first law becomes
ill >• fh ~ /'1
A solution requires a value for h 5 , which can be found by taking a control volume
around the pump:
h s = 188.5 - (-4) = 192.5 kJ/kg
Therefore, for the boiler,
rfi + h = Ai
- 3023.5 - 192.5 - 2831 kJ/kg
EXAMPLE 6.10 The refrigerator shown in Fig. 6.13 uses R-134a as the working fluid. The mass flow rate
through each component is 0.1 kg/s, and the power input to the compressor is 5.0 kW.
The following state data are known, using the state notation of Fig. 6.13,
Pi
= 100 kPa,
7\
= ~20°C
Pi
- 800 kPa,
T 2
= 50°C
= 30°C,
*3
= 0.0
T*
= -25°C
Determine the following;
a. The quality at the evaporator inlet.
b. The rate of heat transfer to the evaporator.
c. The rate of heat transfer from the compressor.
All processes: Steady-state.
Model: R-134a tables.
All analyses: No changes in kinetic or potential energy. The first law in each
case is given by Eq. 6.10,
the transient process H 183
Gcond. to room
FIGURE 6.13
Refrigerator.
Condenser
® Warm liquid
Compressor
Evaporator
Coid vapor ©
WWV4
Expansion valve
or
capillary tube
© Cold liquid + vapor
refrigerated space
Solution
a. For a control volume enclosing the throttle, the first law gives
h^ = h 3 = 241.8 kJ/kg
/i 4 - 241.8 = hp, + x 4 h fg4 = 167.4 + x 4 X 215.6
x 4 - 0.345
b. For a control volume enclosing the evaporator, the first law gives
= 0.1(387.2 - 241.8) = 14.54 kW
c. And for the compressor, the first law gives
- 0.1(435.1 - 387.2) - 5.0 = -0.21 kW
6.5 THE TRANSIENT PROCESS
In Sections 6.3 arid 6.4 we considered the steady-state process and several examples of its
application. Many processes of interest in thermodynamics involve unsteady flow and do
not fit into this category. A certain group of these — for example, filling closed tanks with
a gas or liquid, or discharge from closed vessels— can be reasonably represented to a first
approximation by another simplified model. We call this process the transient process, for
184 @ chafter six first-Law analysis for a control volume
convenience, recognizing that our model includes specific assumptions that are not al-
ways valid. Our transient model assumptions are as follows:
1. The control volume remains constant relative to the coordinate frame.
2. The state of the mass within the control volume may change with time, but at any
instant of time the state is uniform throughout the entire control volume (or over
several identifiable regions that make up the entire control volume),
3. The state of the mass crossing each of the areas of flow on the control surface is
constant with time although the mass flow rates may be time varying.
Let us examine the consequence of these assumptions and derive an expression for the
first law that applies to this process. The assumption that the control volume remains station-
ary relative to the coordinate frame has already been discussed in Section 6.3. The remaining
assumptions lead to the following simplifications for the continuity equation and the first law.
The overall process occurs during time /. At any instant of time during the process,
the continuity equation is
dm C y.
dt
where the summation is over all areas on the control surface through which flow occurs. Inte-
grating over time t gives the change of mass in the control volume during the overall process:
The total mass leaving the control volume during time t is
and the total mass entering the control volume during time / is
Therefore, for this period of time t, we can write the continuity equation for the
transient process as
(m 2 - >»i)c.v. + 2 m e - X m t = (6.15)
In writing the first law of the transient process we consider Eq. 6.7, which applies at any
instant of time during the process:
(V 2 * \ dE ( V 2 \
A, + + gZ,j = - jp + £ m e [h e + f + gZ e J + Wfcv.
Since at any instant of time the state within the control volume is uniform, the first law for
the transient process becomes
Qcv. + 2 m U + ^ + g z] = 2 m e ( h e + ^ + g z e
dt
m \ u + ~ + gZ
c.v.
The Transient Process w 185
Let us now integrate this equation over time t, during which time we have
Jo
dt = 2 m t ( A, + ~- + gZ t
f^C.V.
dt = 2 ( A, + + gZ (
^ = tfc.v.
Therefore, for this period of time /, we can write the first law for the transient process as
= 2 m * { K + y + gZ t
»h [ "2 + "T + sZi 1 - »*i 1 "i + V + S z \
+ ^c.v.
(6.16)
As an example of the type of problem for which these assumptions are valid and Eq.
6.16 is appropriate, let us consider the classic problem of flow into an evacuated vessel.
This is the subject of Example 6.11.
EXAMPLE 6.11 Steam at a pressure of 1.4 MPa and temperature of 300°C is flowing in a pipe (Fig.
6.14). Connected to this pipe through a valve is an evacuated tank. The valve is opened
and the tank fills with steam until the pressure is 1.4 MPa, and then the valve is closed.
The process takes place adiabatically and kinetic energies and potential energies are
negligible. Determine the final temperature of the sieam.
Control volume: Tank, as shown in Fig. 6.14.
Initial state (in tank): Evacuated, mass m l = 0.
Final state: P 2 known.
Met state: P i} T { (in line) known.
Process: Transient,
Model: Steam tables.
186 M Chapter Six First-Law analysis for a control Volume
) 1.4MPa,300 7 C"
FIGURE 6.14 Flow
into an evacuated
vessel — control volume
analysis.
Initially
evacuated
-Control surface
Analysis
From the first law, Eq. 6.16, we have
Sc.v. + E + y +gZi
( \j\
= Z™A K + y + gz<
+
+ Way.
We note that £? cv . = 0, W CmVm = 0, m e = 0, and (wti) c .v. = 0. We further assume
that changes in kinetic and potential energy are negligible. Therefore, the statement of
the first law for this process reduces to
From the continuity equation for this process, Eq. 6.15, we conclude that
m z = m,
Therefore, combining the continuity equation with the first law, we have
hi = u 2
That is, the final internal energy of the steam in the tank is equal to the enthalpy of the
steam entering the tank.
Solution
From the steam tables we obtain
h ( = u 2 = 3040.4 kJ/kg
Since the final pressure is given as 1.4 MPa, we know two properties at the final
state and therefore the final state is detennined. The temperature corresponding to a
pressure of 1.4 MPa and an internal energy of 3040.4 kJ/kg is found to be 452°C.
This problem can also be solved by considering the steam that enters the tank and
the evacuated space as a control mass, as indicated in Fig. 6.15.
The process is adiabatic, but we must examine the boundaries for work. If we
visualize a piston between the steam that is included in the control mass and the
steam that flows behind, we readily recognize that the boundaries move and that the
The transient process m 187
FIGURE 6.15 Flow
into an evacuated
vessel — control mass.
Control mass
1.4 MPa, 300 - c[!
Initially
evacuated
steam in the pipe does work on the steam that comprises the control mass. The
amount of this work is
-W=P l V l = mP^Vy
Writing the first law for the control mass, Eq. 5.1 1, and noting that kinetic and potential
energies can be neglected, we have
X Q 2 =U 2 -U X + X W 2
§ = U 2 -U x -P l V x
= mu 2 — mu x - mP l v l = mu 2 — mlh
Therefore,
u 2 = h x
which is the same conclusion that was reached using a control volume analysis.
The two other examples that follow illustrate further the transient process.
EXAMPLE 6,12 Let the tank of the previous example have a volume of 0.4 m3 and initially contain satu-
rated vapor at 350 kPa. The valve is then opened and steam from the line at 1.4 MPa and
3007C flows into the tank until the pressure is 1 .4 MPa.
Calculate the mass of steam that flows into the tank.
Control volume:
Initial state:
Final state:
Inlet state:
Process:
Model:
Tank, as in Fig. 6,14.
P u saturated vapor; state fixed,
P»
P h 7); state fixed.
Transient.
Steam tables.
Analysis
The situation is the same as in Example 6.10, except that the tank is not evacuated ini-
tially. Again we note that £ c v . = 0, W cs . = 0, and m t = 0, and we assume that changes
in kinetic and potential energy are zero. The statement of the first law for this process,
Eq. 6.16, reduces to
niihj = m 2 ti 2 — >"t»i
188 H chapter Six first-law analysis for a Control Volume
The continuity equation, Eq, 6.15, reduces to
m 2 — mi ~ m i
Therefore, combining the continuity equation with the first law, we have
(»b. ~ "hVh " "h*h ~
m 2 (h; ~ u 2 ) ~ mi(h r «,) (a)
There are two unknowns in this equation — m 2 and a 2 . However, we have one addi-
tional equation:
m 2 v 2 = V= 0.4w 3 (b)
Substituting (b) into (a) and rearranging, we have
V
v 2 Qh ■■ "2) ~ "hVh - «,) = (c)
in which the only unknowns are v 2 and u 2i both functions of T 2 and P 2 . Since T z is un-
known, it means that there is only one value of T 2 for which Eq. (c) will be satisfied, and
we must find it by trial and error.
Solution
We have
v t = 0.5243 m 3 /kg, v h = = 0.763 kg
hi= 3040.4 kJ/kg, « L = 2548.9 kj/kg
Assume that
T 2 -■ 300°C
For this temperature and the known value of P 2 , we get
u 2 = 0.1823 mVkg, u 2 = 2785.2 kJ/kg
Substituting into (c), we obtain
^-^•(3040.4 - 2785.2) - 0.763(3040.4 - 2548.9) = +185.0 kJ
Now assume instead that
T 2 --= 350°C
For this temperature and the known P 2 , we get
v 2 = 0.2003 mVkg, u 2 = 2869.1 kj/kg
The Transient Process H 189
Substituting these values into (c), we pbtain
q|^(3040.4 - 2869.1) - 0.763(3040.4 - 2548.9) = ~32.9kJ
and we find that the actual T 2 must be between these two assumed values, in order that
(c) be equal to zero. By interpolation,
T 2 = 342°C and v 2 = 0.1974 mVkg
The final mass inside the tank is
and the mass of steam that flows into the tank is
m t = m 2 ~ m x = 2.026 - 0.763 - 1.263 kg
EXAMPLE 6. 13 A tank of 2 m 3 volume contains saturated ammonia at a temperature of 40°C. Initially the
tank contains 60% liquid and 50% vapor by volume. Vapor is withdrawn from the top of
the tank until the temperature is 10°C. Assuming that only vapor (i.e., no liquid) leaves
and that the process is adiabatic, calculate the mass of ammonia that is withdrawn.
Control volume: Tank.
Initial state: T Xi V liq> V wv ; state fixed.
Final state: T 2 .
Exit state: Saturated vapor (temperature changing).
Process: Transient,
Model: Ammonia tables.
Analysis
In the first law, Eq. 6.16, we note that Q c y = 0, W c v _ = 0, and m ; = 0, and we assume
that changes in kinetic and potential energy are negligible. However, the enthalpy of sat-
urated vapor varies with temperature, and therefore we cannot simply assume that the
enthalpy of the vapor leaving the tank remains constant. However, we note that at 40°C,
h g = 1470.2 kJ/kg and at 10°C, h g = 1452.0 kJ/kg. Since the change in h g during this
process is small, we may accurately assume that h e is the average of the two values
given above. Therefore,
(h e \v= 1461.1 kJ/kg
and the first law reduces to
m e h e + m 2 u 2 ~ =
and the continuity equation (from Eq. 6.15) becomes
(m 2 ~ mOc.v. + m e =
190 P chapter Six First-law analysis for a Control volume
Combining these two equations we have
m 2 (h e - u 2 ) = mih t ~
Solution
The following values are from the ammonia tables:
Vfi = 0.001 725 nrVkg, v gl = 0.083 13 mVkg
Vf2 = 0.001 60, v m = 0.203 81
u n = 368.7 kJ/kg, m x1 = 1341.0
1/^ = 226.0, % 2 = 1099.7
Calculating first the initial mass, m u in the tank, we find that the mass of the liquid
initially present, m^, is
^ ~ Vfl ~ 0.00 1 725 " 579 ' 7 kg
Similarly, the initial mass of vapor, m gU is
m *~*i~ odWi3 = 12 - 0kg
ni, = m n + m a i = 579.7 + 12.0 = 591.7 kg
m y h 9 = 591.7 X 1461.1 = 864 533 kJ
m x u x = (mu) A + (mu) gl = 579.7 X 368.7 + 12.0 X 1341.0
= 229 827 kJ
Substituting these into the first law, we obtain
m 2 (K ~ "2) = miK ~ JWi«i = 864 533 - 229 827 = 634 706
There are two unknowns, m 2 and u 2t in this equation. However,
= £ = 2.0
™ 2 v 2 0.001 60 +x 2 (0.203 81)
and
« 2 '= 226.0 + ^(1099.7)
and thus both are functions only of x 2 , the quality at the final state. Consequently,
2.0(1461.1 - 226.0 - 1099.7x 2 )
0.001 60 + 0.203 81*2
Solving for.v 2 we get
x 2 = 0.011 057
= 634 706
the Transient process H 191
Therefore,
v 2 = 0.001 60 + 0.011 057 X 0.203 81 - 0.003 853 5 m7kg
m 2
519 kg
"2 0.003 853 5
and the mass of ammonia withdrawn, m e , is
m e = m x - m 2 = 591.7 - 519 = 72.7 kg
EXAMPLE 6.13E A tank of 50 ft 3 volume contains saturated ammonia at a temperature of 100 F. Ini-
tially the tank contains 50% liquid and 50% vapor by volume. Vapor is withdrawn
from the top of the tank until the temperature is 50 F. Assuming that only vapor (i.e.,
no liquid) leaves and that the process is adiabattc, calculate the mass of ammonia that
is withdrawn.
Control volume: Tank.
Initial state: T u K Kq , V vsp ; state fixed.
Final state; T 2 .
Exit state; Saturated vapor (temperature changing).
Process: Transient.
Model: Ammonia tables.
Analysis
In the first law, Eq. 6.16, we note that gcv. = 0) ^c.v. = 0, and m s = 0, and we assume
that changes in kinetic and potential energy are negligible. However, the enthalpy of sat-
urated vapor varies with temperature, and therefore we cannot simply assume that the
enthalpy of the vapor leaving the tank remains constant. We note that at 100 F, h g =
631.8 Btu/lbm and at 50 F, h g = 624.26 Btu/lbm. Since the change in h g during this
process is small, we may accurately assume that h e is the average of the two values
given above. Therefore
QiXv Z = 628 Btu/lbm
and the first law reduces to
?njt e + ni 2 «2 — M\th ~
and the continuity equation (from Eq. 6, 15) is
(m 2 ~ Wi)c.v. + »i e =
192 ffl chapter six first-law analysis for a Control volume
Combining these two equations we have
The following values are from the ammonia tables:
u fl 0.0274 7 ftVlbm, v gl = 1.4168 ftVlbm
v n - 0.0256 4 ftVlbm y /s2 - 3.264 7 ftVlbm
u fx - 153.89 Btu/lbm, » gl = 576.23 Btu/lbm
lip = 97.16 Btu/lbm, u m = 472.78 Btu/lbm
Calculating first the initial mass, m x , in the tank, the mass of the liquid initially
present, m fX , is
"'" v, 0.0274- " [>,mlbm
Similarly, the initial mass of vapor, m gl , is
g 25
17.65 Ibm
gl ~ v gl ~ 1.416 8
ni! = + m gX = 910.08 + 17.65 = 927.73 lbm
m x h e = 921.11 X 628 =582 614 Btu
m x u x = (mu) A + (mu) gX = 915 X 149.9 + 16.6 + 577.0 X 146 700 Btu
Substituting these into the first law,
m z (h e - tt 2 ) - ?n l h e - m x it x = 582 614 - 150 223 = 432 391 Btu
There are two unknowns, m 2 and u 2 , in this equation. However,
„ V _ 50
7 " 2 "2 0.025 64 4- x 2 (3.264 7)
and
«2 =-97.16 + x 2 (472.78)
both functions only of x 2 , the quality at the final state. Consequently,
50(628 - 97.16 -x 2 472.78)
0.025 64 + 3.2647x 2
Solving, x 2 = 0.010 768
= 432 391
Summary ffl 193
Therefore,
v 2 = 0.025 64 4- 0.010 768 X 3.264 7 = 0.060 794 ftVlbm
»'-^ = raf794 = 822 - 4 ' bm
and the mass of ammonia withdrawn, m e> is
m, = m, - m 2 = 927.73 - 822.4 = 105.3 Ibm
iUjVIMARY conservation of mass is expressed as a rate of change of total mass due to mass flows in
or out of the control volume. The control mass energy equation is extended to include
mass flows that also carry energy (internal, kinetic, and potential) and the flow work
needed to push the flow in or out of the control volume against the prevailing pressure.
The conservation of mass (continuity equation) and the conservation of energy (first law)
are applied to a number of standard devices.
A steady-state device has no storage effects, with ail properties constant with time, and
constitutes the majority of all flow-type devices. A combination of several devices forms a
complete system built for a specific purpose, such as a power plant, jet engine, or refrigerator.
A transient process with a change in mass (storage) such as filling or emptying of a
container is considered based on an average description. It is also realized that the start-up
or shutdown of a steady-state device leads to a transient process.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Understand the physical meaning of the conservation equations. Rate = +in - out
• Understand the concepts of mass flow rate, volume flow rate, and local velocity
• Recognize the flow and nonflow terms in the energy equation
• Know how the most typical devices work and if they have heat or work transfers
• Have a sense about devices where kinetic and potential energies are important
• Analyze steady-state single-flow devices such as nozzles, throttles, turbines, or pumps
• Extend the application to a multiple-flow device such as a heat exchanger, mixing
chamber, or turbine, given the specific setup
• Apply the conservation equations to complete systems as a whole or to the individ-
ual devices and recognize their connections and interactions
• Recognize and use the proper form of the equations for transient problems
• Be able to assume a proper average value for any flow term in a transient
• Recognize the difference between storage of energy (dEldf) and flow (mh)
A number of steady-flow devices are listed in Table 6.1 with a very short statement
of the device's purpose, known facts about work and heat transfer, and a common as-
sumption if appropriate. This list is not complete with respect to the number of devices
nor with respect to the facts listed but is meant as a short list of typical devices, some of
which may be unfamiliar to many readers.
194 H CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
Table 6.1
Typical Steady-Flow Devices
Device
Purpose
Given
Assumption
Aftercooler
Cool a flow after a compressor
w =
P =
constant
Boiler
Bring substance to a vapor state
w =
P =
constant
Condenser
Take q out to bring substance to liquid state
w =
p =
constant
Combustor
Bum fuel; acts like heat transfer in
w =
P =
constant
Compressor
Bring a substance to higher pressure
tv in
1 =
Deaerator
Remove gases dissolved in liquids
w =
P =
constant
Dehumidifier
Remove water from air
P =
constant
D esup erheater
Add liquid water to superheated vapor steam to
make it saturated vapor
>v =
P =
constant
Diffuser
Convert KE energy to higher P
w =
q =
Economizer
Low-T, \ow-P heat exchanger
w =
P =
constant
Evaporator
Bring a substance to vapor state
w =
P =
constant
Expander
Similar to a turbine, but may have a q
Fan/blower
Move a substance, typically air
w in, KE up
p =
C,q=
Peed water heater
Heat liquid water with another flow
w =
P =
constant
Flash evaporator
Generate vapor by expansion (throttling)
w =
q =
A device that converts part of heat into work
q in, w out
Heat exchanger
Transfer heat from one medium to another
w =
P^
constant
Heat pump
A device moving a Q from 7] 0W to T^^, requires
a work input, refrigerator
w m
Heater
Heat a substance
w =
P =
constant
Humidifier
Add water to air-water mixture
w =
P =
constant
Intercooler
Heat exchanger between compressor stages
w=
P =
constant
Nozzle
Create KE; P drops
Measure flow rate
, w =
9 =
Mixing chamber
Mix two or more flows
w =
? =
Pump
Same as compressor, but handles liquid
w in, P up
9 =
Reactor
Allow reaction between two or more substances
w =
q =
t P= C
Regenerator
Usually a heat exchanger to recover energy
w =
P =
constant
Steam generator
Same as boiler, heat liquid water to superheat vapor
w =
p =
constant
Supercharger
A compressor driven by engine shaft work
to drive air into an automotive engine
w in
Superheater
A heat exchanger that brings Tvp over 7*^,
w=
P =
constant
Turbine
Create shaft work from high P flow
wout
9 =
Turbocharger
A compressor driven by an exhaust flow
turbine to charge air into an engine
Throttle
Same as valve
Valve
Control flow by restriction; P drops
W =
q =
Concept-Study Guide Problems H 195
Key concepts Volume flow Fate
and Formulas Massflowrate
Flow work rate
Flow direction
V ~ /V dA = AV (using average velocity)
m = Jfp V dA = pAV ~ AViu (using average values)
^fiow = PV = mPv
From higher P to lower P unless significant KE or PE.
Instantaneous Process
Continuity equation
Energy equation
Total enthalpy
>"c.v. = 2 ~ 2 f K
Eqn. = Qcv. ~ Wc.v. + 2 "hKn ~ 2 >»A
Aw = ^ + \ V 2 + gZ - /: stagoatioQ + gZ
Steady State
Continuity equation
Energy equation
Specific heat transfer
Specific work
Steady-state single flow
energy equation
No storage; m cv = 0; £ cv> =
2 yn i = 2 ™e ( m = out )
Qc.v. + 2 >"Aot; = ffc.v. + 2 m./itot,
# = Qcx/m (steady state only)
w = JF c . v /'« (steady state only)
? + *tot / = w + Aiot « (in = out)
(in — out)
Transient Process
Continuity equation
Energy equation
^2 - E\ = \Qi "1^2 + 2 '"Aow - 2 «A
^2
Ami
- ra 2 (« 2 4- 1 V| + g^) - m,(Ki + \ Vi + gZ,)
= A
tot exit average 2 ^hot el ^tot £2)
Concept-Study Guide problems
6.1 A mass flow rate into a control volume requires a
normal velocity component. Why?
6.2 A temperature difference drives a heat transfer.
Does a similar concept apply to ml
6.3 Can a steady-state device have boundary work?
6.4 Can you say something about changes in m and V
through a steady-flow device?
6.5 How does a nozzle or sprayhead generate kinetic
energy?
6.6 Liquid water at 15°C flows out of a nozzle straight
up 15 m. What is the nozzle V Mit ?
6.7 What is the difference between a nozzle flow and a
throttle process?
6.8 If you throttle a saturated liquid, what happens to
the fluid state? What if this is done to an ideal gas?
6.9 R-134a at 30°C, 800 kPa is throttled so that it be-
comes cold at - 10°C. What is the exit PI
6.10 Air at 500 K, 500 kPa is expanded to 100 kPa in
two steady-flow cases. Case 1 is a throttle and case
2 is a turbine. Which has the highest exit T? Why?
6.11 A turbine at the bottom of a dam has a flow of
liquid water through it. How does that produce
196 H Chapter Six First-Law Analysis for a Control Volume
power? Which terms in the energy equation are
important?
6.12 A windmill takes a fraction of the wind kinetic en-
ergy out as power on a shaft. In what manner does
the temperature and wind velocity influence the
power? Hint: Write the power as mass flow rate
times specific work.
6.13 If you compress air the temperature goes up. Why?
When the hot air, high P t flows in long pipes, it
eventually cools to ambient T. How does that
change the flow?
6.14 In a boiler you vaporize some liquid water at 100
kPa flowing at 1 m/s. What is the velocity of the
saturated vapor at 100 kPa if the pipe size is the
same? Can the flow then be constant PI
6.15 A mixing chamber has all flows at the same P, ne-
glecting losses. A heat exchanger has separate
flows exchanging energy, but they do not mix.
Why have both kinds?
6.16 In a coflowing (same direction) heat exchanger, 1
kg/s of air at 500 K flows into one channel and 2
kg/s of air flows into the neighboring channel at
300 K. If the heat exchanger is infinitely long, what
is the exit temperature? Sketch the variation of T in
the two flows.
6.17 Air at 600 K flows with 3 kg/s into a heat exchanger
and out at 100 o C. How much (kg/s) water coming in
at 100 kPa, 20°C can the air heat to the boiling point?
6.18 Steam at 500 kPa, 300°C is used to heat cold water
at 15°C to 75°C for a domestic hot water supply.
How much steam per kilogram of liquid water is
needed if the steam should not condense?
6.19 Air at 20 m/s, 260 K, 75 kPa flows at 5 kg/s into a
jet engine and flows out at 500 m/s, 800 K, 75 kPa.
What is the change (power) in flow of kinetic
energy?
6.20 An initially empty cylinder is filled with air from
20°C, 100 kPa until it is full. Assuming no heat
transfer, is the final temperature larger, equal to, or
smaller than 20°C? Does the final T depend on the
size of the cylinder?
6.21 A cylinder has 0.1 kg of air at 25°C, 200 kPa with a
5 -kg piston on top. A valve at the bottom is opened
to let the air out, and the piston drops 0.25 m to-
ward the bottom. What is the work involved hi this
process? What happens to the energy?
Homework Problems
Continuity Equation and Flow Rates
6.22 Air at 35°C and 105 kPa flows in a 100-mm X
1 50-mm rectangular duct in a heating system. The
volumetric flow rate is 0.015 m 3 /s. What is the ve-
locity of the air flowing in the duct and what is the
mass flow rate?
6.23 A boiler receives a constant flow of 5000 kg/h
liquid water at 5 MPa and 20°C, and it heats the
flow such that the exit state is 450°C with a pres-
sure of 4.5 MPa. Determine the necessary mini-
mum pipe flow area in both the inlet and exit
pipe(s) if there should be no velocities larger than
20 m/s.
6.24 An empty bath tub has its drain closed and is being
filled with water from the faucet at a rate of 10
kg/min. After 10 min the drain is opened and 4
kg/min flows out, and at the same time the inlet
flow is reduced to 2 kg/min. Plot the mass of the
water in the bathtub versus time and determine the
time from the very beginning when the tub will be
empty.
6.25 Nitrogen gas flowing in a 50-mm-diameter pipe at
15°C and 200 kPa, at the rate of 0.05 kg/s, encoun-
ters a partially closed valve. If there is a pressure
drop of 30 kPa across the valve and essentially no
temperature change, what are the velocities up-
stream and downstream of the valve?
6.26 Saturated vapor R-134a leaves the evaporator in a
heat pump system at 10°C } with a steady mass flow
rate of 0.1 kg/s. What is the smallest diameter tub-
ing that can be used at this location if the velocity
of the refrigerant is not to exceed 7 m/s?
6.27 A hot-air home heating system takes 0.25 m 3 /s air
at 100 kPa, 17°C into a furnace, heats it .to 52°C,
and delivers the flow to a square duct 0.2 m by 0.2
m at 1 10 kPa (see Fig. P6.27). What is the velocity
in the duct?
homework Problems H 197
FIGURE P6.27
6.28 Steam at 3 MPa and 400°C enters a turbine with a
volume flow rate of 5 m 3 /s. An extraction of 15%
of the inlet mass flow rate exits at 600 kPa and
200°C. The rest exits the turbine at 20 kPa with a
quality of 90% and a velocity of 20 m/s. Determine
the volume flow rate of the extraction flow and the
diameter of the final exit pipe.
6.29 A household fan of diameter 0.75 m takes air in at
98 kPa, 22°C and delivers it at 105 kPa, 23°C with
a velocity of 1.5 m/s (see Fig. P6.29). What are the
mass flow rate (kg/s), the inlet velocity, and the
outgoing volume flow rate in m 3 /s?
FIGURE P6.29
500 kPa and 350°C, and the flow is adiabatic. Find
- the nozzle exit velocity and the exit area.
6.32 Superheated vapor ammonia enters an insulated
nozzle at 20°C and 800 kPa, as shown in Fig.
P6.32, with a low velocity and at a steady rate of
0.01 kg/s. The ammonia exits at 300 kPa with a ve-
locity of 450 m/s. Determine the temperature (or
quality, if saturated) and the exit area of the nozzle.
FIGURE P6.32
6.33 In a jet engine a flow of air at 1000 K, 200 kPa, and
30 m/s enters a nozzle, as shown in Fig. P6.33,
where the air exits at 850 K, 90 kPa. What is the
exit velocity assuming no heat loss?
Diffuser Compressor Combustor Turbine Nozzle
FIGURE P6.33
Single-Flow, Single-Device Processes
Nozzles, Diffusers
6.30 Nitrogen gas flows into a convergent nozzle at 200
kPa and 400 K and very low velocity. It flows out
of the nozzle at 100 kPa and 330 K. If the nozzle is
insulated, find the exit velocity.
6.31 A nozzle receives 0.1 kg/s of steam at 1 MPa and
400°C with negligible kinetic energy. The exit is at
6.34 In a jet engine a flow of air at 1000 K, 200 kPa, and
40 m/s enters a nozzle, where the air exits at 500
m/s, 90 kPa. What is the exit temperature, assum-
ing no heat loss?
6.35 A sluice gate dams water up 5 m. A 1-cm-diameter
hole at the bottom of the gate allows liquid water at
20°C to come out. Neglect any changes in internal
energy and find the exit velocity and mass flow
rate.
198 ■ Chapter six first-Law analysis for a control Volume
]
6.36 A diffuser, shown in Fig. P6.36, has air entering at
100 kPa and 300 K with a velocity of 200 m/s. The
inlet cross-secttonal area of the diffuser is 100
mm 2 . At the exit, the area is 860 mm 2 , and the exit
velocity is 20 m/s. Detennine the exit pressure and
temperature of the air.
FIGURE P6.36
6.37 A diffuser receives an ideal-gas flow at 100 kPa
and 300 K with a velocity of 250 m/s, and the exit
velocity is 25 m/s. Determine the exit temperature
if the gas is argon, helium, or nitrogen.
6.38 Air flows into a diffuser at 300 mis, 300 K, and
100 kPa. At the exit the velocity is very small but
the pressure is high. Find the exit temperature as-
suming zero heat transfer.
6.39 The front of a jet engine acts as a diffuser, receiv-
ing air at 900 km/h, -5°C, and 50 kPa, bringing it
to 80 mis relative to the engine before entering the
compressor (see Fig. P6.39). If the flow area is re-
duced to 80% of the inlet area, find the temperature
and pressure in the compressor inlet.
FIGURE P6.39
Throttle Flow
6.40 Helium is throttled from 1.2 MPa and 20°C to a
pressure of 100 kPa. The diameter of the exit pipe
is so much larger than that of the inlet pipe that the
inlet and exit velocities are equal. Find the exit
temperature of the helium and the ratio of the pipe
diameters.
6.41 Saturated vapor R-134a at 500 kPa is throttled to
200 kPa in a steady flow through a valve. The ki-
netic energy in the inlet and exit flow is the same.
What is the exit temperature?
6.42 Saturated liquid R-12 at 25°C is throttled to 150.9
kPa in your refrigerator. What is the exit tempera-
ture? Find the percent increase in the volume flow
rate.
6.43 Water is flowing in a line at 400 kPa, and saturated
vapor is taken out through a valve to 100 kPa.
What is the temperature as it leaves the valve, as-
suming no changes in kinetic energy and no heat
transfer?
6.44 Liquid water at 180°C and 2000 kPa is throttled
into a flash evaporator chamber having a pressure
of 500 kPa. Neglect any change in the kinetic en-
ergy. What is the fraction of liquid and vapor in the
chamber?
6.45 Water at 1.5 MPa and 150°C is throttled adiabati-
cally through a valve to 200 kPa. The inlet velocity
is 5 m/s, and the inlet and exit pipe diameters are
the same. Deterrnine the state (neglecting kinetic
energy in the energy equation) and the velocity of
the water at the exit.
6.46 R-134a is throttled in a line flowing at 25°C and
750 kPa with negligible kinetic energy to a pres-
sure of 165 kPa. Find the exit temperature and the
ratio of exit pipe diameter to that of the inlet pipe
(Z^/Dia) so that the velocity stays constant.
6.47 Methane at 3 MPa, 300 K is throttled through a
valve to 100 kPa. Calculate the exit temperature
assuming no changes in the kinetic energy and
ideal-gas behavior. Repeat the answer for real-gas
behavior.
Turbines, Expanders
6.48 A steam turbine has an inlet of 2 kg/s water at 1000
kPa and 350°C with velocity of 15 m/s. The exit is
at 100 kPa, x = 1, and very low velocity. Find the
specific work and the power produced.
6.49 A small, high-speed turbine operating on com-
pressed air produces a power output of 100 W. The
inlet state is 400 kPa, 50°C, and the exit state is
150 kPa, — 30°C. Assuming the velocities to be
low and the process to be adiabatic, find the re-
quired mass flow rate of air through the turbine.
6.50 A liquid water turbine receives 2 kg/s water at
2000 kPa and 20°C with a velocity of 15 m/s. The
exit is at 100 kPa, 20°C, and very low velocity.
Find the specific work and the power produced.
homework Problems H 199
Hoover Dam
Hydraulic
turbine
FIGURE P6.51
6.51 Hoover Dam across the Colorado River dams up 6.56 The compressor of a large gas turbine receives air
Lake Mead 200 m higher than the river down- . from the ambient surroundings at 95 kPa and 20°C
stream (see Fig. P6.51). The electric generators dri- with a low velocity. At the compressor discharge,
ven by water-powered turbines deliver 1300 MW air exits at 1.52 MPa and 430°C with velocity of
of power. If the water is 17.5°C, find the minimum 90 m/s. The power input to the compressor is
amount of water running through the turbines. 5000 kW. Determine the mass flow rate of air
through the unit.
6.57 A compressor brings R-134a from 150 kPa, - 10°C
to 1200 kPa, 50°C. It is water cooled, with a heat
loss estimated as 40 kW } and the shaft work input
is measured to be 150 kW. How much is the mass
flow rate through the compressor?
6.58 An ordinary portable fan blows 0.2 kg/s of room
Colorado River air with a velocity of 18 m/s (see Fig. P6.29). What
is the minimum power electric motor that can drive
it? Hint: Are there any changes in P or T7
6.59 An air compressor takes in ah at 100 kPa and
1 TC, and delivers it at 1 MPa and 600 K to a con-
6.52 A windmill with rotor diameter of 30 m takes 40% start-pressure cooler, which the air exits at 300 K
of the kinetic energy out as shaft work on a day (see Fig. P6. 59). Find the specific compressor work
with 20°C and wind speed of 30 km/h. What power and the specific heat transfer in the cooler.
is produced?
6.53 A small turbine, shown in Fig. P6.53, is operated at
part load by throttling a 0.2 5 -kg/s steam supply at
1.4 MPa and 250°C down to 1.1 MPa before it en-
ters the turbine, and the exhaust is at 10 kPa. If the
turbine produces 1 10 kW, find the exhaust temper-
ature (and quality if saturated).
Compressor section Cooler section
FIGURE P6.59
6.60 A 4-kg/s steady flow of ammonia runs through a
device where it goes through a polytropic process.
The inlet state is 150 kPa, -20°C and the exit state
is 400 kPa, 80°C, where all kinetic and potential
energies can be neglected. The specific work input
has been found to be given as Crz-f) A(Py)
a. Find the polytropic exponent n.
b. Find the specific work and the specific heat
transfer.
6.61 An exhaust fan in a building should be able to
move 2.5 kg/s of air at 98 kPa and 20°C through a
0.4-m-diameter vent hole. How high a velocity
must it generate and how much power is required
to do that?
6.62 How much power is needed to run the fan in Prob-
lem 6.29?
FIGURE P6.53
6.54 A small expander (a turbine with heat transfer) has
0.05 kg/s helium entering at 1000 kPa, 550 K and it
leaves at 250 kPa, 300 K. The power output on the
shaft is measured to 55 kW. Find the rate of heat
transfer neglecting kinetic energies.
Compressors, Fans
6.55 A compressor in a commercial refrigerator receives
R-22 at -25°C and x = 1. The exit is at 800 kPa
and 40°C. Neglect kinetic energies and find the
specific work.
200 m CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL. VOLUME
Heaters, Coolers
6.63 Carbon dioxide enters a steady-state, steady-flow
heater at 300 fcPa and 15°C, and exits at 275 kPa
and 1200°C, as shown in Fig, P6.63. Changes in ki-
netic and potential energies are negligible. Calcu-
late the required heat transfer per kilogram of
carbon dioxide flowing through the heater.
CO,
U
ft
FIGURE P6.63
6,64 A condenser (cooler) receives 0.05 kg/s of R-22 at
800 kPa and 40°C, and cools it to 15°C. There is a
small pressure drop so that the exit state is satu-
rated liquid. What cooling capacity (k\V) must the
condenser have?
6.65. A chiller cools liquid water for air-conditioning
purposes. Assume 2.5 kg/s water at 20°C and 100
kPa is cooled to 5°C in a chiller. How much heat
transfer (kW) is needed?
6.66 Saturated liquid nitrogen at 500 kPa enters a boiler
at a rate of 0.005 kg/s and exits as saturated vapor
(see Fig. P6.66). It then flows into a superheater,
also at 500 kPa, where it exits at 500 kPa and
275 K. Find the rate of heat transfer in the boiler
and in the superheater.
Boiler
Super heater
Q super heater
Q boiler
FIGURE P6.66
6.67 In a steam generator, compressed liquid water at
10 MPa and 30°C enters a 30-mm-diameter tube
at a rate of 3 L/s. Steam at 9 MPa and 400°C
exits the tube. Find the rate of heat transfer to the
water.
6.68 The air conditioner in a house or a car has a cooler
that brings atmospheric air from 30°C to 10°C with
both states at 101 kPa. If the flow rate is 0.5 kg/s,
find the rate of heat transfer.
6.69 A flow of liquid glycerine flows around an en-
gine, cooling it as it absorbs energy. The glycer-
ine enters the engine at 60°C and receives 9 k\V
of heat transfer. What is the required mass flow
rate if the glycerine should come out at maximum
95°C?
6.70 Liquid nitrogen at 90 K and 400 kPa flows into a
probe used in cryogenic survey. In the return line
the nitrogen is then at 160 K and 400 kPa. Find the
specific heat transfer to the nitrogen. If the return
line has a cross-sectional area 100 times larger than
that of the inlet line, what is the ratio of the return
velocity to the inlet velocity?
Pumps, Pipe and Channel Flows
6.71 A small stream with 20°C water runs out over a
cliff, creating a 100-m-taiI waterfall. Estimate the
downstream temperature when you neglect the hor-
izontal flow velocities upstream and downstream
from the waterfall. How fast was the water drop-
ping just before it splashed into the pool at the bot-
tom of the waterfall?
6.72 A small water pump is used in an irrigation system.
The pump takes water in from a river at 10°C and
100 kPa at a rate of 5 kg/s. The exit line enters a
pipe that goes up to an elevation 20 m above the
pump and river, where the water runs into an open
channel. Assume the process is adiabatic and that
the water stays at 10°C. Find the required pump
work.
6.73 A steam pipe for a 300-m-talI building receives su-
perheated steam at 200 kPa at ground level. At the
top floor the pressure is 125 kPa, and the heat loss
in the pipe is 1 10 kJ/kg. What should the inlet tem-
perature be so that no water will condense inside
the pipe?
6.74 The main water line into a tall building has a
pressure of 600 kPa at 5 m below ground level, as
shown in Fig. P6.74. A pump brings the pressure
up so the water can be delivered at 200 kPa at the
top floor 150 m above ground level. Assume a
flow rate of 10 kg/s liquid water at 10°C and ne-
glect any difference in kinetic energy and inter-
nal energy u. Find the pump work.
HOMEWORK PROBLEMS M 201
150 rn
...-i.^Y^V--:^
Top floor
Ground
Water main ' ! — ^ l^'
Pump
FIGURE P6.74
6.75 Consider a water pump that receives liquid water at
15°C and 100 kPa, and delivers it to a same diame-
ter short pipe having a nozzle with exit diameter of
1 cm (0.01 m) to the atmosphere at 100 kPa (see
Fig. P6.75). Neglect the kinetic energy in the pipes
and assume constant u for the water. Find the exit
velocity and the mass flow rate if the pump draws 1
kW of power.
MPa and 350°C, and the rest exits the turbine at
. 75 kPa, with 95% quality. Assuming no heat trans-
fer and no changes in kinetic energy, find the total
turbine power output.
Steam
FIGURE P6.78
6.79 A steam turbine receives steam from two boilers
(see Fig. P6.79). One flow is 5 kg/s at 3 MPa and
700°C, and the other flow is 15 kg/s at 800 kPa
and 50O°C. The exit state is 10 kPa, with a qual-
ity of 96%. Find the total power out of the adia-
batic turbine.
Pump
FIGURE P6.75
6.76 A cutting tool uses a nozzle that generates a high-
speed jet of liquid water. Assume an exit velocity
of 1000 m/s of 20°C liquid water with a jet diame-
ter of 2 mm (0.002 m). How much mass flow rate
is this? What size (power) pump is needed to gen-
erate this from a steady supply of 20°C liquid water
at 200 kPa?
6.77 A pipe flows water at 15°C from one building to
another. In the winter time the pipe loses an esti-
mated 500 W of heat transfer. What is the mini-
mum required mass flow rate that will ensure that
the water does not freeze (i.e., reach 0°C)?
Multiple-Flow, Single-Device Processes
Turbines, Compressors, Expanders
6.78 A steam turbine receives water at 15 MPa and
600°C at a rate of 100 kg/s, as shown in Fig. P6.78.
In the middle section 20 kg/s is withdrawn at 2
© FIGURE P6.79
6.80 Two steady flows of air enter a control volume,
shown in Fig. P6.80. One is 0.025 kg/s of flow at
350 kPa, 150°C (state 1), and the other enters at
450 kPa, 15°C (state 2). A single flow exits at 100
kPa, -40°C (state 3). The control volume rejects
1 kW of heat to the surroundings and produces
4 kW of power output. Neglect kinetic energies
and determine the mass flow rate at state 2.
-^reject ^ W
FIGURE P6.80
202 M CHAPTER SIX FIRST-LAW ANALYSIS FOR A CONTROL VOLUME
6.81 A large steady expansion engine has two low-
velocity flows of water entering. High-pressure
steam enters at point 1 with 2.0 kg/s at 2 MPa and
500°C, and 0.5 kg/s of cooling water at 120 kPa
and 30°C centers at point 2. A single flow exits at
point 3, with 150 kPa and 80% quality, through a
0.15-m-diameter exhaust pipe. There ts a heat loss
of 300 kW. Find the exhaust velocity and the
power output of the engine.
6.82 Cogeneration is often used where a steam supply
is needed for industrial process energy. Assume a
supply of 5 kg/s steam at 0.5 MPa is needed.
Rather than generating this from a pump and
boiler, the setup in Fig. P6.82 is used to extract
the supply from the high-pressure turbine. Find
the power the turbine now cogenerates in this
process.
20 kg/s
supply
>W Tu
15 kg/s
to condenser
FIGURE P6.84
©
Lake waEer
6.85 A cooler in an air conditioner brings 0.5 kg/s of air
at 35°C to 5°C, both at 101 kPa. It then mixes the
output with a flow of 0.25 kg/s air at 20°C and 101
kPa, sending the combined flow into a duct. Find
the total heat transfer in the cooler and the temper-
ature in the duct flow.
6.86 A heat exchanger, shown in Fig. P6.86, is used to
cool an air flow from 800 to 360 K, with both
states at 1 MPa. The coolant is a water flow at 15°C
and 0.1 MPa. If the water leaves as saturated vapor,
find the ratio of the flow rates m w Jrn i]x .
H,0
FIGURE P6.S2
FIGURE P6.86
6.83 A compressor receives 0.1 kg/s of R-134a at 150
kPa, - 10°C and delivers it at 1000 kPa, 40°C. The
power input is measured to be 3 kW. The compres-
sor has heat transfer to air at 100 kPa coming in at
20°C and leaving at 25°C. How much is the mass
flow rate of air?
Heat Exchangers
6.84 A condenser (heat exchanger) brings 1 kg/s water
flow at 10 kPa from 300°C to saturated liquid at 10
kPa, as shown in Fig. P6.84. The cooling is done
by lake water at 20°C that returns to the lake at
30°C. For an insulated condenser, find the flow rate
of cooling water.
6.87 A superheater brings 2.5 kg/s of saturated water
vapor at 2 MPa to 450°C. The energy is provided
by hot air at 1200 K flowing outside the steam tube
in the opposite direction as the water, a setup
known as a counterflowing heat exchanger (similar
to Fig. P6.86). Find the smallest possible mass flow
rate of the air to ensure that its exit temperature is
20°C larger than the incoming water temperature.
6.88 An automotive radiator has glycerine at 95°C enter
and return at 55°C as shown in Fig. P6.88. Air
flows in at 20°C and leaves at 25°C. If the radiator
should transfer 25 kW, what is the mass' flow rate
of the glycerine and what is the volume flow rate of
air in at 100 kPa?
Homework Problems M 203
[ n Glycerine in
X
X
Glycerine out
Air in -
•Air out
FIGURE P6.88
6.89 A two-fluid heat exchanger has 2 kg/s of liquid
ammonia at 20°C, 1003 kPa entering at state 3
and exiting at state 4. It is heated by a flow of
1 kg/s nitrogen at 1500 K, state 1, and leaving at
600 K, state 2, similar to Fig. P6.86. Find the
total rate of heat transfer inside the heat ex-
changer. Sketch the temperature versus distance
for the ammonia and find state 4 (7*, v) of the
ammonia.
6.90 A copper wire has been heat treated to 1000 K and
is now pulled into a cooling chamber that has 1.5
kg/s of air coming in at 20°C; the air leaves the
other end at 60°C. If the wire moves 0.25 kg/s of
copper, how hot is the copper as it comes out?
Mixing Processes
6.91 An open feedwater heater in a power plant heats 4
kg/s water at 45°C and 100 kPa by mixing it with
steam from the turbine at 100 kPa and 250°C as in
Fig. P6.91. Assume the exit flow is saturated liquid
at the given pressure and find the mass flow rate '
from the turbine.
Mixing "
chamber
6.92 A desuperheater mixes superheated water vapor
. with liquid water in a ratio that produces saturated
water vapor as output without any external heat
transfer. A flow of 0.5 kg/s superheated vapor at
5 MPa and 400°C J and a flow of liquid water at
5 MPa and 40°C enters a desuperheater. If satu-
rated water vapor at 4.5 MPa is produced, deter-
mine the flow rate of the liquid water.
6.93 Two air flows are combined to a single flow. One
flow is 1 m 3 /s at 20°C and the other is 2 m 3 /s at
200°C, both at 100 kPa as in Fig. P6.93. They mix
without any heat transfer to produce an exit flow at
100 kPa. Neglect kinetic energies and find the exit
temperature and volume flow rate.
FIGURE P6.93
6,94 A mixing chamber with heat transfer receives 2 kg/s
of R-22 at 1 MPa and 40°C in one line and 1 kg/s of
R-22 at 30°C with a quality of 50% in a line with a
valve. The outgoing flow is at 1 MPa and 60°C.
Find the rate of heat transfer to the mixing chamber.
FIGURE P6.91
© ^
FIGURE P6.94
6.95 Two flows are mixed to form a single flow. Flow at
state 1 is 1.5 kg/s of water at 400 kPa and 200°C,
and flow at state 2 is at 500 kPa and 100°C. Which
mass flow rate at state 2 will produce an exit T 2 =
1 50°C if the exit pressure is kept at 300 kPa? -
6.96 An insulated mixing chamber receives 2 kg/s of R-
134a at 1 MPa and 100°C in a line with low veloc-
ity. Another line with R-134a as saturated liquid at
60°C flows through a valve to the mixing chamber
at 1 MPa after the valve, as shown in Fig. P6.94.
The exit flow is saturated vapor at 1 MPa flowing
at 20 m/s. Find the flow rate for the second line.
6.97 To keep a jet engine cool, some intake air bypasses
the combustion chamber. Assume 2 kg/s of hot air
204 U CHAPTER NUMBER CHAPTER TITLE
at 2000 K and 500 kPa is mixed with 1 .5 kg/s air at
500 K, 500 kPa without any external heat transfer
as in Fig. P6.97. Find the exit temperature using
constant heat capacity from Table A.5.
FIGURE P6.97
6.98 Solve the previous problem using values from
Table A.7.
Multiple Devices, Cycle Processes
6.99 The following data are for a simple steam power
plant as shown in Fig. P6.99. State 6 has x 6 = 0.92
and velocity of 200 m/s. The rate of steam flow is
25 kg/s, with 300 kW of power input to the pump.
Piping diameters are 200 mm from the steam gen-
erator to the turbine and 75 mm from the condenser
to the economizer and steam generator. Determine
the velocity at state 5 and the power output of the
turbine.
State 1 2 3 4 5 6 7
P } kPa 6200 6100 5900 5700 5500 10 9
T^C 45 175 500 490 40
h, kJ/kg 194 744 3426 3404 168
FIGURE P6.99
6.100 For the steam power plant shown in Problem
6.99, assume the cooling water comes from a
lake at 15°C and is returned at 25°C. Deter-
mine the rate of heat transfer in the condenser
and the mass flow rate of cooling water from
the lake.
6.101 For the steam power plant shown in Problem
6.99, determine the rate of heat transfer in the
economizer, which is a low-temperature heat ex-
changer. Also find the rate of heat transfer needed
in the steam generator.
6.102 A somewhat simplified flow diagram for a nuclear
power plant is given in Fig. P6.102. Mass flow
rates and the various states in the cycle are shown
in the accompanying table.
Point
mi, kg/s
i>,kPa
A, kJ/kg
1
75.6
7240
sat vap
2
75.6
6900
2765
3
62.874
345
2517
4
310
5
7
2279
6
75.6
7
33
138
7
415
140
8
2.772
35
2459
9
4.662
310
558
10
35
34
142
11
75.6
380
68
285
12
8.064
345
2517
13
75.6
330
14
349
15
4.662
965
139
584
16.
75.6
7930
565
17
4.662
965
2593
18
75.6
7580
688
19
1386
7240
277
1220
20
1386
7410
1221
21
1386
7310
The cycle includes a number of heaters in
which heat is transferred from steam, taken out of
the turbine at some intermediate pressure, to liq-
uid water pumped from the condenser on its way
to the steam drum. The heat exchanger in the re-
actor supplies 157 MW, and it may be assumed
that there is no heat transfer in the turbines.
HOMEWORK PROBLEMS H 205
pressure Condensate
heater pump FIGURE P6.102
a. Assuming the moisture separator has no heat
transfer between the two turbine sections, de-
termine the enthalpy and quality (7i 4l ,r 4 ).
b. Determine the power output of the low-pressure
turbine.
c. Detennine the power output of the high-pressure
turbine.
d. Find the ratio of the total power output of the
two turbines to the total power delivered by the
reactor.
6.103 Consider the power plant as described in the pre-
vious problem.
a. Determine the quality of the steam leaving the
reactor.
b. What is the power to the pump that feeds water
to the reactor?
6.104 A gas turbine set-up to produce power during peak
demand is shown in Fig. P6. 104. The turbine pro-
vides power to the air compressor and the electric
generator. If the electric generator should provide
5 MW what is the needed air flow at state 1 and
the combustion heat transfer between states 2 and
3? The states are
1. 90kPa,290K
2. 900kPa,560K
3. 900 kPa, 1400 K
4. 100 kPa, 850 K
FIGURE P6.104
206 9 CHAPTER SIX first- Law analysis for a CONTROL VOLUME
6.105 A proposal is made to use a geothermal supply
of hot water to operate a steam turbine, as shown
in Fig. P6.105. The high-pressure water at 1.5
MPa and 180°C is throttled into a flash evapora-
tor chamber, which forms liquid and vapor at a
lower pressure of 40 kPa, The liquid is discarded
while the saturated vapor feeds the turbine and
exits at 10 kPa with a 90% quality. If the turbine
should produce 1 MW, find the required mass
flow rate of hot geothermal water in kilograms
per hour.
Condenser
©
Hot water
Gevap from cold
outside air
FIGURE P6.106
// /• V : V V N
® vapor out
/■/-VFtasK.: \ 'i
(evaporator.
4
Turbine
1
Saturated
liquid out
w
Exhaust
FIGURE P6.105
6.106 An R-12 heat pump cycle shown in Fig. P6.106
has an R-12 flow rate of 0.05 kg/s with 4 kW into
the compressor. The following data are given:
State
1
P, kPa 1250
T, °C 120
h, kJ/kg 260
1230
110
253
1200
45
79.7
320
300
188
290
5
191
6.107 A modern jet engine has a temperature after com-
bustion of about 1500 K at 3200 kPa as it enters
the turbine section (see state 3, Fig. P6.107). The
compressor inlet is at 80 kPa, 260 K (state 1) and
the outlet (state 2) is at 3300 kPa, 780 K; the tur-
bine outlet (state 4) into the nozzle is at 400 kPa,
900 K and the nozzle exit (state 5) is at 80 kPa f
640 K. Neglect any heat transfer and neglect ki-
netic energy except out of the nozzle. Find the
compressor and turbine specific work terms and
the nozzle exit velocity.
Compressor
Combustors
Turbine
Diffuser
FIGURE P6.107
Nozzle
Calculate the heat transfer from the compres-
sor, the heat transfer from the R-12 in the con-
denser, and the heat transfer to the R-12 in the
evaporator.
Transient Processes
6.108 A l-m 3 , 40-kg rigid steel tank contains air at 500
kPa, and both tank and air are at 20°C. The tank is
Homework, problems I! 207
connected to a line flowing air at 2 MPa and
20°C. The valve is opened, allowing air to flow
into the tank until the pressure reaches 1.5 MPa,
and is then closed. Assume the air and tank are al-
ways at the same temperature and the final tem-
perature is 35 C. Find the final air mass and the
heat transfer.
6.109 An evacuated 150-L tank is connected to a line
flowing air at room temperature, 25°C, and 8 MPa
pressure. The valve is opened, allowing air to
flow into the tank until the pressure inside is
6 MPa. At this point the valve is closed. This fill-
ing process occurs rapidly and is essentially adia-
batic. The tank is then placed in storage where it
eventually returns to room temperature. What is
the final pressure?
6.U0 An initially empty bottle is filled with water from
a line at 0.8 MPa and 350°C. Assume no heat
transfer and that the bottle is closed when the
pressure reaches the line pressure. If the final
mass is 0.75 kg, find the final temperature and the
volume of the bottle.
6.111 A 25-L tank, shown in Fig. P6.ll 1, that is initially
evacuated is connected by a valve to an air supply
line flowing air at 20°C and 800 kPa. The valve is
opened, and air flows into the tank until the pres-
sure reaches 600 kPa. Determine the final temper-
ature and mass inside the tank, assuming the
process is adiabatic. Develop an expression for the
relation between the line temperature and the final
temperature using constant specific heats.
Air
supply
line
mm?////////////*
FIGURE P6.111
6.112 Helium in a steel tank is at 250 kPa, 300 K with a
volume of 0.1 m 3 . It is used to fill a balloon.
When the tank pressure drops to 1 50 kPa, the flow
of helium stops by itself. If all the helium still is
at 300 K how big a balloon did I get? Assume the
pressure in the balloon varies linearly with vol-
• ume from 100 kPa (V = 0) to the final 150 kPa.
How much heat transfer took place?
6.113 A rigid I00-L tank contains air at 1 MPa and
200°C. A valve on the tank is now opened, and air
flows out until the pressure drops to 100 kPa.
During this process, heat is transferred from a
heat source at 200°C, such that when the valve is
closed, the temperature inside the tank is 50°C.
What is the heat transfer?
6.114 A 1-m 3 tank contains ammonia at 150 kPa and
25°C. The tank is attached to a line flowing am-
monia at 1200 kPa and 60°C. The valve is
opened, and mass flows in until the tank is half
full of liquid (by volume) at 25°C. Calculate the
heat transferred from the tank during this process.
6.115 An empty canister of volume 1 L is filled with
R-I34a from a line flowing saturated liquid
R-134a at 0°C. The filling is done quickly, so it is
adiabatic. How much mass of R-134a is there
after filling? The canister is placed on a storage
shelf, where it slowly heats up to room tempera-
ture 20°C. What is the final pressure?
6.116 A piston/cyclinder assembly contains 1 kg of
water at 20°C with a constant load on the piston
such that the presstire is 250 kPa. A nozzle in a
line to the cylinder is opened to enable a flow to
the outside atmosphere at 100 kPa. The process
continues until half the mass has flowed out and
there is no heat transfer. Assume constant water
temperature and find the exit velocity and total
work done in the process.
6.117 A 200-L tank (see Fig. P6.117) initially contains
water at 100 kPa and a quality of 1%. Heat is
■Vapor
Q Q
FIGURE P6.117
208 ffl Chapter six first-Law Analysis fob. a control Volume
transferred to the water, thereby raising its pres-
sure and temperature. At a pressure of 2 MPa a
safety valve opens and saturated vapor at 2 MPa
flows out. The process continues, maintaining
2 MPa inside until the quality in the tank is 90%,
then stops. Determine the total mass of water that
flowed out and the total heat transfer.
6.118 A 100-L rigid tank contains carbon dioxide gas at
1 MPa and 300 K. A valve is cracked open, and
carbon dioxide escapes slowly until the tank pres-
sure has dropped to 500 kPa. At this point the
valve is closed. The gas remaining inside the tank
may be assumed to have undergone a polytropic
expansion, with polytropic exponent n = 1.15.
Find the final mass inside and the heat transferred
to the tank during the process.
6.119 A nitrogen line, at 300 K and 0.5 MPa, shown in
Fig. P6.119, is connected to a turbine that ex-
hausts to a closed initially empty tank of 50 m 3 .
The turbine operates to a tank pressure of 0.5
MPa, at which point the temperature is 250 K.
Assuming the entire process is adiabatic, deter-
mine the turbine work.
=$>w T
Tank
FIGURE P6.119
6.120 A 2-m-tall cyclinder has a small hole in the bot-
tom as in Fig. P6.120. It is filled with liquid water
1 m high, on top of which is a 1-m-high air col-
umn at atmospheric pressure of 100 kPa. As the
liquid water near the hole has a higher P than
100 kPa, it runs out. Assume a slow process with
constant T. Will the flow ever stop? When?
6.121 A 2-m 3 insulated vessel, shown in Fig. P6.121,
contains saturated vapor steam at 4 MPa. A valve
on the top of the tank is opened, and steam is al-
lowed to escape. During the process any liquid
formed collects at the bottom of the vessel, so
only saturated vapor exits. Calculate the total
mass that has escaped when the pressure inside
reaches 1 MPa.
'////////////. V//////////A
HO
FIGURE P6.121
6.122 A 750-L rigid tank, shown in Fig. P6.122, ini-
tially contains water at 250 o C, which is 50% liq-
uid and 50% vapor, by volume. A valve at the
bottom of the tank is opened, and liquid is slowly
withdrawn. Heat transfer takes place such that the
temperature remains constant. Find the amount of
heat transfer required to reach the state where half
the initial mass is withdrawn.
1 m
1 m
FIGURE P6.120
FIGURE P6.122
6.123 Consider the previous problem but let the line and
valve be located in the top of the tank. Now satu-
rated vapor is slowly withdrawn while heat trans-
fer keeps the temperature inside constant. Find
the heat transfer required to reach a state where
half the original mass is withdrawn.
HOMEWORK PROBLEMS 13 209
Review Problems
6.124 A flow of 2 kg/s of water at 500 kPa and 20°C is
heated in a constant-pressure process to 1700°C.
Find the best estimate for the rate of heat transfer
needed.
6.125 In a glass factory a 2-m-wide sheet of glass at
1500 K comes out of the final rollers, which fix
the thickness at 5 mm with a speed of 0.5 m/s (see
Fig. P6.I25). Cooling air in the amount of 20 kg/s
conies in at 17°C from a slot 2 m wide and flows
parallel with the glass. Suppose this setup is very
long so the glass and air comes to nearly the same
temperature (a coflowing heat exchanger) what is
the exit temperature?
Air out
f
Moving glass sheet
FIGURE P6.125
b. Find the power to the condensate pump.
c. Do the energy terms balance for the low-
pressure heater or is there a heat transfer not
shown?
6.130 A 500-L insulated tank contains air at 40°C and
2 MPa. A valve on the tank is opened, and air es-
capes until half the original mass is gone, at
which point the valve is closed. What is the pres-
sure inside then?
6.131 A steam engine based on a turbine is shown in
Fig. P6.13 1 . The boiler tank has a volume of 100
L and initially contains saturated liquid with a
very small amount of vapor at 100 kPa. Heat is
now added by the burner. The pressure regulator,
which keeps the pressure constant, does not open
before the boiler pressure reaches 700 kPa. The
saturated vapor enters the turbine at 700 kPa and
is discharged to the atmosphere as saturated vapor
at 100 kPa. The burner is turned off when no
more liquid is present in the boiler. Find the total
turbine work and the total heat transfer to the
boiler for this process.
6.126 Assume a setup similar to the previous problem
but with the air flowing in the opposite direction
as the glass— it comes in where the glass goes
out. How much air flow at 17°C is required
to cool the glass to 450 K assuming the air must
be at least 120 K cooler than the glass at any
location?
6.127 Three air flows, all at 200 kPa, are connected to
the same exit duct and mix without external heat
transfer. Flow 1 has 1 kg/s at 400 K, flow 2 has
3 kg/s at 290 K, and flow 3 has 2 kg/s at 700 K.
Neglect kinetic energies and find the volume flow
rate in the exit flow.
6.128 Consider the power plant as described in Problem '
6.102.
a. Determine the temperature of the water leaving
the intermediate pressure heater, T n , assuming
no heat transfer to the surroundings.
b. Determine the pump work between states 13
and 16.
6.129 Consider the power plant as described in Problem
6.102.
a. Find the power removed in the condenser by
the cooling water (not shown).
5
FIGURE P6.131
6.132 An insulated spring-loaded piston/cyclinder de-
vice, shown in Fig. P6.132, is connected to an air
Air
supply
line
FIGURE P6.132
210 B chapter six first-law Analysis for a control volume
6.133
line flowing air at 600 kPa and 700 K by a valve.
Initially the cylinder is empty and the spring force
is zero. The valve is then opened until the
cylinder pressure reaches 300 kPa. By noting that
»2 = «iine + C V (T 2 - and h ]bK - u ]im = RTfa,
find an expression for T 2 as a function of P 2> P 0>
and 7* line . With P = 100 kPa, find T 2 .
A mass-loaded piston/cylinder shown in Fig.
P6.133, containing air is at 300 kPa, 17°C with a
volume of 0.25 m 3 , while at the stops V = 1 m 3 . An
air line, 500 kPa, 600 K, is connected by a valve
that is then opened until a final inside pressure of
400 kPa is reached, at which point T = 350 K. Find
the air mass that enters, the work, and the heat
transfer.
Air
Air
supply
line
FIGURE P6.133
6.134 A 2-m 3 storage tank contains 95% liquid and 5%
vapor by volume of liquified natural gas (LNG) at
160 K, as shown in Fig. P6.134. It may be as-
sumed that LNG has the same properties as pure
methane. Heat is transferred to the tank and satu-
rated vapor at 160 K flows into the steady flow
Pressure
regulator
Heater
Storage
tank
FIGURE P6.134
heater, which tt leaves at 300 K. The process con-
tinues until all the liquid in the storage tank is
gone. Calculate the total amount of heat transfer
to the tank and the total amount of heat trans-
ferred to the heater.
Heat Transfer Problems
6.135 Liquid water at 80°C flows with 0.2 kg/s inside a
square duct 2 cm on a side, insulated with a 1 -cm-
thick layer of foam, k = 0. 1 W/m K. If the outside
foam surface is at 25°C, how much has the water
temperature dropped for 10 m length of duct?
Neglect the duct material and any comer effects
(A - 4 sL).
6.136 A counterflowing heat exchanger conserves energy
by heating cold outside fresh air at 10°C with the
outgoing combustion gas (air) at 100°C as in Fig.
P6. 136. Assume both flows are 1 kg/s and the tem-
perature difference between the flows at any point
is 50°C. What is the incoming fresh air tempera-
ture after the heat exchanger? What is the equiva-
lent (single) convective heat transfer coefficient
between the flows if the interface area is 2 m 2 ?
1
Hot gas — |— 1»-
4
Outside air
Wall
FIGURE P6.136
6.137 Saturated liquid water at 1000 kPa flows at 2 kg/s
inside a 10-cm-outer-diameter steel pipe, and out-
side of the pipe is a flow of hot gases at 1000 K
with a convection coefficient of h = 150 W/m 2 K.
Neglect any AT in the steel and any inside con-
vection h t and find the length of pipe needed to
bring the water to saturated vapor,
6.138 A flow of 1000 K, 100 kPa air with 0.5 kg/s in a
furnace flows over a steel plate of surface temper-
ature 400 K. The flow is such that the convective
heat transfer coefficient is A = 125 W/m 2 K. How
much of a surface area does the air have to flow
over to exit with a temperature of 800 K? How
about 600 K?
ENGLISH UNIT PROBLEMS M 211
English Unit Problems
6.139E Liquid water at 60 F flows out of a nozzle
straight up 40 ft. What is the nozzle V Mit ?
6.140E R-134a at 90 F, 125 psia is throttled so that it
becomes cold at 10 F, What is the exit P?
6.141E In a boiler you vaporize some liquid water at
103 psia flowing at 3 ft/s. What is the velocity of
the saturated vapor at 103 psia if the pipe size is
the same? Can the flow then be constant F?
6.142E Air at 60 ft/s, 480 R, 11 psia flows at 10 Ibm/s
into a jet engine and flows out at 1500 ft/s, 1440
R, 1 1 psia. What is the change (power) in flow
of kinetic energy?
6.143E An initially empty cylinder is filled with air from
70 F, 15 psia until it is full. Assuming no heat
transfer, is the final temperature larger, equal to
or smaller than 70 F? Does the final T depend on
the size of the cyclinder?
6.144E Air at 95 F, 16 lbf/in 2 flows in a 4-in. X 6-in.
rectangular duct in a heating system. The volu-
metric flow rate is 30 cfm (ftVmin). What is the
velocity of the air flowing in the duct?
6.145E A hot-air home heating system takes 500 ftVmin
(cfm) air at 14.7 psia, 65 F into a furnace and
heats it to 130 F and delivers the flow to a
square duct 0.5 ft by 0.5 ft at 15 psia. What is
the velocity in the duct?
6.14 6E Saturated vapor R-134a leaves the evaporator
in a heat pump at 50 F, with a steady mass
flow rate of 0.2 lbm/s. What is the smallest di-
ameter tubing that can be used at this location
if the velocity of the refrigerant is not to ex-
ceed 20 ft/s?
6.147E A pump takes 40 F liquid water from a river at
14 lbf/in 2 and pumps it up to an irrigation canal
60 ft higher than the river surface. All pipes
have a diameter of 4 in., and the flow rate is
35 lbm/s. Assume the pump exit pressure is just
enough to carry a water column of the 60 ft
height with 15 lbf/in 2 at the top. Find the flow
work into and out of the pump and the kinetic
energy in the flow.
6.148E In a jet engine a flow of air at 1800 R, 30 psia,
and 90 ft/s enters a nozzle, where the air exits at
1500 R, 13 psia, as shown in Fig. P6.33. What is
the exit velocity assuming no heat loss?
6.149E Nitrogen gas flows into a convergent nozzle at
30 lbf/in 2 , 600 R and very low velocity. It flows
out of the nozzle at 15 lbf/in 2 , 500 R. If the noz-
zle is insulated, find the exit velocity.
6.150E A diffuser, shown in Fig. P6.36, has air entering
at 14.7 lbf/in 2 , 540 R, with a velocity of 600 ft/s.
The inlet cross-sectional air of the diffuser is 0.2
in 2 . At the exit, the area is 1.75 in 2 , and the exit
velocity is 60 ft/s. Determine the exit pressure
and temperature of the air,
6.151E Helium is throttled from 175 lbf/in 2 , 70 F, to a
pressure of 15 lbf/in 2 . The diameter of the exit
pipe is so much larger than the inlet pipe that the
inlet and exit velocities are equal. Find the exit
temperature of the helium and the ratio of the
pipe diameters.
6.1 52E Water flowing in a line at 60 lbf/in 2 , saturated
vapor is taken out through a valve to 14.7
lbf/in 2 . What is the temperature as it leaves the
valve assuming no changes in kinetic energy
and no heat transfer?
6.153E A small, high-speed turbine operating on com-
pressed air produces a power output of 0. 1 hp.
The inlet state is 60 lbf/in 2 , 120 F, and the exit
state is 14.7 lbf/in 2 , -20 F. Assuming the veloc-
ities to be low and the process to be adiabatic,
find the required mass flow rate of air through
the turbine.
6.154E Hoover Dam, across the Colorado River, dams
up Lake Mead 600 ft higher than the river
downstream, as shown in Fig. P6.51. The elec-
tric generators driven by water-powered tur-
bines deliver 1.2 X 10 6 Btu/s. If the water is 65
F, find the minimum amount of water ru nnin g
through the turbines.
6.155E A small expander (a turbine with heat transfer)
has 0.1 lbm/s of helium entering at 160 psia,
1000 R and leaving at 40 psia, 540 R. The
power output on the shaft is measured as 55
Bru/s. Find the rate of heat transfer, neglecting
kinetic energies.
6.156E An exhaust fan in a building should be able to
move 5 Ibm/s of air at 14.4 psia, 68 F through a
1.25-ft-diameter vent hole. How high a velocity
must the fan generate and how much power is
required to do that?
212 M CHAPTER SIX FlRST-L AW ANALYSIS FOR A CONTROL VOLUME
6.157 E In a steam generator, compressed liquid water at
1500 lbf/in 2 , 100 F enters a 1-in-diameter tube
at the rate of 5 ftVmin. Steam at 1250 lbf/in 2 ,
750 F exits the tube. Find the rate of heat trans-
fer to the water.
6.158E Carbon dioxide gas enters a steady-state, steady-
flow heater at 45 lbf/in 2 , 60 F and exits at 40
lbf/in 2 , 1800 F. It is shown in Fig. P6.63, where
changes in kinetic and potential energies are neg-
ligible. Calculate the required heat transfer per
Ibm of carbon dioxide flowing through the heater.
6,159 E A flow of liquid glycerine flows around an engine,
cooling it as it absorbs energy. The glycerine en-
ters the engine at 140 F and receives 13 hp of heat
transfer. What is the required mass flow rate if the
glycerine should come out at a maximum 200 F?
6.160E A small water pump is used in an irrigation sys-
tem. The pump takes water in from a river at 50
F, 1 arm at a rate of 10 Ibm/s. The exit line en-
ters a pipe that goes up to an elevation 60 ft
above the pump and river, where the water runs
into an open channel. Assume that the process is
adiabatic and that the water stays at 50 F. Find
the required pump work.
6.161 E A steam turbine receives water at 2000 lbf/in 2 ,
1200 F at a rate of 200 Ibm/s, as shown in Fig.
P6.78. In the middle section 40 Ibm/s is with-
drawn at 300 lbf/in 2 , 650 F and the rest exits the
turbine at 10 lbf/in 2 , 95% quality. Assuming no
heat transfer and no changes in kinetic energy,
find the total turbine power output.
6.162 E A condenser, as the heat exchanger shown in
Fig. P6.84, brings 1 lbm/s water flow at 1 lbf/in 2
from 500 F to saturated liquid at 1 lbf/in 2 . The
cooling is done by lake water at 70 F that re-
turns to the lake at 90 F. For an insulated con-
denser, find the flow rate of cooling water.
6.163E A heat exchanger is used to cool an air flow
from 1400 to 680 R, both states at 150 lbf/in 2 .
The coolant is a water flow at 60 F, 15 lbf/in 2 ,
and it is shown in Fig. P6.86. If the water leaves
as saturated vapor, find the ratio of the flow
rates 7H watet /m air
6.164 E An automotive radiator has glycerine at 200 F
enter and return at 130 F as shown in Fig. P6,88.
Air flows in at 68 F and leaves at 77 F. If the ra-
diator should transfer 33 hp, what is the mass
flow rate of the glycerine and what is the vol-
ume flow rate of air in at 15 psia?
6.165E An insulated mixing chamber as shown in Fig.
P6.94 receives 4 lbm/s of R-134a at 150 lbf/in 2 ,
220 F in a line with low velocity. Another line
with R-134a of saturated liquid at 130 F flows
through a valve to the mixing chamber at 150
lbf/in 2 after the valve. The exit flow is saturated
vapor at 150 lbf/in 2 flowing at 60 ft/s. Find the
mass flow rate for the second line.
6.166E An air compressor as shown in Fig. P6.59 takes
in air at 14 lbf/in 2 , 60 F and delivers it at 140
lbf/in 2 , 1080 R to a constant-pressure cooler,
which the air exits at 560 R. Find the specific
compressor work and the specific heat transfer.
6.1 67 E The following data are for a simple steam power
plant as shown in Fig. P6.99.
State
1
2
3
4
5
6
7
P lbf/in 2
900
890
860
830
800
i.5
1.4
TF
115
350
920
900
110
h, Btu/lbm
85.3
323
1468
1456
1029
78
State 6 has x 6 = 0.92 and velocity of 600 ft/s.
The rate of steam flow is 200 000 lbm/h, with
400 hp input to the pump. Piping diameters are
8 in. from the steam generator to the turbine and
3 in. from the condenser to the steam generator.
Determine the power output of the turbine and
the heat transfer rate in the condenser.
6.168E For the same steam power plant as shown in Fig.
P6.99 and Problem 6.167 determine the rate of
heat transfer in the economizer, which is a low-
temperature heat exchanger, and the steam gen-
erator. Determine also the flow rate of cooling
water through the condenser, if the cooling water
increases from 55 to 75 F in the condenser.
6il69E A geothermal supply of hot water operates a
steam turbine, as shown in Fig. P6.105. The
high-pressure water at 200 lbf/in 2 , 350 F is
throttled into a flash evaporator chamber, which
forms liquid and vapor at a lower pressure of 60
lbf/in 2 . The liquid is discarded while .the satu-
rated vapor feeds the turbine and exits at 1
lbf/in 2 , 90% quality. If the turbine should pro-
duce 1000 hp, find the required mass flow rate
of hot geothermal water.
Computer, Design, and Open-ended Problems H 213
6.170E A 1-ft 3 tank, shown in Fig. P6.111, that is ini-
tially evacuated is connected by a valve to an air
supply line flowing air at 70 F, 120 lbf/in 2 . The
valve is opened, and air flows into the tank until
the pressure reaches 90 lbf/in 2 . Determine the
final temperature and mass inside the tank, as-
suming the process is adiabatic. Develop an ex-
pression for the relation between the line
temperature and the final temperature using con-
stant specific heats.
6.171E Helium in a steel tank is at 40 psia, 540 R with a
volume of 4 ft 3 . It is used to fill a balloon. When
the tank pressure drops to 24 psia, the flow of
helium stops by itself. If all the helium still is at
540 R, how big a balloon did I get? Assume the
pressure in the balloon varies linearly with vol-
ume from 14.7 psia (V = 0) to the final 24 psia.
How much heat transfer took place?
6.172E A 20-ft 3 tank contains ammonia at 20 lbf/in 2 , 80
F. Tile tank is attached to a line flowing ammo-
nia at 180 lbf/in 2 , 140 F. The valve is opened,
and mass flows in until the tank is half full of
liquid, by volume at 80 F. Calculate the heat
transferred from the tank during this process.
6.173E An initially empty bottle, V = 10 ft 3 , is filled
with water from a line at 120 lbf/in 2 , 500 F. As-
sume no heat transfer and that the bottle is
closed when the pressure reaches line pressure.
Find the final temperature and mass in the bottle.
6.1 74E A nitrogen line, 540 R, 75 lbf/in 2 is connected to
a turbine that exhausts to a closed initially
empty tank of 2000 ft 3 , as shown in Fig. P6.1 19.
The turbine operates to a tank pressure of 75
lbf/in 2 , at which point the temperature is 450 R.
Assuming the entire process is adiabatic, deter-
mine the turbine work.
6.17SE A mass-loaded piston/cylinder containing air is
at 45 lbf/in 2 , 60 F with a volume of 9 ft 3 , while
at the stops V = 36 ft 3 . An air line, 75 lbf/in 2 ,
1100 R is connected by a valve, as shown in
Fig. P6.133. The valve is then opened until a
final inside pressure of 60 lbf/in 2 is reached, at
which point T = 630 R. Find the air mass that
enters, work done, and heat transfer.
Computer, design, and Open-Ended problems
6.176 Fit a polynomial expression of degree n in the
temperature for ideal-gas specific heat. Use the
ideal-gas enthalpy values for one of the sub-
stances listed in Table A. 8 as data. The accuracy
of the correlation should be studied as a function
of the temperature range of the fit as well as of the
polynominal degree n.
6.177 An insulated tank of volume V contains a speci-
fied ideal gas (with constant specific heat) as P h
Tj. A valve is opened, allowing the gas to flow
out until the pressure inside drops to P 2 . Deter-
mine T 2 and m 2 using a stepwise solution in incre-
ments of pressure between P x and P 2 \ the number •
of increments is variable.
6.178 We wish to solve Problem 6.121, using a stepwise
solution, whereby the process is subdivided into
several parts to minimize the effects of a linear
average enthalpy approximation. Divide the
process into two or three steps so that you can get
a better estimate for the mass times enthalpy leav-
ing the tank.
6.179 The air-water counterflowing heat exchanger
given hi Problem 6.86 has an air exit temperature
of 360 K. Suppose the air exit temperature is
listed as 300 K, then a ratio of the mass flow fates
is found from the energy equation to be 5. Show
that this is an impossible process by looking at air
and water temperatures at several locations inside
the heat exchanger. Discuss how this puts a limit
on the energy that can be extracted from the air.
6.180 A coflowing heat exchanger receives air at 800 K,
I MPa and liquid water at 15°C, 100 kPa, as
shown in Fig. P6.180, The air line heats the water
so that at the exit the air temperature is 20°C
above the water temperature. Investigate the lim-
its for the air and water exit temperatures as a
function of the ratio of the two mass flow rates.
Plot the temperatures of the air and water inside
the heat exchanger along the flow path.
-VvWA
-vWM
©
FIGURE P6.180
The Second law of
THERMODYNAMICS
The first law of thermodynamics states that during any cycle that a system undergoes, the
cyclic integral of the heat is equal to the cyclic integral of the work. The first law, how-
ever, places no restrictions on the direction of flow of heat and work. A cycle in which a
given amount of heat is transferred from the system and an equal amount of work is done
on the system satisfies the first law just as well as a cycle in which the flows of heat and
work are reversed. However, we know from our experience that because a proposed cycle
does not violate the first law does not ensure that the cycle will actually occur. It is this
kind of experimental evidence that led to the formulation of the second law of thermody-
namics. Thus, a cycle will occur only if both the first and second laws of thermodynamics
are satisfied.
In its broader significance, the second law acknowledges that processes proceed in
a certain direction but not in the opposite direction. A hot cup of coffee cools by virtue of
heat transfer to the surroundings, but heat will not flow from the cooler surroundings to
the hotter cup of coffee. Gasoline is used as a car drives up a hill, but the fuel level in the
gasoline tank cannot be restored to its original level when the car coasts down the hill.
Such familiar observations as these, and a host of others, are evidence of the validity of
the second law of thermodynamics.
In this chapter, we consider the second law for a system undergoing a cycle, and in
the next two chapters we extend the principles to a system undergoing a change of state
and then to a control volume.
7.1 Heat Engines and refrigerators
Consider the system and the surroundings previously cited in the development of the first
law, as shown in Fig. 7.1. Let the gas constitute the system and, as in our discussion of
the first law, let this system undergo a cycle in which work is first done on the system by
the paddle wheel as the weight is lowered. Then let the cycle be completed by transfer-
ring heat to the surroundings.
We know from our experience that we cannot reverse this cycle. That is, if we
transfer heat to the gas, as shown by the dotted arrow, the temperature of the gas will in-
crease, but the paddle wheel will not turn and raise the weight. With the given surround-
ings (the container, the paddle wheel, and the weight), this system can operate in' a cycle
in which the heat transfer and work are both negative, but it cannot operate in a cycle in
which both the heat transfer and work are positive, even though this would not violate the
first law.
214
Heat Engines and refrigerators ■ 215
FIGURE 7.1 A
system that undergoes a
cycle involving work and
heat.
Gas
Vl
Consider another cycle, known from our experience to be impossible actually to
complete. Let two systems, one at a high temperature and the other at a low temperature,
undergo a process in which a quantity of heat is transferred from the high-temperature
system to the low-temperature system. We know that this process can take place. We also
know that the reverse process, in which heat is transferred from the low-temperature sys-
tem to the high-temperature system, does not occur, and that it is impossible to complete
the cycle by heat transfer only. This impossibility is illustrated in Fig. 7.2.
These two examples lead us to a consideration of the heat engine and the refrigera-
tor, which is also referred to as a heat pump. With the heat engine we can have a system
that operates in a cycle and performs a net positive work and a net positive heat transfer.
With the heat pump we can have a system that operates in a cycle and has heat transferred
to it from a low-temperature body and heat transferred from it to a high-temperature body,
though work is required to do this. Three simple heat engines and two simple refrigerators
will be considered.
The first heat engine is shown in Fig. 7,3. It consists of a cylinder fitted with appro-
priate stops and a piston. Let the gas in the cylinder constitute the system. Initially the pis-
ton rests on the lower stops, with a weight on the platform. Let the system now undergo a
process in which heat is transferred from some high-temperature body to the gas, causing
it to expand and raise the piston to the upper stops. At this point the weight is removed.
Now let the system be restored to its initial state by transferring heat from the gas to a
low-temperature body, thus completing the cycle. Since the weight was raised during the
cycle, it is evident that work was done by the gas during the cycle. From the first law we
conclude that the net heat transfer was positive and equal to the work done during the
cycle.
Such a device is called a heat engine, and the substance to which and from which
heat is transferred is called the working substance or working fluid. A heat engine may
be defined as a device that operates in a thermodynamic cycle and does a certain amount
of 'net positive work through the transfer of heat from a high-temperature body to a low-
temperature body. Often the term heat engine is used in a broader sense to include all de-
vices that produce work, either through heat transfer or through combustion, even though
the device does not operate in a thermodynamic cycle. The internal combustion engine
FIGURE 7.2 An
example showing the
impossibility of
completing a cycle by
transferring heat from a
low-temperature body to a
high-temperature body.
High temperature ^ .
Is
Q \\ Q
■ A
I ■■ ■■ ■■ |i
/ Low temperature n
216 M Chapter Seven the Second Law of Thermodynamics
u
Gas
A
ilTTfl
Qh
FIGURE 7.3 A simple
heat engine.
High-temperature
body
11 1 1 II
r-
Ql
V
Low- temperature
body
and the gas turbine are examples of such devices, and calling them heat engines is an ac-
ceptable use of the term. In this chapter, however, we are concerned with the more re-
stricted form of heat engine, as just defined, one that operates on a thermodynamic cycle.
A simple steam power plant is an example of a heat engine in this restricted sense-
Each component in this plant may be analyzed individually as a steady-state, steady-flow
process, but as a whole it may be considered a heat engine (Fig. 7.4) in which water
(steam) is the working fluid. An amount of heat, Q Hi is transferred from a high-temperature
body, which may be the products of combustion in a furnace, a reactor, or a secondary
fluid that in turn has been heated in a reactor. In Fig. 7.4 the turbine is shown schemati-
cally as driving the pump. What is significant, however, is the net work that is delivered
during the cycle. The quantity of heat Q L is rejected to a low-temperature body, which is
usually the cooling water in a condenser. Thus, the simple steam power plant is a heat en-
gine in the restricted sense, for it has a working fluid, to which and from which heat is
transferred, and which does a certain amount of work as it undergoes a cycle.
Another example of a heat engine is the thermoelectric power generation device that
was discussed in Chapter 1 and shown schematically in Fig. 1.8b. Heat is transferred from
a high-temperature body to the hot junction (Q H X and heat is transferred from the cold
junction to the surroundings (Q L ). Work is done in the form of electrical energy. Since
there is no working fluid, we do not usually think of this as a device that operates in a
Boiler
A Pump
Condenser
FIGURE 7,4 A heat
engine involving steady-
state processes.
Power
net
■System
boundary
HEAT ENGINES AND REFRIGERATORS M 217
cycle. However, if we adopt, a microscopic point of view, we could regard a cycle as the
flow of electrons. Furthermore, as with the steam power plant, the state at each point in the
thermoelectric power generator does not change with time under steady-state conditions.
Thus, by means of a heat engine, we are able to have a system operate in a cycle and
have hoth the net work and the net heat transfer positive, which we were not able to do
with the system and surroundings of Fig. 7.1.
We note that in using the symbols Q H and Q L , we have departed from our sign connota-
tion for heat, because for a heat engine Q L is negative when the working fluid is considered as
the system. In this chapter it will be advantageous to use the symbol Q H to represent the heat
transfer to or from the high-temperature body, and Q L to represent the heat transfer to or from
the low-temperature body. The direction of the heat transfer will be evident from the context.
At this point, it is appropriate to introduce the concept of thermal efficiency of a
heat engine. In general, we say that efficiency is the ratio of output, the energy sought, to
input, the energy that costs, but the output and input must be clearly defined. At the risk of
oversimplification, we may say that in a heat engine the energy sought is the work, and
the energy that costs money is the heat from the high-temperature source (indirectly, the
cost of the fuel). Thermal efficiency is defined as
_ ^(energy sought) _ Q H - Q L Q L
VM &(energy that costs) Q H Q H ^ /A)
Heat engines vary greatly in size and shape, from large steam engines, gas turbines,
or jet engines, to gasoline engines for cars and diesel engines for trucks or cars, to much
smaller engines for lawn mowers or hand-held devices such as chain saws or trimmers.
Typical values for the thermal efficiency of real engines are about 35-50% for large power
plants, 30-35% for gasoline engines, and 35-40% for diesel engines. Smaller utility-type
engines may have only about 20% efficiency, owing to their simple carburetion and con-
trols and to the fact that some losses scale differently with size and therefore represent a
larger fraction for smaller machines.
EXAMPLE 7.1 An automobile engine produces 136 hp on the output shaft with a thermal efficiency of
30%. The fuel it burns gives 35 000 kJ/kg as energy release. Find the total rate of energy
rejected to the ambient and the rate of fuel consumption in kg/s.
Solution
From the definition of a heat engine efficiency, Eq. 7.1, and the conversion of hp from
Table A.l we have:
W = V^Qh = 136 hp X 0.7355 kW/hp = 100 fcW
Qn -■ Wi]c, s = 100/0.3 - 333 kW
The energy equation for the overall engine gives:
Qi-Qh- - 0.3)4, - 233 kW
From the energy release in the burning we have: Q H = mq H , so
218 m CHAPTER SEVEN THE SECOND LAW OF THERMODYNAMICS
FIGURE 7.5 Sketch for Example 7.1.
An actual engine shown in Fig. 7.5 rejects energy to the ambient through the radiator
cooled by atmospheric air, as heat transfer from the exhaust system and the exhaust flow
ofhotgases.
The second cycle that we were not able to complete was the one indicating the im-
possibility of transferring heat directly from a low-temperature body to a high-temperature
body. This can of course be done with a refrigerator or heat pump. A vapor-compression
refrigerator cycle, which was introduced in Chapter 1 and shown in Fig. 1.7, is shown again
in Fig. 7.6. The working fluid is the refrigerant, such as R-134a or ammonia, which goes
through a thermodynamic cycle. Heat is transferred to the refrigerant in the evaporator,
where its pressure and temperature are low. Work is done on the refrigerant in the compres-
sor, and heat is transferred from it in the condenser, where its pressure and temperature are
high. The pressure drops as the refrigerant flows through the throttle valve or capillary tube.
Thus, in a refrigerator or heat pump, we have a device that operates in a cycle, that re-
quires work, and that accomplishes the objective of transferring heat from a low-temperature,
body to a high-temperature body.
The thermoelectric refrigerator, which was discussed in Chapter 1 and shown
schematically in Fig. 1.8a, is another example of a device that meets our definition of a re-
frigerator. The work input to the thermoelectric refrigerator is in the form of electrical en-
ergy, and heat is transferred from the refrigerated space to the cold junction {Q L ) and from
the hot junction to the surroundings {Q H ).
System boundary
FIGURE 7.6 A simple
refrigeration cycle.
Heat Engines and Refrigerators M 219
The "efficiency" of a. refrigerator is expressed in terms of the coefficient of perfor-
mance, which we designate with the symbol j8. For a refrigerator the objective, that is, the
energy sought, is Q Li the heat transferred from the refrigerated space. The energy that
costs is the work W. Thus, the coefficient of performance, is
a = 6l (energy sought) = Q L _ 1 ( 2)
P ^(energy that costs) Q H - Q L Q^Ql ~ 1
A household refrigerator may have a coefficient of performance (often referred to as
COP) of about 2.5, whereas a deep freeze unit will be closer to 1.0. Lower cold tempera-
ture space of higher warm temperature space will result in lower values of COP, as will be
found in Section 7.6. For a heat pump operating over a moderate temperature range, a
value of its COP can be around 4, with this value decreasing sharply as the heat pump's
operating temperature range is broadened.
EXAMPLE 7.2 The refrigerator in a kitchen shown in Fig. 7.7 receives an electrical input power of
150 W to drive the system, and it rejects 400 W to the kitchen air. Find the rate of en-
ergy taken out of the cold space and the coefficient of performance of the refrigerator.
FIGURE 7.7 Sketch
for Example 7.2.
5 It should be noted that a refrigeration or heat pump cycle can be used with either of two objec-
tives. It can be used as a refrigerator, in which case the primary objective is Q L , the heat trans-
ferred to the refrigerant from the refrigerated space. It can also be used as a heating system (in
which case it is usually referred to as a heat pump), the objective being Q H , the heat transferred
from the refrigerant to the high-temperature body, which is the space to be heated. Q L is trans-
ferred to the refrigerant from the ground, the atmospheric air, or well water. The coefficient of
performance for this case, jS', is
01 - (energy sought) = Q H = I
15 W (energy that costs) Qh~Ql \ " Ql'Qh
It also follows that for a given cycle.
0'-/J=l
Unless otherwise specified, the term coefficient of performance will always refer to a refrigerator as
defined by Eq. 7.2.
220 a Chapter Seven the second law of thermodynamics
Solution
C.V. refrigerator. Assume steady state so there is no storage of energy. The information
provided is W = 150 W, and the heat rejected is Q tl - 400 W.
The energy equation gives:
Qi. - Q„ - W=4W- 150 = 250 W ;
This is also the rate of energy transfer into the cold space from the warmer kitchen due
to heat transfer and exchange of cold air inside with warm air when you open the door.
From the definition of the coefficient of performance, Eq. 7.2
a Q.l _ 250 . . . Kn :■
Prefrig - v • 150 - 1-67
Before we state the second law, the concept of a thermal reservoir should be intro-
duced. A thermal reservoir is a body to which and from which heat can be transferred in-
definitely without change in the temperature of the reservoir. Thus, a thermal reservoir
always remains at constant temperature. The ocean and the atmosphere approach this def-
inition very closely. Frequently it will be useful to designate a high-temperature reservoir
and a low-temperature reservoir. Sometimes a reservoir from which heat is transferred is
called a source, and a reservoir to which heat is transferred is called as sink.
7.2 The Second Law of Thermodynamics
On the basis of the matter considered in the previous section, we are now ready to state
the second law of thermodynamics. There are two classical statements of the second law,
known as the Kelvin-Planck statement and the Clausius statement.
The Kelvin-Planck statement: It is impossible to construct a device that will operate
in a cycle and produce no effect other than the raising of a weight and the exchange
of heat with a single reservoir. See Fig. 7.8.
This statement ties in with our discussion of the heat engine. In effect, it states that
it is impossible to construct a heat engine that operates in a cycle, receives a given amount
of heat from a high-temperature body, and does an equal amount of work. The only alter-
FIGURE 7,8 The
Kelvin-Planck statement.
Impossible
the second Law of thermodynamics
m 221
FIGURE 7.9 The
Clausius statement.
Impossible
native is that some heat must be transferred from the working fluid at a lower temperature
to a low-temperature body. Thus, work can be done by the transfer of heat only if there
are two temperature levels, and heat is transferred from the high-temperature body to the
heat engine and also from the heat engine to the low-temperature body. This implies that
it is impossible to build a heat engine that has a thermal efficiency of 100%.
The Clausius statement: It is impossible to construct a device that operates in a
cycle and produces no effect other than the transfer of heat from a cooler body to a
hotter body. See Fig. 7.9.
This statement is related to the refrigerator or heat pump. In effect, it states that it is
impossible to construct a refrigerator that operates without an input of work. This also im-
plies that the coefficient of performance is always less than infinity.
Three observations should be made about these two statements. The first observa-
tion is that both are negative statements. It is of course impossible to "prove" these nega-
tive statements. However, we can say that the second law of thermodynamics (like every
other law of nature) rests on experimental evidence. Every relevant experiment that has
been conducted either directly or indirectly verifies the second law, and no experiment has
ever been conducted that contradicts the second law. The basis of the second law is there-
fore experimental evidence.
A second observation is that these two statements of the second law are equivalent.
Two statements are equivalent if the truth of each statement implies the truth of the
other, or if the violation of each statement implies the violation of the other. That a viola-
tion of the Clausius statement implies a violation of the Kelvin-Planck statement may be
shown. The device at the left in Fig. 7. 1 is a refrigerator that requires no work and thus
violates the Clausius statement. Let an amount of heat Q L be transferred from the low-
temperature reservoir to this refrigerator, and let the same amount of heat Q L be trans-
ferred to the high-temperature reservoir. Let an amount of heat Q H that is greater than Q L
be transferred from the high-temperature reservoir to the heat engine, and let the engine
reject the amount of heat Q L as it does an amount of work IV, which equals Q H ~ Q L . Be-
cause there is no net heat transfer to the low-temperature reservoir, the low-temperature
reservoir, along with the heat engine and the refrigerator, can be considered together as a
device that operates in a cycle and produces no effect other than the raising of a weight
(work) and the exchange of heat with a single reservoir. Thus, a violation of the Clausius
statement implies a violation of the Kelvin-Planck statement. The complete equivalence
222 H Chapter Seven The Second Law of Thermodynamics
FIGURE 7,10
Demonstration of the
equivalence of the two
statements of the second
law.
W=0
High-temperature reservoir
A
Ql
o
~7T
17
Qh
O
Ql
Ql
Low-temperature reservoir
'System boundary
W=Q U - Q L
of these two statements is established when it is also shown that a violation of the
Kelvin-Planck statement implies a violation of the Clausius statement. This is left as an
exercise for the student.
The third observation is that frequently the second law of thermodynamics has
been stated as the impossibility of constructing a perpetual-motion machine of the sec-
ond kind. A perpetual-motion machine of the first kind would create work from noth-
ing or create mass or energy, thus violating the first law, A perpetual-motion machine
of the second kind would extract heat from a source and then convert this heat com-
pletely into other forms of energy, thus violating the second law. A perpetual-motion
machine of the third kind would have no friction, and thus would run indefinitely but
produce no work,
A heat engine that violated the second law could be made into a perpetual-motion
machine of the second kind by taking the following steps. Consider Fig. 7.11, which
might be the power plant of a ship. An amount of heat Q L is transferred from the ocean to
a high-temperature body by means of a heat pump. The work required is W, and the heat
transferred to the high-temperature body is Q H , Let the same amount of heat be trans-
ferred to a heat engine that violates the Kelvin-Planck statement of the second law and
does an amount of work W = Q H . Of this work an amount Q H — Q L is required to drive
the heat pump, leaving the net work (}V nei — Q L ) available for driving the ship. Thus, we
• System boundary
FIGURE 7.11 A
perpetual-motion machine
of the second kind.
High-temperature body
W MX =W-W'l
W=Q H
Qh
O
Qh
Heat
pump
"7Y
Ql
Ocean
THE REVERSIBLE PROCESS H 223
have a perpetual-motion machine in the sense that work is done by utilizing freely avail-
able sources of energy such as the ocean or atmosphere.
:i <j
TumohET
7.3 The Reversible Process
The question that can now logically be posed is this: If it is impossible to have a heat
engine of 100% efficiency, what is the maximum efficiency one can have? The first
step in the answer to this question is to define an ideal process, which is called a re-
versible process.
A reversible process for a system is defined as a process that once having taken
place can be reversed and in so doing leave no change in either system or surroundings.
Let us illustrate the significance of this definition for a gas contained in a cylinder
that is fitted with a piston. Consider first Fig. 7. 12, in which a gas, which we define as the
system, is restrained at high pressure by a piston that is secured by a pin. When the pin is
removed, the piston is raised and forced abruptly against the stops. Some work is done by
the system, since the piston has been raised a certain amount. Suppose we wish to restore
the system to its initial state. One way of doing this would be to exert a force on the piston
and thus compress the gas until the pin can be reinserted in the piston. Since the pressure
on the face of the piston is greater on the return stroke than on the initial stroke, the work
done on the gas in this reverse process is greater than the work done by the gas in the ini-
tial process. An amount of heat must be transferred from the gas during the reverse stroke
so that the system has the same internal energy as it had originally. Thus, the system is re-
stored to its initial state, but the surroundings have changed by virtue of the fact that work
was required to force the piston down and heat was transferred to the surroundings. The
initial process therefore is an irreversible one because it could not be reversed without
leaving a change in the surroundings.
In Fig. 7.13 let the gas in the cylinder comprise the system, and let the piston be
loaded with a number of weights. Let the weights be slid off horizontally one at a time, al-
lowing the gas to expand and do work in raising the weights that remain on the piston. As
the size of the weights is made smaller and their number is increased, we approach a
process that can be reversed, for at each level of the piston during the reverse process
there will be a small weight that is exactly at the level of the platform and thus can be
placed on the platform without requiring work. In the limit, therefore, as the weights be-
come very small, the reverse process can be accomplished in such a manner that both the
FIGURE 7.12 An
example of an irreversible
process.
Gas
-Work
5L
Initial process
Reverse process
-Q
224 M Chapter seven the Second law of thermodynamics
FIGURE 7.13 An
example of a process that
approaches being
reversible.
i
system and surroundings are in exactly the same state they were initially. Such a process
is a reversible process.
7.4 Factors that Mender processes
Irreversible
There are many factors that make processes irreversible. Four of those factors — friction,
unrestrained expansion, heat transfer through a finite temperature difference, and mixing
of two different substances — are considered in this section.
Friction
It is readily evident that friction makes a process irreversible, but a brief illustration may
amplify the point. Let a block and an inclined plane make up a system, as in Fig. 7.14, and
let the block be pulled up the inclined plane by weights that are lowered. A certain
amount of work is needed to do this. Some of this work is required to overcome the fric-
tion between the block and the plane, and some is required to increase the potential en-
ergy of the block. The block can be restored to its initial position by removing some of the
weights and thus allowing the block to slide back down the plane. Some heat transfer
from the system to the surroundings will no doubt be required to restore the block to its
initial temperature. Since the surroundings are not restored to their initial state at the con-
clusion of the reserve process, we conclude that friction has rendered the process irre-
Factors that Render Processes Irreversible U 225
versible. Another type of factional effect is that associated with the flow of viscous fluids
in pipes and passages and in the movement of bodies through viscous fluids.
Unrestrained Expansion
The classic example of an unrestrained expansion, as shown in Fig. 7.15, is a gas separated
from a vacuum by a membrane. Consider what happens when the membrane breaks and the
gas fills the entire vessel. It can be shown that this is an irreversible process by considering
what would he necessary to restore the system to its original state. The gas would have to be
compressed and heat transferred from the gas until its initial state is reached. Since the work
and heat transfer involve a change in the surroundings, the surroundings are not restored to
their initial state, indicating that the unrestrained expansion was an irreversible process. The
process described in Fig. 7.12 is also an example of an unrestrained expansion.
In the reversible expansion of a gas, there must be only an infinitesimal difference
between the force exerted by the gas and the restraining force, so that the rate at which the
boundary moves will be infinitesimal. In accordance with our previous definition, this is a
quasi-equihbrium process. However, actual systems have a finite difference in forces,
which causes a finite rate of movement of the boundary, and thus the processes are irre-
versible in some degree.
Heat Transfer through a Finite Temperature Difference
Consider as a system a high-temperature body and a low-temperature body, and let heat
be transferred from the high-temperature body to the low-temperature body. The only
way in which the system can be restored to its initial state is to provide refrigeration,
which requires work from the surroundings, and some heat transfer to the surroundings
will also be necessary. Because of the heat transfer and the work, the surroundings are not
restored to their original state, indicating that the process was irreversible.
An interesting question is now posed. Heat is defined as energy that is transferred
through a temperature difference. We have just shown that heat transfer through a temper-
ature difference is an irreversible process. Therefore, how can we have a reversible heat-
transfer process? A heat-transfer process approaches a reversible process as the
temperature difference between the two bodies approaches zero. Therefore, we define a
reversible heat-transfer process as one in which the heat is transferred through an infini-
tesimal temperature difference. We realize of course that to transfer a finite amount of
heat through an infinitesimal temperature difference would require an infinite amount of
time or infinite area. Therefore, all actual heat transfers are through a finite temperature
difference and hence are irreversible, with the greater the temperature difference, the
greater the irreversibility. We will find, however, that the concept of reversible heat trans-
fer is very useful in describing ideal processes.
FIGURE 7.15
Demonstration of the fact
that unrestrained
expansion makes
processes irreversible.
1
I Gas
^
Vacuum
System
] f boundary
Initial state
Gas
-w
Reverse process
226 M Chapter Seven The Second Law of thermodynamics
FIGURE 7.16
Demonstration of the fact
that the mixing of two
different substances is an
irreversible process.
2 + N 2
Mixing of Two Different Substances
Figure 7.16 illustrates the process of mixing two different gases separated by a mem-
brane. When the membrane is broken, a homogeneous mixture of oxygen and nitrogen
fills the entire volume, This process will be considered in some detail in Chapter 12. We
can say here that this may be considered a special case of an unrestrained expansion, for
each gas undergoes an unrestrained expansion as it fills the entire volume. A certain
amount of work is necessary to separate these gases. Thus, an air separation plant such as
described in Chapter 1 requires an input of work to accomplish the separation.
Other Factors
A number of other factors make processes irreversible, but they will not be considered in
detail here. Hysteresis effects and the i 2 R loss encountered in electrical circuits are both fac-
tors that make processes irreversible. Ordinary combustion is also an irreversible process.
It is frequently advantageous to distinguish between internal and external irre-
versibility. Figure 7.17 shows two identical systems to which heat is transferred. Assum-
ing each system to be a pure substance, the temperature remains constant during the
heat-transfer process. In one system the heat is transferred from a reservoir at a tempera-
ture T + dT, and in the other the reservoir is at a much higher temperature, T + AT, than
the system. The first is a reversible heat-transfer process, and the second is an irreversible
heat-transfer process. However, as far as the system itself is concerned, it passes through
exactly the same states in both processes, which we assume are reversible. Thus, we can
say for the second system that the process is internally reversible but externally irre-
versible because the irreversibility occurs outside the system.
We should also note the general interrelation of reversibility, equilibrium, and time.
In a reversible process, the deviation from equilibrium is infinitesimal, and therefore it oc-
FIGURE 7.17
Illustration of the
difference between an
internally and externally
reversible process.
■Temperature = T-
T+dT
T + AT
r
j
THE CARNOT CYCLE W 227
curs at an infinitesimal rate. Since it is desirable that actual processes proceed at a finite
rate, the deviation from equilibrium must be finite, and therefore the actual process is irre-
versible in some degree. The greater the deviation from equilibrium, the greater the irre-
versibility, and the more rapidly the process will occur. It should also be noted that the
quasi-equilibrium process, which was described in Chapter 2, is a reversible process, and
hereafter the term reversible process will be used.
7.5 THE CARNOT CYCLE
Having defined the reversible process and considered some factors that make processes ir-
reversible, let us again pose the question raised in Section 7.3. If the efficiency of all heat
engines is less than 100%, what is the most efficient cycle we can have? Let us answer
this question for a heat engine that receives heat from a high- temperature reservoir and re-
jects heat to a low-temperature reservoir. Since we are dealing with reservoirs, we recog-
nize that both the high temperature and the low temperature of the reservoirs are constant
and remain constant regardless of the amount of heat transferred.
Let us assume that this heat engine, which operates between the given high-
temperature and low-temperature reservoirs, does so in a cycle in which every process
is reversible. If every process is reversible, the cycle is also reversible; and if the cycle
is reversed, the heat engine becomes a refrigerator. In the next section we will show
that this is the most efficient cycle that can operate between two constant-temperature
reservoirs. It is called the Carnot cycle and is named after a French engineer, Nicolas
Leonard Sadi Carnot (1796^1832), who expressed the foundations of the second law
of thermodynamics in 1824.
We now turn our attention to the Carnot cycle. Figure 7.18 shows a power plant that
is similar in many respects to a simple steam power plant and, we assume, operates on the
Carnot cycle. Consider the working fluid to be a pure substance, such as steam. Heat is
transferred from the high-temperature reservoir to the water (steam) in the boiler. For this
FIGURE 7.18
Example of a heat engine
that operates on a Camot
cycle.
High-temperature reservoir
: Condenser
(evaporator)
Ql
Low-temperature reservoir
228 H Chapter Seven the second law of thermodynamics
process to be a reversible heat transfer, the temperature of the water (steam) must be only
infinitesimally lower than the temperature of the reservoir. This result also implies, since
the temperature of the reservoir remains constant, that the temperature of the water must
remain constant. Therefore, the first process in the Carnot cycle is a reversible isothermal
process in which heat is transferred from the high-temperature reservoir to the working
fluid. A change of phase from liquid to vapor at constant pressure is of course an isother-
mal process for a pure substance.
The next process occurs in the turbine without heat transfer and is therefore adiabatic.
Since all processes in the Camot cycle are reversible, this must be a reversible adiabatic
process, during which the temperature of the working fluid decreases from the temperature
of the high-temperature reservoir to the temperature of the low-temperature reservoir.
In the next process heat is rejected from the working fluid to the low-temperature
reservoir. This must be a reversible isothermal process in which the temperature of the
working fluid is infinitesimally higher than that of the low-temperature reservoir. During
this isothermal process some of the steam is condensed.
The final process, which completes the cycle, is a reversible adiabatic process in
which the temperature of the working fluid increases from the low temperature to the high
temperature. If this were to be done with water (steam) as the working fluid, a mixture of
liquid and vapor would have to be taken from the condenser and compressed. (This would
be very inconvenient in practice, and therefore in all power plants the working fluid is
completely condensed in the condenser. The pump handles only the liquid phase.)
Sine the Carnot heat engine cycle is reversible, every process could be reversed, in
which case it would become a refrigerator. The refrigerator is shown by the dotted lines
and parentheses in Fig. 7.18. The temperature of the working fluid in the evaporator
would be infinitesimally lower than the temperature of the low-temperature reservoir, and
in the condenser it is infinitesimally higher than that of the high-temperature reservoir.
It should be emphasized that the Carnot cycle can, in principle, be executed in many
different ways. Many different working substances can be used, such as a gas or a thermo-
electric device such as described in Chapter 1. There are also various possible arrange-
ments of machinery. For example, a Carnot cycle can be devised that takes place entirely
within a cylinder, using a gas as a working substance, as shown in Fig. 7.19.
The important point to be made here is that the Camot cycle, regardless of what the
working substance may be, always has the same four basic processes. These processes are;
1. A reversible isothermal process in which heat is transferred to or from the high-
temperature reservoir.
2. A reversible adiabatic process in which the temperature of the working fluid de-
creases from the high temperature to the low temperature.
FIGURE 7.19
Example of a gaseous
system operating on a
Camot cycle.
1-2
Isothermal
expansion
3-4
Isothermal
compression
4-1 1
Adiabatic
compression
TWO PROPOSITIONS REGARDING THE EFFICIENCY OF A CARNOT CYCLE ffl 229
3. A reversible isothermal process in which heat is transferred to or from the low-
temperature reservoir.
4. A reversible adiabatic process in which the temperature of the working fluid in-
creases from the low temperature to the high temperature.
7.6 TWO PROPOSITIONS REGARDING THE
EFFICIENCY OF A CARNOT CYCLE
There are two important propositions regarding the efficiency of a Carnot cycle.
First Proposition
It is impossible to construct an engine that operates between two given reservoirs and is
more efficient than a reversible engine operating between the same two reservoirs.
The proof of this statement is accomplished through a "thought experiment." An
initial assumption is made, and it is then shown that this assumption leads to impossible
conclusions. The only possible conclusion is that the initial assumption was incorrect.
Let us assume that there is an irreversible engine operating between two given reser-
voirs that has a greater efficiency than a reversible engine operating between the same two
reservoirs. Let the heat transfer to the irreversible engine be Q H> the heat rejected be Q' L , and
the work be W m (which equals Q H - Q' L ) as shown in Fig. 7.20. Let the reversible engine
operate as a refrigerator (this is possible since it is reversible). Finally, let the heat transfer
with the low-temperature reservoir be Q L , the heat transfer with the high-temperature reser-
voir be Q H , and the work required be (which equals Q H — Q L ).
Since the initial assumption was that the irreversible engine is more efficient, it fol-
lows (because Q H is the same for both engines) that Q' L < Q L and W m > W^. Now the ir-
reversible engine can drive the reversible engine and still deliver the net work W ntt , which
equals lV iE — = Q L — Q' L . If we consider the two engines and the high-temperature
reservoir as a system, as indicated in Fig. 7.20, we have a system that operates in a cycle,
exchanges heat with a single reservoir, and does a certain amount of work. However, this
■ System boundary
FIGURE 7.20
Demonstration of the fact
that the Camot cycle is
the most efficient cycle
operating between two
fixed-temperature
reservoirs.
W iE = Q s -Qi
i H]gh;temperatUre ^eservaiHsi
Qh
V
Q'l
Qh
Irreversible
Reversible
engine
engine
7\
q_ 4! ,
Low-temperature reservoir
230 U chapter seven The second law of thermodynamics
would constitute a violation of the second law, and we conclude that our initial assump-
tion (that the irreversible engine is more efficient than a reversible engine) is incorrect.
Therefore, we cannot have an irreversible engine that is more efficient than a reversible
engine operating between the same two reservoirs.
Second Proposition
All engines that operate on the Camot cycle between two given constant-temperature
reservoirs have the same efficiency. The proof of this proposition is similar to the proof
just outlined, which assumes that there is one Carnot cycle that is more efficient than an-
other Carnot cycle operating between the same temperature reservoirs. Let the Carnot
cycle with the higher efficiency replace the irreversible cycle of the previous argument,
and let the Carnot cycle with the lower efficiency operate as the refrigerator. The proof
proceeds with the same line of reasoning as in the first proposition. The details are left as
an exercise for the student.
7,7 The Thermodynamic
Temperature Scale
In discussing the matter of temperature in Chapter 2, we pointed out that the zeroth law of
thermodynamics provides a basis for temperature measurement, but that a temperature
scale must be defined in terms of a particular thermometer substance and device, A tem-
perature scale that is independent of any particular substance, which might be called an
absolute temperature scale, would be most desirable. In the last paragraph we noted that
the efficiency of a Carnot cycle is independent of the working substance and depends only
on the temperature. This fact provides the basis for such an absolute temperature scale,
which we call the thermodynamic scale.
The concept of this temperature scale may be developed with the help of Fig. 7.21,
which shows three reservoirs and three engines that operate on the Carnot cycle. T } is
the highest temperature, r 3 is the lowest temperature, and T 2 is an intermediate tempera-
ture, and the engines operate between the various reservoirs as indicated. Q l is the same
FIGURE 7.21
Arrangement of heat
engines to demonstrate
the thermodynamic
temperature scale.
3f
3f
5
e 2
1
Qz
e 3
V
THE THERMODYNAMIC TEMPERATURE SCALE M 231
for both A and C and, since we are dealing with reversible cycles, Q 3 is the same for B
and C.
Since the efficiency of a Camot cycle is a function only of the temperature, we can
write
iJ^i = * ~ |r = 1 ~ Wl> Tid (73)
where tp designates a functional relation.
Let us apply this functional relation to the three Camot cycles of Fig. 7.21:
g = iKr 2 ,r 3 )
Since
it follows that
Qi _ QxQi
W» Ti) = tfTu T 2 ) X iP(T 2 , r 3 ) (7.4)
Note that the left side is a function of T x and T 3 (and not of T 2 \ and therefore the
right side of this equation must also be a function of T, and T 3 (and not of T 2 ). From'this
fact we can conclude that the form of the function \p must be such that
Wu r 2 ) = fiTl)
f(T 2 )
f(T 2 )
Ah)
for in this wayj{T 2 ) will cancel from the product of ip(T u T 2 ) X ijf(T 2i 7 3 ). Therefore, we
conclude that
| = ^,r 3 ) = ^ (7.5)
In general terms,
Ql At l ) w
Now there are several functional relations that will satisfy this equation. For the
thermodynamic scale of temperature, which was originally proposed by Lord Kelvin, the
selected relation is
232 II Chapter Seven The Second Law of thermodynamics
With absolute temperatures so defined, the efficiency of a Camot cycle may be expressed
in terms of the absolute temperatures.
This means that if the thermal efficiency of a Carnot cycle operating between two given
constant-temperature reservoirs is known, the ratio of the two absolute temperatures is
also known.
It should be noted that Eq. 7.7 gives us a ratio of absolute temperatures, but it does
not give us information about the magnitude of the degree. Let us first consider a qualita-
tive approach to this matter and then a more rigorous statement.
Suppose we had a heat engine operating on the Camot cycle that received heat at
the temperature of the steam point and rejected heat at the temperature of the ice point.
(Because a Camot cycle involves only reversible processes, it is impossible to construct
such a heat engine and perform the proposed experiment. However, we can follow the
reasoning as a "thought experiment" and gain additional understanding of the thermody-
namic temperature scale.) If the efficiency of such an engine could be measured, we
would find it to be 26.80%. Therefore, from Eq. 7.8,
T L Tfce point
rj A = 1 - jr = 1 ~ t " °' 2680
1 H steam point
J ice point
—
L steam pcint
0.7320
This gives us one equation concerning the two unknowns T H and T L . The second
equation comes from an arbitrary decision regarding the magnitude of the degree on the
thermodynamic temperature scale. If we wish to have the magnitude of the degree on the
absolute scale correspond to the magnitude of the degree on the Celsius scale, we can
write
T - — T- =100
steam point J ice point 1XJV
Solving these two equations simultaneously, we find
^steam point 373.15 K, ^Icepoint — 273.15 K
It follows that
I( D C) + 273.15 = 7/(K)
The absolute scale Telated to the Fahrenheit scale is the Rankine scale, designated
by R. On both these scales there are 180 degrees between the ice point and the steam
point. Therefore, for a Camot cycle heat engine operating between the steam point and the
ice point, we would have the two relations
team point -^icepoint ' 180
[ ice point
= 0.7320
1 steam point
Solving these two equations simultaneously, we find
repaint = 671.67 R, r iC£point = 491.67 R
THE IDEAL- GAS TEMPERATURE SCALE ffl 233
It follows that temperatures on the Fahrenheit and Rankine scales are related as
follows:
As already noted, the measurement of efficiencies of Camot cycles is, however, not
a practical way to approach the problem of temperature measurement on the thermody-
namic scale of temperature. The actual approach is based on the ideal-gas thermometer
and an assigned value for the triple point of water. At the tenth Conference on Weights
and Measures, which was held in 1954, the temperature of the triple point of water was
assigned the value 273.16 K. [The triple point of water is approximately O.Orc above the
ice point. The ice point is defined as the temperature of a mixture of ice and water at a
pressure of 1 arm (101.3 kPa) of air that is saturated with water vapor.] The ideal-gas ther-
mometer is discussed in the following section.
7.8 The Ideal-Gas Temperature Scale
In this section we reconsider in greater detail the ideal-gas temperature scale introduced in
Section 3.6. This scale is based on the observation that as the pressure of a real gas ap-
proaches zero, its equation of state approaches that of an ideal gas:
Pv = RT
It will be shown that the ideal-gas temperature scale satisfies the definition of thermody-
namic temperature given in the preceding section by Eq. 7.7, but first let us consider how
an ideal gas might be used to measure temperature in a constant- volume gas thermometer,
shown schematically in Fig. 7.22.
Let the gas bulb be placed in the location where the temperature is to be measured,
and let the mercury column be adjusted so that the level of mercury stands at the reference
mark A. Thus, the volume of the gas remains constant. Assume that the gas in the capil-
lary tube is at the same temperature as the gas in the bulb. Then the pressure of the gas,
which is indicated by the height L of the mercury column, is a measure of the temperature!
T(F) + 459.67 = T(K)
B
4
L
Capillary tube
3
Mercury column
t
FIGURE 7.22
Schematic diagram of a
constant-volume gas
thermometer.
234 H chapter Seven the second law of thermodynamics
Let the pressure that is associated with the temperature of the triple point of water
(273.1 6 K) also be measured, and let us designate this pressure P t? . Then, from the defin-
ition of an ideal gas, any other temperature T could be deterrnined from a pressure mea-
surement P by the relation
From a practical point of view, we have the problem that no gas behaves exactly
like an ideal gas. However, we do know that as the pressure approaches zero, the behavior
of all gases approaches that of an ideal gas. Suppose then that a series of measurements is
made with varying amounts of gas in the gas bulb. This means that the pressure measured
at the triple point, and also the pressure at any other temperature, will vary. If the indi-
cated temperature T, (obtained by assuming that the gas is ideal) is plotted against the
pressure of gas with the bulb at the triple point of water, a curve like the one shown in
Fig. 7.23 is obtained. When this curve is extrapolated to zero pressure, the correct ideal-
gas temperature is obtained. Different curves might result from different gases, but they
would all indicate the same temperature at zero pressure.
We have outlined only the general features and principles for measuring tempera- .
ture on the ideal-gas scale of temperatures. Precision work in this field is difficult and la-
borious, and there are only a few laboratories in the world where such work is carried on.
The International Temperature Scale, which was mentioned in Chapter 2, closely approxi-
mates the thermodynamic temperature scale and is much easier to work with in actual
temperature measurement.
We now demonstrate that the ideal-gas temperature scale discussed earlier is, in
fact, identical to the thermodynamic temperature scale, which was defined in the discus-
sion of the Carnot cycle and the second law. Our objective can be achieved by using an
ideal gas as the working fluid for a Carnot-cycle heat engine and analyzing the four
processes that make up the cycle. The four state points, 1, 2, 3, and 4, and the four
processes are as shown in Fig. 7.24, For convenience, let us consider a unit mass of gas
inside the cylinder. Now for each of the four processes, the reversible work done at the
moving boundary is given by Eq. 4.2:
Sw = Pdv
Similarly, for each process the gas behavior is, from the ideal-gas relation, Eq. 3,5,
T= 273.16
Pv=RT
CD
B
o
FIGURE 7.23 Sketch
showing how the ideal-
gas temperature is
determined.
Pressure at triple point, P tp .
the Ideal-Gas Temperature Scale @ 235
P
H
V
FIGURE 7.24 The
ideal-gas Camot cycle.
Ideal gas
and the internal energy change, from Eq. 5.20, is
du = C c0 dT
Assuming no changes in kinetic or potential energies, the first law is, from Eq. 5.7 at unit
mass,
Substituting the three previous expressions into this equation, we have for each of the four
processes
The shape of the two isothermal processes shown in Fig. 7.24 is known, since Pv is
constant in each case. The process 1-2 is an expansion at T H , such that v 2 is larger than i? t .
Similarly, the process 3~4 is a compression at a lower temperature, T L , and v A is smaller
than v y The adiabatic process 2-3 is an expansion from T H to T L , with an increase in spe-
cific volume, while the adiabatic process 4-1 is a compression from T L to T H , with a de-
crease in specific volume. The area below each process line represents the work for that
process, as given by Eq. 4.2.
We now proceed to integrate Eq. 7.9 for each of the four processes that make up the
Camot cycle. For the isothermal heat addition process 1-2, we have
Sq — du + Sw
Sq^C v0 dT+^-dv
(7.9)
9h=i^2 = + RT h ln^
(7.10)
For the adiabatic expansion process 2-3,
(7.11)
For the isothermal heat rejection process 3-4,
? i = -3?4 = -o-i?r i in^
= +RT L \a~
(7.12)
236 CHAPTER SEVEN THE SECOND LAW OF THERMODYNAMICS
and for the adiabatic compression process 4—1,
0=^ j^"~dT+ Rln~ (7.13)
From Eqs. 7.11 and 7.13, we get
J,
Therefore,
T T v 2 u 4
v 3 u 4 y 3 u 2
or Jri (7J4)
Thus, from Eqs. 7.10 and 7.12 and substituting Eq. 7.14, we find that
which is Eq. 7.7, the definition of the thermodynamic temperature scale in connection
with the second law.
7.9 Ideal versus real machines
Following the definition of the thermodynamic temperature scale by Eq. 7.7, it was noted
that the thermal efficiency of a Camot cycle heat engine is given by Eq. 7.8. It also fol-
lows that a Camot cycle operating as a refrigerator or heat pump will have a coefficient of
performance expressed as
" - 1 (7.15)
Qh Ql Camot T H T L
\lH yi Camot l H 1 L
For all three "efficiencies" in Eqs. 7.8, 7.15, and 7.16, the first equality sign is the defini-
tion with the use of the energy equation and thus is always true. The second equality sign
is valid only if the cycle is reversible, that is, a Carnot cycle. Any real heat engine, refrig-
erator, or heat pump will be less efficient, such that
= 1 _ <; 1 -Ik
''Jrea! thermal r\ — ^ t
n. r.
Pteai =
#ea! =
Qh-Ql t h -t l
Qh _ T H
Qh~Ql T h ~T l
A final point needs to be made about the significance of absolute-zero temperature in
connection with the second law and the thermodynamic temperature scale. Consider a
Carnot-cycle heat engine that receives a given amount of heat from a given high-temperature
Ideal versus Real Machines H 237
reservoir. As the temperature at which heat is rejected from the cycle is lowered, the net
work output increases and the amount of heat rejected decreases. In the limit, the heat re-
jected is zero, and the temperature of the reservoir corresponding to this limit is absolute
zero.
Similarly, for a Carnot-cycle refrigerator, the amount of work required to produce a
given amount of refrigeration increases as the temperature of the refrigerated space de-
creases. Absolute zero represents the limiting temperature that can be achieved, and the
amount of work required to produce a finite amount of refrigeration approaches infinity as
the temperature at which refrigeration is provided approaches zero.
EXAMPLE 7.3 Let us consider the heat engine, shown schematically in Fig. 7.25, that receives a heat-
transfer rate of 1 MW at a high temperature of 550°C and rejects energy to the ambient
surroundings at 300 K. Work is produced at a rate of 450 kW. We would like to know
how much energy is discarded to the ambient surroundings and the engine efficiency
and compare both of these to a Carnot heat engine operating between the same two
reservoirs.
Solution
If we take the heat engine as a control volume, the energy equation gives
Qi. ■- Q H ~ W= 1000 - 450 = 550 kW
and from the definition of the efficiency
dermal = WQu = 450/1000 = 0.45
For the Carnot heat engine, the efficiency is given by the temperature of the reservoirs :
= 1-^=1-
300
= 0.635
T H 550 + 273
The rates of work and heat rejection become
W^ycimo&H = 0-635 X 1000 = 635 kW
Ql = Qh~W= 1000 - 635 = 365 kW
FIGURE 7,25 A heat
engine operating between
two constant temperature
energy reservoirs for
Example 7.1.
238 a CHAPTER SEVEN THE SECOND LAW OF THERMODYNAMICS
The actual heat engine thus has a lower efficiency than the Camot (ideal) heat engine,
with a value of 45% typical for a modem steam power plant. This also implies that the
actual engine rejects a larger amount of energy to the ambient surroundings (55%) com-
pared with the Carnot heat engine (36%).
Example 7.4
Solution
The coefficient of performance (COP) is
so the rate of work or power input will be
W= Q L /p = 4/27 = 0.15 kW
Since the power was estimated assuming a Carnot refrigerator, it is the smallest amount
possible. Recall also the expressions for heat-transfer rates in Chapter 4. If the refrigera-
tor should push 4.15 kW out to the atmosphere at 35°C, the high-temperature side of it
should be at a higher temperature, maybe 45°C, to have a reasonably small-sized heat
As one mode of operation of an air conditioner is the cooling of a room on a hot day, it
works as a refrigerator, shown in Fig. 7.26. A total of 4 kW should be removed from a
room at 24°C to the outside atmosphere at 35°C. We would like to estimate the magni-
tude of the required work. To do this we will not analyze the processes inside the refrig-
erator, which is deferred to Chapter U, but we can give a lower limit for the rate of
work assuming it is a Carnot-cycle refrigerator.
FIGURE 7.26 An air
conditioner in cooling
mode where T L is the
room.
An air conditioner In cooling mode
Summary U 239
exchanger. As tt cools the room, a flow of air of less than, say, 18°C would be needed.
Redoing the COP with a high of 45°C and a low of 18°C gives 10.8, which is more real-
istic. A real refrigerator would operate with a COP of the order of 5 or less.
In the previous discussion and examples we considered the constant-temperature
energy reservoirs and used those temperatures to calculate the Carnot-cycle efficiency.
However, if we recall the expressions for the rate of heat transfer by conduction, convec-
tion, or radiation in Chapter 4, they can all be shown as
Q = CM (7.17)
The constant C depends on the mode of heat transfer as
lc4
Conduction: C = — Convection: C = hA
Ax
Radiation: C = eaA(T s 2 + Tl)(T s + T n )
For more complex situations with combined layers and modes, we also recover the form
in Eq. 7.17, but with a value of C that depends on the geometry, materials, and modes of
heat transfer. To have a heat transfer, we therefore must have a temperature difference so
that the working substance inside a cycle cannot attain the reservoir temperature unless
the area is infinitely large.
NUMMARY
The classical presentation of the second law of thermodynamics starts with the concept of
heat engines and refrigerators. A heat engine produces work from a heat transfer obtained
from a thermal reservoir, and its operation is limited by the Kelvin-Planck statement. Re-
frigerators are functionally the same as heat pumps, and they drive energy by heat transfer
from a colder environment to a hotter environment, something that will not happen by it-
self. The Clausius statement says in effect that the refrigerator or heat pump does need
work input to accomplish the task. To approach the limit of these cyclic devices, the idea
of reversible processes is discussed and further explained by the opposite, namely, irre-
versible processes and impossible machines. A perpetual motion machine of the first kind
violates the first law (energy equation), and a perpetual machine of the second kind vio-
lates the second law of thermodynamics.
The limitations for the performance of heat engines (thermal efficiency) and heat
pumps or refrigerators (coefficient of performance or COP) are expressed by the corre-
sponding Camot-cycie device. Two propositions about the Carnot cycle device are an-
other way of expressing the second law of thermodynamics instead of the statements of
Kelvin-Planck or Clausius. These propositions lead to the establishment of the thermody-
namic absolute temperature, done by Lord Kelvin, and the Camot-cycle efficiency. We
show this temperature to be the same as the ideal-gas temperature introduced in Chapter 3,
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Understand the concepts of heat engines, heat pumps, and refrigerators.
• Have an idea about reversible processes.
• Know a number of irreversible processes and recognize them.
240 H chapter Seven the second Law of thermodynamics
• Know what a Camot-cycle is.
• Understand the definition of thermal efficiency of a heat engine.
• Understand the definition of coefficient of performance of a heat pump.
• Know the difference between the absolute and relative temperature.
• Know the limits of thermal efficiency as dictated by the thermal reservoirs and the
Camot-cycle device.
• Have an idea about the thermal efficiency of real heat engines.
• Know the limits of coefficient of performance as dictated by the thermal reservoirs
and the Carnot-cycle device.
• Have an idea about the coefficient of performance of real refrigerators.
Key Concepts
and Formulas
(All IV, Q can also be rates W, Q)
Heat engine
^HE
= Qh~
Ql,
T?HE ~
^HE
Qh
Heat pump
= Qh~
Ql,
jShp =
Qh
w m
Refrigerator
^REF
= Qh~
Ql>
/3ref =
Ql
Factors that make
processes irreversible
Camot cycle
Proposition I
Proposition II
Absolute temperature
Real heat engine
Real heat pump
Real refrigerator
Heat-transfer rates
_Ql
Qh
Qh
Friction, unrestrained expansion (W = 0), Q over AT,
mixing, current through a resistor, combustion, or valve
flow (throttle).
1- 2 Isothermal heat addition Q H in at T H
2- 3 Adiabatic expansion process T does down
3- 4 Isothermal heat rejection Q L out at T L
4- 1 Adiabatic compression process Tgoes up
Same T H) T L
Same T H , T L
^lany ~ ^Jreversible
'fcanvot 1 ~ 1 7camot 2
T H Qh
£he
Qh
''fcainot HE * 7*
l H
i?he -
J8hp =
/^REF ~ 7}7 — ^CamotREF
"REF
Q= CAT
''HP 1 H J t
T„ - Tr
Concept-Study Guide problems
7.1 Electrical appliances (TV, stereo) use electric
power as input. What happens to the power? Are
those heat engines? What does the second law say
about those devices?
7.2 A gasoline engine produces 20 hp using 35 kW of
heat transfer from burning fuel. What is its thermal
efficiency, and how much power is rejected to the
ambient?
7.3 A refrigerator removes 1.5 kj from the cold space
using 1 kJ work input. How much energy goes
into the kitchen, and what is its coefficient of
performance?
HOMEWORK PROBLEMS U 241
7.4 Assume we have a refrigerator operating at steady
state using 500 W of electric power with a COP of
2.5. What is the net effect on the kitchen air?
7.5 A window air-conditioner unit is placed on a lab-
oratory bench and tested in cooling mode using
750 W of electric power with a COP of 1.75. What
is the cooling power capacity, and what is the net
effect on the laboratory?
7.6 Geothermal underground hot water or steam can be
used to generate electric power. Does that violate
the second law?
7.7 A car engine takes atmospheric air in at 20°C, no
fuel, and exhausts the air at -20°C producing work
in the process. What do the first and second laws
say about that?
7.8 A windmill produces power on a shaft taking ki-
netic energy out of the wind. Is it a heat engine? Is
it a perpetual machine? Explain.
7.9 Ice cubes in a glass of liquid water will eventually
melt and all the water will approach room tempera-
ture. Is this a reversible process? Why?
7.10 A room is heated with a 1500 W electric heater.
How much power can be saved if a heat pump with
a COP of 2.0 is used instead?
7.11 If the efficiency of a power plant goes up as the
low temperature drops, why do they not just reject
energy at say -40°C?
Homework problems
Heat Engines and Refrigerators
7.18 Calculate the thermal efficiency of the steam power
plant given in Example 6.9.
7.19 Calculate the coefficient of performance of the
R-134a refrigerator given in Example 6.10.
7.20 Calculate the thermal efficiency of the steam power
plant cycle described in Problem 6.99.
7.21 Calculate the coefficient of performance of the
R-12 heat pump cycle described in Problem 6.106.
7.22 A farmer runs a heat pump with a 2 kW motor. It
should keep a chicken hatchery at 30°C, which
loses energy at a rate of 10 kW to the colder ambt-
ent T'amb- What is the minimum coefficient of per-
formance that will be acceptable for the heat
pump?
7.12 If the efficiency of a power plant goes up as the
■low temperature drops, why not let the heat rejec-
tion go to a refrigerator at, say, — 10°C instead of
ambient 20°C?
7.13 A coal-fired power plant operates with a high T of
600°C } whereas a jet engine has about 1400 K.
Does that mean we should replace all power plants
with jet engines?
7.14 A heat transfer requires a temperature* difference
(see Chapter 4) to push the Q, What implications
does that have for a real heat engine? a refrigerator?
7.15 A large stationary diesel engine produces 15 MW
with a thermal efficiency of 40%. The exhaust gas,
which we assume is air, flows out at 800 K, and the
intake is 290 K. How large a mass flow rate is that
if that accounts for half the Q{i Can the exhaust
flow energy be used?
7.16 Hot combustion gas (air) at 1500 K is used as the
heat source in a heat engine where the gas is
cooled to 750 K and the ambient is at 300 K. This
is not a constant T source. How does that affect the
efficiency?
7.17 A remote location without electricity operates a re-
frigerator with a bottle of propane feeding a burner
to create hot gases. Sketch the setup in terms of
cyclic devices and give a relation for the ratio of Q L
in the refrigerator to £> fcel in the burner in terms
of the various reservoir temperatures.
7.23 A power plant generates 150 MW of electrical
power. It uses a supply of 1000 MW from a geo-
thermal source and rejects energy to the atmos-
phere. Find the power to the air and how much air
should be flowed to the cooling tower (kg/s) if its
temperature cannot be increased more than 10°C.
7.24 A car engine delivers 25 hp to the driveshaft with a
thermal efficiency of 30%. The fuel has a heating
value of 40 000 kj/kg. Find the rate of fuel con-
sumption and the combined power rejected through
the radiator and exhaust.
7.25 For each of the cases below, determine if the heat
engine satisfies the first law (energy equation) and
if it violates the second law.
a. Q H = 6 kW Q L = 4 kW W = 2 kW
b. Q H = 6 kW Q L = kW W = 6 kW
242 H CHAPTER SEVEN THE SECOND LAW OF THERMODYNAMICS
c. fiff = 6kW fii = 2kW IF=5kW
d. g„=6kW L = 6kW JF=0kW
7.26 In a steam power plant 1 MW is added in the
boiler, 0.58 MW is taken out in the condenser, and
the pump work is 0.02 MW. Find the plant thermal
efficiency. If everything could be reversed, find the
coefficient of performance as a refrigerator?
7.27 Electric solar cells can produce electricity with
15% efficiency. Compare that to a heat engine dri-
ving an electric generator of efficiency 80%. What
should the heat engine efficiency be to have the
same overall efficiency as the solar cells?
7.28 For each of the cases in Problem 7.25 determine if
a heat pump satisfies the first law (energy equation)
and if it violates the second law.
7.29 An air conditioner discards 5.1 kW to the ambient
surroundings with a power input of 1,5 kW. Find the
rate of cooling and the coefficient of performance.
w=\J5WN
FIGURE P7.29
7.30 Calculate the amount of work input a refrigerator
needs to make ice cubes out of a tray of 0.25 kg
liquid water at 10°C. Assume the refrigerator has
— 3.5 and a motor-compressor of 750 W. How
much time does it take if this is the only cooling
load?
7.31 A house needs to be heated by a heat pump, with
j3' = 2.2, and maintained at 20°C at all times. It is
estimated that it loses 0.8 kW per degree the ambi-
ent temperature is lower than the 20°C, Assume an
outside temperature of — 10°C and find the needed
power to drive the heat pump.
7.32 Refrigerant R-12 at 95°C with x- 0.1 flowing at
2 kg/s is brought to saturated vapor in a constant-
pressure heat exchanger. The energy is supplied by
a heat pump with a coefficient of performance of
(3 f = 2.5. Find the required power to drive the heat
pump.
Second Law and Processes
7.33 Prove that a cyclic device that violates the
Kelvfn-Planck statement of the second law also vi-
olates the Clausius statement of the second law.
7.34 Discuss the factors that would make the power
plant cycle described in Problem 6.99 an irre-
versible cycle.
7.35 Assume a cyclic machine that exchanges 6 kW
with a 250°C reservoir and has
a. Q L = kW W = 6 kW
b. Qi = 6kw W= OkW
and Q L is exchanged with a 30°C ambient surround-
ings. What can you say about the processes in the
two cases a and b if the machine is a heat engine?
Repeat the question for the case of a heat pump.
7.36 Discuss the factors that would make the heat pump
cycle described in Problem 6.106 an irreversible
cycle.
7.37 The water in a shallow pond heats up during the
• day and cools down during the night. Heat transfer
by radiation, conduction, and convection with the
ambient surroundings thus cycles the water tem-
perature. Is such a cyclic process reversible or irre-
versible?
7.38 Consider a heat engine and heat pump connected as
shown in Fig. P7.38, Assume T m ~T m > and
determine for each of the three cases if the setup
satisfies the first law and/or violates the second
law.
HOMEWORK PROBLEMS M 243
Qm
Qli
Qm
012 ^2
a
6
4
2
3
2 1
b
6
4
2
5
4 1
c
3
2
1
4
3 1
FIGURE P7.38
7.39 Consider the four cases of a heat engine in Problem
7.25 and detenrtine if any of those are perpetual-
motion machines of the first or second kind.
Carnot Cycles and Absolute Temperature
7.40 Calculate the thermal efficiency of a Carnot-
cycle heat engine operating between reservoirs at
300°C and 45°C. Compare the result to that of
Problem 7.18,
7.41 At a few places where the air is very cold in the
winter, for example, — 30°C, it is possible to find a
temperature of 13°C down below ground. What ef-
ficiency will a heat engine have operating between
these two thermal reservoirs?
7.42 Calculate the thermal efficiency of a Camot-cycle
heat pump operating between reservoirs at 0°C and
45°C. Compare the result to that of Problem 7.21.
7.43 Find the power output and the low T heat rejec-
tion rate for a Carnot-cycle heat engine that re-
ceives 6 kW at 250°C and rejects heat at 30°C as
in Problem 7.35.
7.44 A car engine burns 5 kg of fuel (equivalent to
adding Q H ) at 1500 K and rejects energy to the ra-
diator and exhaust at an average temperature of
750 K. Assume the fuel has a heating value of
40 000 kJ/kg and find the maximum amount of
work the engine can provide.
7.45 Differences in surface water and deep water tem-
• perature can be utilized for power generation. It is
proposed to construct a cyclic heat engine that will
operate near Hawaii, where the ocean temperature
is 20°C near the surface and 5°C at some depth.
What is the possible thermal efficiency of such a
heat engine?
7.46 Find the maximum coefficient of performance for
the refrigerator in your kitchen, assuming it runs in
a Carnot cycle.
7.47 An air conditioner provides 1 kg/s of air at 15°C
cooled from outside atmospheric air at 35°C. Esti-
mate the amount of power needed to operate the air
conditioner. Clearly state all assumptions made.
7.48 We propose to heat a house in the winter with a
heat pump. The house is to be maintained at 20°C
at all times. When the ambient temperature outside
drops to — 10°C, the rate at which heat is lost from
the house is estimated to be 25 kW. What is the
minimum electrical power required to drive the
heat pump?
FIGURE P7.48
7.49 A salesperson selling refrigerators and deep freezers
will guarantee a minimum coefficient of perfor-
mance of 4.5 year round. How would the perfor-
mance of these machines compare? Would it be
steady throughout the year?
7.50 A cyclic machine, shown in Fig. P7.50, receives
325 kJ from a 1000 K energy reservoir. It rejects
Cyclic
machine
4
o
fi H = 325 kJ
ZZ^> If =200 kJ
fi L =125kJ
/ T,, = 400 K \
FIGURE P7.50
244 H
Chapter Seven The Second Law of thermodynamics
125 kJ to a 400 K energy reservoir, and the cycle
produces 200 kJ of work as output. Is this cycle re-
versible, irreversible, or impossible?
7.51 An inventor has developed a refrigeration unit that
maintains the cold space at — 1 0°C, while operating
in a 25°C room. A coefficient of performance of
8.5 is claimed. How do you evaluate this?
7.52 A household freezer operates in a room at 20°C.
Heat must be transferred from the cold space at a
rate of 2 kW to maintain its temperature at -30°C.
What is the theoretically smallest (power) motor
required to operate this freezer?
7.53 In a cryogenic experiment you need to keep a con-
tainer at -125°C although it gains 100 W due to
heat transfer. What is the smallest motor you would
need for a heat pump absorbing heat from the con-
tainer and rejecting heat to the room at 20°C?
7.54 A temperature of about 0.01 K can be achieved by
magnetic cooling. In this process a strong magnetic
field is imposed on a paramagnetic salt, maintained
at 1 K. by transfer of energy to liquid helium boil-
ing at low pressure. The salt is then thermally
isolated from the helium, the magnetic field is re-
moved, and the salt temperature drops. Assume
that 1 mj is removed at an average temperature of
0.1 K to the helium by a Carnot-cycle heat pump.
Find the work input to the heat pump and the coef-
ficient of performance with an ambient at 300 K.
7.55 The lowest temperature that has been achieved is
about 1 X 10~ 6 K. To achieve this an additional
stage of cooling is required beyond that described in
the previous problem, namely, nuclear cooling. This
process is similar to magnetic cooling, but it in-
volves the magnetic moment associated with the nu-
cleus rather than that associated with certain ions in
the paramagnetic salt. Suppose that 10 juJ is to be re-
moved from a specimen at an average temperature of
10~ 5 K (10 microjoules is about the potential energy
loss of a pin dropping 3 mm). Find the work input to
a Carnot heat pump and its coefficient of perfor-
mance to do this assuming the ambient is at 300 K.
7.56 A certain solar-energy collector produces a maxi-
mum temperature of 100°C. The energy is used in a
cyclic heat engine that operates in a 10°C environ-
ment. What is the maximum thermal efficiency?
What is it if the collector is redesigned to focus the
incoming light to produce a maximum temperature
of300°C?
7.57 Helium has the lowest normal boiling point of any
of the elements at 4,2 K. At this temperature the
enthalpy of evaporation is 83.3 kJ/kmol. A Camot
refrigeration cycle is analyzed for the production of
1 kmol of liquid helium at 4.2 K from saturated
vapor at the same temperature. What is the work
input to the refrigerator and the coefficient of per-
formance for the cycle with an ambient tempera-
ture at 300 K?
7.58 Calculate the amount of work input a refrigerator
needs to make ice cubes out of a tray of 0.25 kg
liquid water at 10°C. Assume the refrigerator
works in a Camot cycle between — 8°C and 35°C
with a motor-compressor of 750 W. How much
time does it take if this is the only cooling load?
7.59 A steel bottle of V - 0.1 m 3 contains R-134a at
20°C and 200 kPa. It is placed in a deep freezer
where it is cooled to —20°C. The deep freezer sits
in a room with ambient temperature of 20°C and
has an inside temperature of — 20°C. Find the
amount of energy the freezer must remove from
the R-134a and the extra amount of work input to
the freezer to do the process.
7.60 Liquid sodium leaves a nuclear reactor at 800°C
and is used as the energy source in a steam power
plant. The condenser cooling water comes from a
cooling tower at 15°C. Determine the maximum
thermal efficiency of the power plant. Is it mislead-
ing to use the temperatures given to calculate this
value?
7.61 A thermal storage device is made with a rock
(granite) bed of 2 m 3 that is heated to 400 K using
solar energy. A heat engine receives a Q H from the
bed and rejects heat to the ambient surroundings at
290 K. The rock bed therefore cools down, and as
it reaches 290 K the process stops. Find the energy
the rock bed can give out. What is the heat engine
efficiency at the beginning of the process, and what
is it at the end of the process?
FIGURE P7.61
HOMEWORK PROBLEMS M 245
7 62 A heat engine has a solar collector receiving 0,-2
kW/m 2 , inside of which a transfer media is heated
to 450 K. The collected energy powers a heat en-
gine that rejects heat at 40°C. If the heat engine
should deliver 2.5 kW, what is the minimum size
(area) solar collector?
7.63 Sixty kilograms per hour of water runs through a
heat exchanger, entering as saturated liquid at 200
kPa and leaving as saturated vapor. The heat is
supplied by a Carnot heat pump operating from a
low-temperature reservoir at 16°C. Find the rate of
work into the heat pump.
7.64 A heat pump is driven by the work output of a heat
engine as shown in Figure P7.64. If we assume ideal .
devices, find the ratio of the total power Q Ll + Q m
that heats the house to the power from the hot en-
ergy source Q m in terms of the temperatures.
J \ amb /
Gi2
w
Qhz
House T„
FIGURE P7.64
7.65 it is proposed to build a 1000 MW electric power
plant with steam as the working fluid. The con-
densers are to be cooled with river water (see Fig.
P7.65). The maximum steam temperature is 550°C,
and the pressure in the condensers will be 10 kPa.
. Estimate the temperature rise of the river down-
stream from the power plant.
7.66 Two different fuels can be used in a heat engine
operating between the fuel burning tempera-
ture and a low temperature of 350 K. VuelA burns
at 2200 K delivering 30 000 kJ/kg and costs
St.50/kg. Fuel B burns at 1200 K, delivering 40
000 kJ/kg and costs $1.30/kg. Which fuel would
you buy and why?
Finite AJHeat Transfer
7.67 A refrigerator keeping 5°C inside is located in a
30°C room. It must have a high temperature A T
above room temperature and a low temperature AT 1
below the refrigerated space in the cycle to actually
transfer the heat. For a A7/ of 0, 5, and 10°C, re-
spectively, calculate the COP assuming a Carnot
cycle.
7.68 A refrigerator uses a power input of 2.5 kW to cool
a 5°C space with the high temperature in the cycle
as 50°C. The Q H is pushed to the ambient air at
35°C in a heat exchanger where the transfer coeffi-
cient is 50 W/m 2 K. Find the required minimum
heat transfer area.
7.69 A house is heated by a heat pump driven by an
electric motor using the outside as the low-
temperature reservoir. The house loses energy in
direct proportion to the temperature difference as
fioss ~ K(Th ~ Ti). Determine the minimum
electric power required to drive the heat pump as
a function of the two temperatures.
FIGURE P7.65
FIGURE P7.69
7.70 A farmer runs a heat pump with a motor of 2 k\V.
It should keep a chicken hatchery at 3Q°C, which
loses energy at a rate of 0.5 kW per degree differ-
ence to the colder ambient T^. The heat pump has
a coefficient of performance that is 50% of a
Camot heat pump. What is the minimum ambient
temperature for which the heat pump is sufficient?
1
246 H CHAPTER SEVEN THE SECOND LAW OF THERMODYNAMICS
7.71 Consider a Carnot-cycle heat engine operating in
outer space. Heat can be rejected from this engine
only by thermal radiation which is proportional to
the radiator area and the fourth power of absolute
temperature, Q Ri = KAT A . Show that for a given
engine work output and given T H , the radiator area
will be minimum when the ratio T L IT H = 3 / A .
7.72 A house is heated by an electric heat pump using
the outside as the low-temperature reservoir. For
several different winter outdoor temperatures, esti-
mate the percent savings in electricity if the house
is kept at 20°C instead of 24°C. Assume that the
house is losing energy to the outside as in Eq. 7. 17.
7.73 A house is cooled by an electric heat pump using
the outside as the high-temperature reservoir. For
several different summer outdoor temperatures,
estimate the percent savings in electricity if the
house is kept at 25°C instead of 20°C. Assume
that the house is gaining energy from the outside
in direct proportion to the temperature difference,
as in Eq. 7.17.
FIGURE P7.73
7.74 A heat pump has a COP of J3' = 0.5 P'carnot and
maintains a house at T H = 20°C, while it leaks en-
ergy out as Q = Q.6(T H - 7i)[kW]. For a maxi-
mum of 1.0 kW power input, find the minimum
outside temperature, T L , for which the heat pump is
a sufficient heat source.
7.75 An air conditioner cools a house at T L = 20°C with
a maximum of 1.2 kW power input. The house
gains energy as Q = 0.6{T H - T L )[k\V] and the re-
frigeration COP is j3 = 0.6 0CARNOT- Find the maxi "
mum outside temperature, T H , for which the air
conditioner unit provides sufficient cooling.
7.76 A Carnot heat engine, shown in Fig. P7.76 receives
energy from a reservoir at T re3 through a heat ex-
changer where the heat transferred is proportional
to the temperature difference as Qh ~ ~~
It rejects heat at a given low temperature T L . To de-
sign the heat engine for maximum work output,
show that the high temperature, T H , in the cycle
should be selected as T H = (T L T KS ) m .
Qh
V
' t l V FIGURE P7.76
Ideal-Gas Carnot Cycles
7.77 Hydrogen gas is used in a Camot cycle having an
efficiency of 60% with a low temperature of 300 K.
During the heat rejection the pressure changes
from 90 kPa to 120 kPa. Find the high- and low-
temperature heat transfer and the net cycle work
per unit mass of hydrogen.
7.78 An ideal-gas Carnot cycle with air in a piston cylin-
der has a high temperature of 1200 K and a heat re-
jection at 400 K. During the heat addition, the
volume triples. Find the two specific heat transfers
(q) in the cycle and the overall cycle efficiency.
7.79 Air in a piston/cylinder setup goes through a Carnot
cycle with the P-u diagram shown in Fig. 7.24. The
high and low temperatures are 600 K and 300 K, re- .
spectively. The heat added at the high temperature
is 250 kJ/kg, and the lowest pressure in the cycle is
75 kPa. Find the specific volume and pressure after
heat rejection and the net work per unit mass-
Review Problems
7.80 A car engine operates with a thermal efficiency of
35%. Assume the air conditioner has a coefficient
of performance of j3 = 3 working as a refrigerator
- cooling the inside using engine shaft work to drive
it. How much extra fuel energy should be spent to
remove 1 kJ from the inside?
7.81 An air conditioner with a power input of 1 .2 kW is
working as a refrigerator (fi = 3) or as a heat pump
(j8' = 4). It maintains an office at 20°C year round,
which exchanges 0.5 kW per degree temperature
difference with the atmosphere. Find the maximum
and the minimum outside temperature for which
this unit is sufficient.
homework Problems ffl 247
7 82 A rigid insulated container has two rooms separated
by a membrane. Room A contains 1 kg of air at
200°C, and room B has 1.5 kg of air at 20°C; both
rooms are at 100 kPa. Consider two different cases:
1. Heat transfer between A and B creates a final
uniform T.
2. The membrane breaks, and the air comes to a
uniform state.
For both cases find the final temperature. Are the
two processes reversible and different? Explain.
7.83 At a certain location, geothermal energy in under-
ground water is available and used as an energy
source for a power plant. Consider a supply of sat-
urated liquid water at 150°C. What is the maximum
possible thermal efficiency of a cyclic heat engine
using this source as energy with the ambient sur-
roundings at 20°C? Would it be better to locate a
source of saturated vapor at 150°C than to use the
saturated liquid?
7.84 We wish to produce refrigeration at — 30°C. A
reservoir, shown in Fig. P7.84, is available at
200°C, and the ambient temperature is 30°C. Thus,
work can be done by a cyclic heat engine operat-
ing between the 200°C reservoir and the ambient
surroundings. This work is used to drive the refrig-
erator. Determine the ratio of the heat transferred
from the 20O°C reservoir to the heat transferred
from the -30°C reservoir, assuming alt processes
are reversible.
\ T ho\ I \ ^ambient j
Qh
m
w
©
Ql
FIGURE P7.84
"A r
7.85 A 4 L jug of milk at 25°C is placed in your refrig-
erator where it is cooled down to 5°C. The high
temperature in the Carnot refrigeration cycle is
45°C, and the properties of milk are the same as for
liquid water. Find the amount of energy that must
be removed from the milk and the additional work
needed to drive the refrigerator.
7.86 A combination of a heat engine driving a heat
; pump (see Fig. P7.86) takes waste energy at 50°C
as a source Q wU to the heat engine rejecting heat at
30°C. The remainder, Q^, goes into the heat pump
that delivers a Q H at lSO'C. If the total waste en-
ergy is 5 MW, find the rate of energy delivered at
the high temperature.
\ Waste energy 50°C j
' A-m j \j2wz
W
I 30°C \ It h =VSWC\ FIGURE P7.86
7.87 Air in a rigid 1 m 3 box is at 300 K and 200 kPa. It
is heated to 600 K by heat transfer from a re-
versible heat pump that receives energy from the
ambient surroundings at 300 K besides the work
input. Use constant specific heat at 300 K. Since
the coefficient of performance changes write SQ =
m^C„ dT and find 8W. Integrate SJFwith tempera-
ture to find the required heat pump work.
7.88 Consider the rock bed thermal storage in Problem
7.61. Use the specific heat so you can write 8Q H in
terms of dT mtk and find the expression for SJFout of
the heat engine. Integrate this expression over tem-
perature and find the total heat engine work output.
7.89 A heat pump heats a house in the winter and then re-
verses to cool it in the summer. The interior temper-
ature should be 20°C in the winter and 25°C in the
summer. Heat transfer through the walls and ceilings
is estimated to be 2400 kJ per hour per degree tem-
perature difference between the inside and outside.
a. If the outside winter temperature is 0°C, what is
the minimum power required to drive the heat
pump?
b. For the same power as in part (a), what is the
maximum outside summer temperature for which
the house can be maintained at 25°C?
7.90 A furnace, shown in Fig. P7.90, can deliver heat,
Q m , at T m , and it is proposed to use this to drive a
heat engine with a rejection at T itm instead of direct
room heating. The heat engine drives a heat pump
248 U Chapter seven the Second law of thermodynamics
FIGURE P7.90
English Unit problems
Concept Problems
7.92E A gasoline engine produces 20 hp using 35
Btu/s of heat transfer from burning fuel. What is
its thermal efficiency, and how much power is
rejected to the ambient?
7.93E A refrigerator removes 1.5 Btu from the cold
space using 1 Btu work input. How much en-
ergy goes into the kitchen, and what is its coeffi-
cient of performance?
7.94E A window air-conditioner unit is placed on a
laboratory bench and tested in cooling mode
using 0.75 Btu/s of electric power with a COP
of 1,75. What is the cooling power capacity, and
what is the net effect on the laboratory?
7.95E A car engine takes atmospheric air in at 70 F, no
fuel, and exhausts the air at F producing work
in the process. What do the first and second laws
say about that?
7.96E A large stationary diesel engine produces 20 000
hp with a thermal efficiency of 40%. The exhaust
gas, which we assume is air, flows out at 1400 R,
and the intake is 520 R. How large a mass flow
rate is that if that accounts for half the Q L 7 Can
the exhaust flow energy be used?
English Unit Problems
7.97E Calculate the thermal efficiency of the steam
power plant described in Problem 6. 1 67.
7.98E A farmer runs a heat pump with a 2 kW motor.
It should keep a chicken hatchery at 90 F which
that delivers Q m at T mom using the atmosphere as
the cold reservoir. Find the ratio Q m !Q m as a func-
tion of the temperatures. Is this a better setup than
direct room heating from the furnace?
7.91 A 10 m 3 tank of air at 500 kPa and 600 K acts as the
high-temperature reservoir for a Carnot heat engine
that rejects heat at 300 K. A temperature difference
of 25°C between the air tank and the Camot-cycle
high temperature is needed to transfer the heat. The
heat engine runs until the air temperature has
dropped to 400 K and then stops. Assume constant
specific heat capacities for air and determine how
much work is given out by the heat engine.
loses energy at a rate of 10 Btu/s to the colder
T^, What is the minimum coefficient of perfor-
mance that will be acceptable for the heat
pump?
7.99E Calculate the amount of work input a refrig-
erator needs to make ice cubes out of a tray of
0.5 Ibm liquid water at 50 F. Assume the refrig-
erator has (3 = 3.5 and a motor-compressor of
750 W. How much time does it take if this is the
only cooling load?
7.100E In a steam power plant 1000 Btu/s is added at
1200 F in the boiler, 580 Btu/s is taken out at
100 F in the condenser, and the pump work is 20
Btu/s. Find the plant thermal efficiency. Assum-
ing the same pump work and heat transfer to the
boiler as given, how much turbine power could
be produced if the plant were mrining in a
Camot cycle?
7.101E Calculate the thermal efficiency of a Carnot-
cycle heat engine operating between reservoirs
at 920 F and 1 10 F. Compare the result with that
of Problem 7.97E.
7.102E A car engine bums 10 lbm of fuel (equivalent to
addition of Q H ) at 2600 R and rejects energy to
the radiator and the exhaust at an average tem-
perature of 1300 R. If the fuel provides 17 200
Btu/lbm, what is the maximum amount of work
the engine can provide?
7.103E An air-conditioner provides 1 Ibm/s of air at
60 F cooled from outside atmospheric air at 95 F.
Estimate the amount of power needed to operate
ENGLISH UNIT PROBLEMS 11 249
the air-conditioner. Clearly state all assumptions
made.
7.104E We propose to heat a house in the winter with a
heat pump. The house is to be maintained at 68
F at all times. When the ambient temperature
outside drops to 15 F, the rate at which heat is
lost from the house is estimated to be 80 000
Btu/h. What is the minimum electrical power re-
quired to drive the heat pump?
7.105E An inventor has developed a refrigeration unit
that maintains the cold space at 14 F, while op-
erating in a 77 F room. A coefficient of perfor-
mance of 8.5 is claimed. How do you evaluate
this?
7.106E Liquid sodium leaves a nuclear reactor at 1500
F and is used as the energy source in a steam
power plant. The condenser cooling water
comes from a cooling tower at 60 F. Determine
the maximum thermal efficiency of the power
plant. Is it misleading to use the temperatures
given to calculate this value?
7.107E A house is heated by an electric heat pump
using the outside as the low-temperature reser-
voir. For several different winter outdoor tem-
peratures, estimate the percent savings in
electricity if the house is kept at 68 F instead of
75 F. Assume that the house is losing energy to
the outside directly proportional to the tempera-
ture difference as Q l0%5 = K(T H - T L ),
7.108E Refrigerant-22 at 180 F, x = 0.1 flowing at 4
lbm/s is brought to saturated vapor in a con-
stant-pressure heat exchanger. The energy is
supplied by a heat pump with a low temperature
of 50 F. Find the required power input to the
heat pump.
7.109E A heat engine has a solar collector receiving
600 Btu/h per square foot inside which a trans-
fer media is heated to 800 R. The collected en-
ergy powers a heat engine that rejects heat at
100 F. If the heat engine should deliver 8500
Btu/h, what is the minimum size (area) solar
collector?
7.1 10E Six-hundred pound-mass per hour of water
runs through a heat exchanger, entering as sat-
urated liquid at 250 F and leaving as saturated
vapor. The heat is supplied by a Carnot heat
pump operating from a low-temperature reser-
voir at 60 F. Find the rate of work into the heat
pump.
7. HIE A car engine operates with a thermal efficiency
of 35%. Assume the air-conditioner has a coeffi-
cient of performance that is one-third the theo-
retical maximum and it is mechanically pulled
by the engine. How much fuel energy should
you spend extra to remove I Btu at 60 F when
the ambient is at 95 F?
7.112E A heat pump cools a house at 70 F with a maxi-
mum of 4000 Btu/h power input. The house gains
2000 Btu/h per degree temperature difference to
the ambient, and the refrigerator coefficient of
performance is 60% of the theoretical maximum.
Find the maximum outside temperature for which
the heat pump provides sufficient cooling.
7.113E A house is cooled by an electric heat pump
using the outside as the high-temperature reser-
voir. For several different summer outdoor tem-
peratures estimate the percent savings in
electricity if the house is kept at 77 F instead of
68 F. Assume that the house is gaining energy
from the outside directly proportional to the
temperature difference.
7.114E A thermal storage is made with a rock (granite)
bed of 70 ft 3 , which is heated to 720 R using
solar energy. A heat engine receives a Q H from
the bed and rejects heat to the ambient at 520 R.
The rock bed therefore cools down, and as it
reaches 520 R the process stops. Find the energy
the rock bed can give out. What is the heat en-
gine efficiency at the beginning of the process,
and what is it at the end of the process?
7.115E We wish to produce refrigeration at -20 F. A
reservoir is available at 400 F and the ambient
temperature is 80 F, as shown in Fig. P7.84.
Thus, work can be done by a cyclic heat engine
operating between the 400 F reservoir and the
ambient. This work is used to drive the refrigera-
tor. Determine the ratio of the heat transferred
from the 400 F reservoir to the heat transferred
from the -20 F reservoir, assuming all processes
are reversible.
7.1I6E Air in a rigid 40 ft 3 box is at 540 R, 30 lbf/in. 2 .
It is heated to 1 100 R by heat transfer from a
reversible heat pump that receives energy
from the ambient at 540 R besides the work
input. Use constant specific heat at 540 R.
250 M Chapter Seven the second Law of thermodynamics
Since the coefficient of performance changes,
write 5Q = m ail C u dT and find 8 W. Integrate
SlVwith temperature to find the required heat
pump work.
7.117E A 350 ft 3 tank of air at 80 lbf/in. 2 , 1080 R acts as
the high-temperature reservoir for a Carnot heat
engine that rejects heat at 540 R. A temperature
difference of 45 F between the air tank and the
Carnot cycle high temperature is needed to
transfer the heat. The heat engine runs until the
air temperature has dropped to 700 R and then
stops. Assume constant specific heat capacities
for air, and find how much work is given out by
the heat engine.
7.118E Air in a piston/cylinder goes through a Carnot
cycle with the P-u diagram shown in Fig. 7.24.
The high and low temperatures are 1200 R and
600 Rj respectively. The heat added at the high
temperature is 100 Btu/lbm, and the lowest
pressure in the cycle is 10 lbf/in. 2 . Find the spe-
cific volume and pressure at all four states in the
cycle assuming constant specific heats at 80 F.
Entropy
Up to this point in our consideration of the second law of thermodynamics, we have dealt
only with thermodynamic cycles. Although this is a very important and useful approach,
we are often concerned with processes rather than cycles. Thus, we might be interested in
the second-law analysis of processes we encounter daily, such as the combustion process
in an automobile engine, the cooling of a cup of coffee, or the chemical processes that
take place in our bodies. It would also be beneficial to be able to deal with the second law
quantitatively as well as qualitatively.
In our consideration of the first law, we initially stated the law in terms of a cycle,
but we then defined a property, the internal energy, that enabled us to use the first law
quantitatively for processes. Similarly, we have stated the second law for a cycle, and we
now find that the second law leads to a property, entropy, that enables us to treat the sec-
ond law quantitatively for processes. Energy and entropy are both abstract concepts that
help to describe certain observations. As we noted in Chapter 2, thermodynamics can be
described as the science of energy and entropy. The significance of this statement will be-
come increasingly evident.
The fist step in our consideration of the property we call entropy is to establish the in-
equality of Clausius, which is
The inequality of Clausius is a corollary or a consequence of the second law of thermody-
namics. It will be demonstrated to be valid for all possible cycles, including both re-
versible and irreversible heat engines and refrigerators. Since any reversible cycle can be
represented by a series of Camot cycles, in this analysis we need consider only a Carnot
cycle that leads to the inequality of Clausius.
Consider first a reversible (Camot) heat engine cycle operating between reservoirs
at temperatures T H md T L , as shown in Fig. 8.1. For this cycle, the cyclic integral of the
heat transfer, f 8Q, is greater than zero.
Since T H and T L are constant, from the definition of the absolute temperature scale
and from the fact this is a reversible cycle, it follows that
8.1 The inequality of Clausius
251
252 H Chapter eight entropy
FIGURE 8.1
Reversible heat engine
cycle for demonstration of
the inequality of Clausius.
Qh
o
Qi
w H
If $ $Q> the cyclic integral of SQ, approaches zero (by making T H approach and
the cycle remains reversible, the cyclic integral of 5Q/T remains zero. Thus, we conclude
that for all reversible heat engine cycles
cj> SQ>0
and
82
T =°
Now consider an irreversible cyclic heat engine operating between the same T H
and T L as the reversible engine of Fig. 8.1 and receiving the same quantity of heat Q H .
Comparing the irreversible cycle with the reversible one, we conclude from the second
law that
Since Q H — Q L = IF for both the reversible and irreversible cycles, we conclude that
and therefore
Qi in > Ql rev
Consequently, for the irreversible cyclic engine,
j>SQ=Q H ~Q Lilz >0
18Q^Q H Q Lin
J T T H T L
Suppose that we cause the engine to become more and more irreversible, but
keep Q H , T H , and T L fixed. The cyclic integral of SQ then approaches zero, and that for
SQ/T becomes a progressively larger negative value. In the limit, as the work output
goes to zero,
C SQ =
8Q n
-f<0
The inequality of Clausius M 253
Thus, we conclude that for all irreversible heat engine cycles
j>8Q>0
ft
<0
To complete the demonstration of the inequality of Clausius, we must perform simi-
lar analyses for both reversible and irreversible refrigeration cycles. For the reversible re-
frigeration cycle shown in Fig. 8.2,
and
ft
Qh i Ql „ n
As the cyclic integral of SQ approaches zero reversibly (T H approaches T L ), the cyclic in-
tegral of SQIT remains at zero. In the limit,
j> 5g=0
Thus, for all reversible refrigeration cycles,
j> SQ <
ft-
Finally, let an irreversible cyclic refrigerator operate between temperatures T H and
T L and receive the same amount of heat Q L as the reversible refrigerator of Fig. 8.2. From
the second law, we conclude that the work input required will be greater for the irre-
versible refrigerator, or
W- > W
FIGURE 8.2
Reversible refrigeration
cycle for demonstration of
the inequality of Clausius.
o <=
254 M CHAPTER EIGHT ENTROPY
Since Qh~ Ql~ ft 7 for each cycle, it follows that
and therefore,
Qh irr > Qh ttv
That is, the heat rejected by the irreversible refrigerator to the high-temperature reservoir
is greater than the heat rejected by the reversible refrigerator. Therefore, for the irre-
versible refrigerator,
j>SQ=~Q Iiiir +Q L <0
As we make this machine progressively more irreversible, but keep Q L , T H , and T L
constant, the cyclic integrals of BQ and BQ/Tboth become larger in the negative direction.
Consequently, a limiting case as the cyclic integral of BQ approaches zero does not exist
for the irreversible refrigerator.
Thus, for all irreversible refrigeration cycles,
BQ <0
Summarizing, we note that, in regard to the sign of f BQ, we have considered all
possible reversible cycles (i.e., $ BQ ^ 0), and for each of these reversible cycles
We have also considered all possible irreversible cycles for the sign of f BQ (that is,
§ 8Q ^ 0), and for all these irreversible cycles
f*<0
Thus, for all cycles we can write
^f^O (8.1)
where the equality holds for reversible cycles and the inequality for irreversible cycles.
This relation, Eq. 8.1, is known as the inequality of Clausius.
The significance of the inequality of Clausius may be illustrated by considering the
simple steam power plant cycle shown in Fig. 8.3. This cycle is slightly different from the
usual cycle for steam power plants in that the pump handles a mixture of liquid and vapor
in such proportions that saturated liquid leaves the pump and enters the boiler. Suppose
that someone reports that the pressure and quality at various points in the cycle are as
given in Fig. 8.3. Does this cycle satisfy the inequality of Clausius?
Heat is transferred in two places, the boiler and the condenser. Therefore,
condenser
entropy— A property of a system H 255
Saturated vapor, 0.7 MPa
Tufbine — © g % quality, 15 kPa
FIGURE 8.3 A simple
steam power plant that
demonstrates the Pump (3
inequality of Clausius.
- -© Saturated liquid, 0.7 MPa
10% quality, 15kPa
Condenser
Since the temperature remains constant in both the boiler and condenser, this may be inte-
grated as follows:
4?e = ±
7t r,
Let us consider a 1 kg mass as the working fluid. We have then
x q 2 = A 2 - ^ = 2066.3 kJ/kg, T x = 164.97°C
3 ?4 = h - h 3 = 463.4 - 236 1.8 - - 1898.4 kJ/kg, T 3 = 53.9TC
Therefore,
2066.3 1898.4
J T "" 164.97 + 273.15 53.97 + 273.15
- -1.087 kJ/kg-K
Thus, this cycle satisfies the inequality of Clausius, which is equivalent to saying that it
does not violate the second law of thermodynamics.
8.2 Entropy— A Property of a System
By applying Eq. 8.1 and Fig. 8.4, we can demonstrate that the second law of thermo-
dynamics leads to a property of a system that we call entropy. Let a system (control mass)
•undergo a reversible process from state 1 to state 2 along a path A, and let the cycle be
completed along path 5, which is also reversible.
Because this is a reversible cycle, we can write
Now consider another reversible cycle, which proceeds first along path C and is then
completed along path B. For this cycle we can write
Subtracting the second equation from the first, we have
256 B Chapter eight ErfraoPY
FIGURE 8.4 Two
reversible cycles
demonstrating the fact
that entropy is a property
of a substance.
Since the f 8Q/T is the same for all reversible paths between states 1 and 2, we conclude
that this quantity is independent of the path and it is a function of the end states only; it is
therefore a property. This property is called entropy and is designated S. It follows that
entropy may be defined as a property of a substance in accordance with the relation
(8.2)
Entropy is an extensive property, and the entropy per unit mass is designated s. It is
important to note that entropy is defined here in terms of a reversible process.
The change in the entropy of a system as it undergoes a change of state may be
found by integrating Eq. 8.2. Thus,
(8.3)
To perform this integration, we must know the relation between T and Q, and illustrations
will be given subsequently. The important point is that since entropy is a property, the
change in the entropy of a substance in going from one state to another is the same for all
processes, both reversible and irreversible, between these two states. Equation 8.3 enables
us to find the change in entropy only along a reversible path. However, once the change
has been evaluated, this value is the magnitude of the entropy change for all processes be-
tween these two states.
Equation 8.3 enables us to calculate changes of entropy, but it tells us nothing about
absolute values of entropy. From the third law of thermodynamics, which is based on ob-
servations of low-temperature chemical reactions, it is concluded that the entropy of all
pure substances (in the appropriate structural form) can be assigned the absolute value of
zero at the absolute zero of temperature. It also follows from the subject of statistical ther-
modynamics that all pure substances in the (hypothetical) ideal-gas state at absolute zero
temperature have zero entropy.
However, when there is no change of composition, as would occur in a chemical re-
action, for example, it is quite adequate to give values of entropy relative to some arbitrar-
ily selected reference state, such as was done earlier when tabulating values of internal
energy and enthalpy. In each case, whatever reference value is chosen, it will cancel out
when the change of property is calculated between any two states. This is the procedure
followed with the thermodynamic tables to be discussed in the following section.
A word should be added here regarding the role of T as an integrating factor. We noted
in Chapter 4 that Q is a path function, and therefore 8Q in an inexact differential. However,
since (8QfT)^ is a thermodynamic property, it is an exact differential. From a mathematical
The Entropy of a pure Substance H 257
perspective, we note that an inexact differential may be converted to an exact differential by
the introduction of an integrating factor. Therefore, IIT serves as the integrating factor in
converting the inexact differential SQ to the exact differential SQ/T for a reversible process.
8.3 The Entropy of a Pure substance
Entropy is an extensive property of a system. Values of specific entropy (entropy per unit
mass) are given in tables of thermodynamic properties in the same manner as specific volume
and specific enthalpy. The units of specific entropy in the steam tables, refrigerant tables, and
ammonia tables are kJ/kg K, and the values are given relative to an arbitrary reference state.
In the steam tables the entropy of saturated liquid at 0.01°C is given the value of zero. For
many refrigerants, the entropy of saturated liquid at -40°C is assigned the value of zero.
In general, we use the term entropy to refer to both total entropy and entropy per
unit mass, since the context or appropriate symbol will clearly indicate the precise mean-
ing of the term.
In the saturation region the entropy may be calculated using the quality. The rela-
tions are similar to those for specific volume, internal energy and enthalpy.
s = (1 — x)Sf+ xs s
s =s f +xs /jl (8.4)
The entropy of a compressed liquid is tabulated in the same manner as the other
properties. These properties are primarily a function of the temperature and are not greatly
different from those for saturated liquid at the same temperature. Table 4 of the steam ta-
bles, which is -summarized in Table B. 1.4, give the entropy of compressed liquid water in
the same manner as for other properties.
The thennodynamic properties of a substance are often shown on a temperature-
entropy diagram and on an enthalpy-entropy diagram, which is also called a Moltier
diagram, after Richard Mollier (1863-1935) of Germany. Figures 8.5 and 8.6 show the
258 chapter eight entropy
FIGURE 8.6
Enthalpy-entropy
diagram for steam.
3000
2000
1000
Critical
point
"Saturated liquid
4 5 6
Entropy, kJ/kg K
FIGURE 8.7
Temperature-entropy
diagram to show
properties of a
compressed liquid, water.
Entropy Change in Reversible Processes H 259
essential elements of temperature-entropy and enthalpy-entropy diagrams for steam. The
general features of such diagrams are the same for all pure substances. A more complete
temperature-entropy diagram for steam is shown in Fig. E.l in Appendix E.
These diagrams are valuable both because they present thermodynamic data and be-
cause they enable us to visualize the changes of state that occur in various processes. As
our study progresses, the student should acquire facility in visualizing thermodynamic
processes on these diagrams. The temperature-entropy diagram is particularly useful for
this purpose.
For most substances the difference in the entropy of a compressed liquid and a satu-
rated liquid at the same temperature is so small that a process in which liquid is heated at
constant pressure nearly coincides with the saturated-liquid line until the saturation temper-
ature is reached (Fig. 8.7). Thus, if water at 10 MPa is heated from 0°C to the saturation
temperature, it would be shown by line ABD, which coincides with the saturated-liquid line.
8.4 ENTROPY CHANGE IN
REVERSIBLE PROCESSES
Having established that entropy is a thermodynamic property of a system, we now
consider its significance in various processes. In this section we will limit ourselves to
systems that undergo reversible processes and consider the Carnot cycle, reversible heat-
transfer processes, and reversible adiabatic processes.
Let the working fluid of a heat engine operating on the Carnot cycle make up the
system. The first process is the isothermal transfer of heat to the working fluid from the
high- temperature reservoir. For this process we can mite
*- s '-/'(t)„
Since this is a reversible process in which the temperature of the working fluid remains
constant, the equation can be integrated to give
This process is shown in Fig. 8.8a, and the area under line 1-2, area 1-2-&-A-1, repre-
sents the heat transferred to the working fluid during the process.
FIGURE 8.8 The
Carnot cycle on the
temperature-entropy
diagram.
(a)
b S
■Ql
260 H CHAPTER EIGHT ENTROPY
The second process of a Carnot cycle is a reversible adiabatic one. From the defini-
tion of entropy,
it is evident that the entropy remains constant in a reversible adiabatic process. A con-
stant-entropy process is called an isentropic process. Line 2-3 represents this process, and
this process is concluded at state 3 when the temperature of the working fluid reaches T L .
The third process is the reversible isothermal process in which heat is transferred
from the working fluid to the low-temperature reservoir. For this process we can write
Because during this process the heat transfer is negative (in regard to the working fluid),
the entropy of the working fluid decreases. Moreover, because the final process 4-1,
which completes the cycle, is a reversible adiabatic process (and therefore isentropic), it is
evident that the entropy decrease in process 3-4 must exactly equal the entropy increase
in process 1-2. The area under line 3-4, area 3-4-0-6-3, represents the heat transferred
from the working fluid to the low-temperature reservoir.
Since the net work of the cycle is equal to the net heat transfer, then area 1-2-3-4-1
must represent the net work of the cycle. The efficiency of the cycle may also be ex-
pressed in terms of areas:
= Hj* = area 1-2-3-4-1
Vm Q H area 1-2-fr-o-I
Some statements made earlier about efficiencies may now be understood graphically. For
example, increasing T H while T L remains constant increases the efficiency. Decreasing T L
as T H remains constant increases the efficiency. It is also evident that the efficiency ap-
proaches 100% as the absolute temperature at which heat is rejected approaches zero.
If the cycle is reversed, we have a refrigerator or heat pump. The Carnot cycle for a
refrigerator is shown in Fig. 8.86. Notice that the entropy of the working fluid increases at
T Li since heat is transferred to the working fluid at T L . The entropy decreases at T H be-
cause of heat transfer from the working fluid.
Let us next consider reversible heat-transfer processes. Actually, we are concerned
here with processes that are internally reversible, that is, processes that have no irreversibil-
ities within the boundary of the system. For such processes the heat transfer to or from a
system can be shown as an area on a temperature-entropy diagram. For example, consider
the change of state from saturated liquid to saturated vapor at constant pressure. This
process would correspond to the process 1-2 on the T-s diagram of Fig. 8.9 (note that ab-
solute temperature is required here), -and the area 1-2-6-a-l represents the heat transfer.
Since this is a constant-pressure process, the heat transfer per unit mass is equal to h /g . Thus,
s,-s -s - 1 ( 2 ( SQ ) - 1 [\n-& h *
This relation gives a clue about how s fg is calculated for tabulation in tables of thermody-
namic properties. For example, consider steam at 10 MPa. From the steam tables we have
h fs = 1317.1 kJ/kg
T = 311.06 + 273.15 = 584.21 K
ENTROPY CHANGE IN REVERSIBLE PROCESSES M 261
T
FIGURE 8.9 A
temperature-entropy
diagram to show areas
that represent heat transfer
for art internally reversible
process.
a
b
c
s
Therefore,
1317.1
584.21
- 2.2544 fcJ/kg K
This is the value listed for s fg in the steam tables.
If heat is transferred to the saturated vapor at constant pressure, the steam is super-
heated along line 2-3. For this process we can write
Since T is not constant, this equation cannot be integrated unless we know a relation be-
tween temperature and entropy. However, we do realize that the area under line 2-3, area
2~3~c~b~2 i represents j\T ds and therefore represents the heat transferred during thisre-
versible process.
The important conclusion to draw here is that for processes that are internally re-
versible, the area underneath the process line on a temperature-entropy diagram repre-
sents the quantity of heat transferred. This is not true for irreversible processes, as will be
demonstrated later.
EXAMPLE 8.1 Consider a Carnot-cycle heat pump with R-134a as the working fluid. Heat is absorbed
into the R-134a at 0°C, during which, process it changes from a two-phase state to satu-
rated vapor. The heat is rejected from the R-134a at 60°C so that it ends up as saturated
liquid. Find the pressure after compression, before the heat rejection process, and deter-
mine the coefficient of performance for the cycle.
From the definition of the Camot cycle we have two constant-temperature (isothermal)
processes that involve heat transfer and two adiabatic processes in which the tempera-
ture changes. The variation in s follows from Eq. 8.2
Solution
ds = BqlT
262 H Chapter Eight Entropy
FIGURE 8,10
Diagram for Example 8.1.
1682
294
60
-
and the Camot cycle is shown in Fig. 8.8 and for this case in Fig, 8.10. We therefore
have
State 4 Table B.5.1
State 1 Table B.5.1
State 2 Table B.5.2
s 4 = s 3 = JflgsMeg = 1 .2857 kj/kg K
si = s 2 = s m03eg = 1 .7262 kJ/kg K
60°C, S 2 = Si = -S^godeg
Interpolate between 1400 kPa and 1 600 kPa in Table B.5.2:
P 2 = 1400 + (1600 - 1400) _ = W87.1 kPa
From the fact that it is a Carnot cycle the COP becomes, from Eq. 7.13,
333.15
= 5.55
win T h - T L 60
Remark. Notice how much the pressure varies during the heat rejection process. Because
this process is very difficult to accomplish in a real device, no heat pump or refrigerator
is designed to attempt to approach a Carnot cycle.
EXAMPLE 8.2 A cylinder/piston setup contains 1 L of saturate liquid refrigerant R-12 at 20°C. The piston
now slowly expands, mainta inin g constant temperature to a final pressure of 400 kPa in a
reversible process. Calculate the required work and heat transfer to accomplish this process.
Solution
C.V. The refrigerant R-12, which is a control mass.
Continuity Eq.: m 2 - ;; m x = m\
Energy Eq. 5.11: m(\i 2 - u{) = X Q 2 - X W 2
Entropy Eq. 8.3: m(s 2 ~ s t ) s j 8Q/T
Process: T ~ constant, reversible so equal sign applies in entropy equation.
State 1 (r,P)TableB.3.1: u x = 54.45kJ/kg, i, = 0.2078kJ/kgK
m = Vfvi = 0.001/0.000752 = 1.33 kg
State 2 (T,P) Table B.3.2: u 2 = 180.57kJ/kg, s 2 = 0,7204 kj/kg K
The thermodynamic Property Relation S3 263
Diagram for Example 8.2. v
As Tis constant we have fSQIT = \Q 2 iT, so from the entropy equation
i& - mT(s 2 - s{) = 1.33 X 293.15 X (0.7204 - 0.2078) = 200 kJ V V ■ :
The work is then, from the energy equation,
iW 2 = m(u x - u 2 ) + 3 g 2 = 1.33 X (54.45 - 180.57) + 200 = 32.3 kJ
Note from Fig. 8.11 that it would he difficult to calculate the work as the area in the P-v
diagram due to the shape of the process curve. The heat transfer is the area in the T-s.
diagram.
8.5 The thermodynamic
Property relation
At this point we derive two important thermodynamic relations for a simple compressible
substance. These relations are
TdS = dU+ PdV
TdS = dH- VdP
The first of these relations can be derived by considering a simple compressible
substance in the absence of motion or gravitational effects. The first law for a change of
state under these conditions can be written
8Q = dU+8W
The equations we are deriving here deal first with the changes of state in which the
state of the substance can ,be identified at all times. Thus, we must consider a quasi-equi-
librium process or, to use the term introduced in the last chapter, a reversible process. For
a reversible process of a simple compressible substance, we can write
8Q = TdS and 8W = P dV
Substituting these relations into the first-law equation, we have
TdS = dU+ PdV (8.5)
which is the first equation we set out to derive. Note that this equation was derived by as-
suming a reversible process. This equation can therefore be integrated for any reversible
process, for during such a process the state of the substance can be identified at any point
264 B Chapter Eight Eotropy
during the process. We also note that Eq. 8.5 deals only with properties. Suppose we have
an irreversible process taking place between the given initial and final states. The proper-
ties of a substance depend only on the state, and therefore the changes in the properties
during a given change of state are the same for an irreversible process as for a reversible
process. Therefore, Eq, 8.5 is often applied to an irreversible process between two given
states, but the integration of Eq. 8.5 is performed along a reversible path between the
same two states.
Since enthalpy is defined as
H=U + PV
it follows that
dH=dU+ PdV + VdP
Substituting this relation into Eq. 8.5, we have
TdS = dH~VdP (8.6)
which is the second relation that we set out to derive. These two expressions, Eqs. 8.5 and
8.6, are two forms of the thermodynamic property relation and are frequently called Gibbs
equations.
These equations can also be written for a unit mass,
Tds = du + Pdo (8.7)
Tds = dh~v dP (8.8)
The Gibbs equations will be used extensively in certain subsequent sections of this book.
If we consider substances of fixed composition other than a simple compressible
substance, we can write "T dS" equations other than those just given for a simple com-
pressible substance. In Eq. 4.15 we noted that for a reversible process we can write the
following expression for work:
bW=PdV-$dL -3 > dA~%dZ+--<
It follows that a more general expression for the thermodynamic property relation would be
TdS = dU+PdV-VdL-&dA-%dZ + --- (8.9)
8.6 entropy change of a control mass
During an irreversible process
Consider a control mass that undergoes the cycles shown in Fig. 8.12. The cycle made up
of the reversible processes A and B is a reversible cycle. Therefore, we can write
The cycle made of the irreversible process C and the reversible process B is an irre-
versible cycle. Therefore, for this cycle the inequality of Clausius may be applied, giving
the result
ENTROPY CHANGE OF A CONTROL MASS DURING AN IRREVERSIBLE PROCESS H 265
FIGURE 8.12
Entropy change of a
control mass during an
irreversible process.
Subtracting the second equation from the first and rearranging, we have
/,m>/,m
Since path A is reversible, and since entropy is a property,
Therefore,
As path C was arbitrary, the general result is
s 7 -s t >
(8.10)
In these equations the equality holds for a reversible process and the inequality for an irre-
versible process.
This is one of the most important equations of thermodynamics. It is used to de-
velop a number of concepts and definitions. In essence, this equation states the influence
of irreversibility on the entropy of a control mass. Thus, if an amount of heat SQ is trans-
ferred to a control mass at temperature T in a reversible process, the change of entropy is
given by the relation
dS
If any irreversible effects occur while the amount of heat SQ is transferred to the control
mass at temperature T, however, the change of entropy will be greater than for the re-
versible process. We would then write
dS>
SQ
Equation 8.10 holds when 8Q - 0, when SQ < 0, and when SQ > 0. If SQ is negative,
the entropy will tend to decrease as a result of the heat transfer. However, the influence of
266 II Chapter eight entropy
irreversibilities is still to increase the entropy of the mass, and from the absolute numeri-
cal perspective we can still write for 8Q:
8.7 Entropy Generation
The conclusion from the previous considerations is that the entropy change for an irre-
versible process is larger than the change in a reversible process for the same 8Q and T,
This can be written out in a common form as an equality
dS = ^f + 8S gia (8.11)
provided the last term is positive,
SS E ^Q (8.12)
The amount of entropy, SS giai is the entropy generation in the process due to irreversibil-
ities occurring inside the system, a control mass for now but later extended to the more
general control volume. This internal generation can be caused by the processes men-
tioned in Section 7.4, such as friction, unrestrained expansions, and the internal transfer
of energy (redistribution) over a finite temperature difference. In addition to this internal
entropy generation, external irreversibilities are possible by heat transfer over finite
temperature differences as the 8Q is transferred from a reservoir or by the mechanical
transfer of work.
Equation 8,12 is then valid with the equal sign for a reversible process and the
greater than sign for an irreversible process. Since the entropy generation is always posi-
tive and the smallest in a reversible process, namely zero, we may deduce some limits for
the heat transfer and work terms.
Consider a reversible process, for which the entropy generation is zero, and the heat
transfer and work terms therefore are
8Q-=TdS and SW=PdV
For an irreversible process with a nonzero entropy generation, the heat transfer from Eq.
8.11 becomes
8Q ilt =TdS-T8S gea
and thus is smaller than that for the reversible case for the same change of state, dS. we
also note that for the irreversible process, the work is no longer equal to P dV but is
smaller. Furthermore, since the first law is
and the property relation is valid,
TdS = dU + PdV
it is found that
8W in = PdV~ T8S^
(8.13)
Entropy Generation M 267
showing that the work is reduced by an amount proportional to the entropy generation.
For this reason the term TSS sea is often called "lost work," although it is not a real work
or energy quantity lost but rather a lost opportunity to extract work.
Equation 8.11 can be integrated between initial and final states to
Thus, we have an expression for the change of entropy for an irreversible process as an
equality, whereas in the last section we had an inequality. In the limit of a reversible
process, with a zero-entropy generation, the change in S expressed in Eq. 8.14 becomes
identical toEq. 8.10 as the equal sign applies and the work term becomes fPdF. Equation
8.14 is now the entropy balance equation for a control mass in the same form as the en-
ergy equation in Eq. 5.5, and it could include several subsystems. The equation can also
be written in the general form
expressing that we can generate but not destroy entropy. This is in contrast to energy
which we can neither generate nor destroy-
Some important conclusions can be drawn from Eqs. 8.11 to 8.14. First, there are
two ways in which the entropy of a system can be increased — by transferring heat to it
and by having an irreversible process. Since the entropy generation cannot be less than
zero, there is only one way in which the entropy of a system can be decreased, and that is
to transfer heat from the system. These changes are illustrated in a T-s diagram in Fig.
8.13 showing the halfplane into which the state moves due to a heat transfer or an entropy
generation.
Second, as we have already noted for an adiabattc process, 8Q = 0, and therefore
the increase in entropy is always associated with the irreversibilities.
Finally, the presence of irreversibilities will cause the work to be smaller than the
reversible work. This means less work out in an expansion process and more work into
the control mass (S W < 0) in a compression process.
One other point concerning the representation of irreversible processes on P-V and
TS diagrams should be made. The work for an irreversible process is not equal to jP dV,
and the heat transfer is not equal to fT dS. Therefore, the area underneath the path does
not represent work and heat on the P-Kand T-S diagrams, respectively. In fact, in many
situations we are not certain of the exact state through which a system passes when it un-
dergoes an irreversible process. For this reason it is advantageous to show irreversible
processes as dashed lines and reversible processes as solid lines. Thus, the area under-
neath the dashed line will never represent work or heat. Figure 8.14a shows an irre-
versible process, and, because the heat transfer and work for this process are zero, the area
underneath the dashed line has no significance. Figure S.14& shows the reversible process,
(8.14)
A Entropy = + in — out 4- gen
T
Gout
FIGURE 8.13 Change
of entropy due to heat
transfer and entropy
generation.
'gen
268 H Chapter Eight Entropy
FIGURE 8.14
Reversible and
irreversible processes on
pressure-volume and
temperature-entropy
diagrams.
x 2
b v a
b s
and area \—2—b—a-l represents the work on the P—V diagram and the heat transfer on the
TS diagram.
8.8 PRINCIPLE OF THE INCREASE OF ENTROPY
In the previous section, we considered irreversible processes in which the irreversibilities
occurred inside the system or control mass. We also found that the entropy change of a
control mass could be either positive or negative, since entropy can be increased by inter-
nal entropy generation and either increased or decreased by heat transfer, depending on
the direction of that transfer. In this section, we examine the effect of heat transfer on the
change of state in the surroundings, as well as on the control mass itself.
Consider the process shown in Fig. 8.15 in which a quantity of heat SQ is trans-
ferred from the surroundings at temperature r to the control mass at temperature T. Let
the work done during this process be BW. For this process we can apply Eq. 8,10 to the
control mass and write
SQ
""cm. — j
For the surroundings at r , SQ is negative, and we assume a reversible heat extraction so
■*0
The total net change of entropy is therefore
dS atl - dS^ + dS smT -Tf
SQ SQ
T T c
>50[~-™) (8.15)
FIGURE 8.15
Entropy change for the
control mass plus
surroundings.
=J> SW
SQ ) cm. 2
Surroundings
temperature = T
Principle of the increase of Entropy H 269
Since T > T, the quantity [(1/7) - (1/T )] is positive, and we conclude that
dS Mt = dS^ + dSnn s
If T > T 0i the heat transfer is from the control mass to the surroundings, and both SQ and
the quantity [(VT) - (1/T )] are negative, thus yielding the same result.
It should be noted that the right-hand side of Eq. 8.15 represents an external entropy
generation due to heat transfer through a finite temperature difference. To amplify this
point, take as a control mass the system that connects the surroundings at T with the pre-
vious control mass at T, which typically is the walls. This mass does not experience any
change of state, yet heat transfer causes fluxes of entropy to flow in and out in an irre-
versible process. For this mass, Eq. 8.11 gives
and we see that the difference in the two SQ/T terms (fluxes of S) is the entropy generated
in this control mass.
This is also precisely the location in space where the heat transfer takes place over the fi-
nite temperature difference T - T, This term is always positive (or zero for an adiabatic
process), but as the temperature difference is made to approach zero, this term also ap-
proaches zero.
There could also be additional entropy-generation terms in the surroundings of the
types discussed in the previous section, if those factors were also present in the surround-
ings, and those will be positive, as well. Thus, we conclude that the net entropy change is
the sum of a number of terms, each of which is positive, due to a specific cause of irre-
versible entropy generation, such that the net entropy change could also be termed the
total entropy generation:
dS aet = dS^ + dS su „ = 2 SS gin > (8.16)
where the equality holds for reversible processes and the inequality for irreversible
processes. This is a very important equation, not only for thermodynamics but also for
philosophical thought. This equation is referred to as the principle of the increase of en-
tropy. The great significance is that the only processes that can take place are those in
which the net change in entropy of the control mass plus its surroundings increases (or in
the limit, remains constant). The reverse process, in which both the control mass and sur-
roundings are returned to their .original state, can never be made to occur. In other words,
Eq. 8.16 dictates the single direction in which any process can proceed. Thus, the principle
of the increase of entropy can be considered a quantitative general statement of the second
law from the macroscopic point of view and applies to the combustion of fuel in our auto-
mobile engines, the cooling of our coffee, and the processes that take place in our body.
Sometimes this principle of the increase of entropy is stated in terms of an isolated
system, one in which there is no interaction between the system and its surroundings.
Then there is no change in the surroundings, and we then conclude that
isolated system = ^gen, system — (8-17)
That is, in an isolated system, the only processes that can occur are those that have an as-
sociated increase in entropy.
I_ jL
t r
270 1 CHAPTER EIGHT ENTROPY
The development of Eq. 8.16 as the principle of the increase of entropy was made
for an infinitesimal change of state. When we wish to test a claimed process to see
whether it satisfies the second law of thermodynamics, it will necessarily be for a finite
change of state. Consider a control mass undergoing a process from initial state 1 to final
state 2, with an associated heat transfer l Q 2i which may be known or calculated from the
first law. The heat transfer is to or from a reservoir at temperature T . For this process,
A^cm. ~ $2 ~~ $U AS suir —
kS^^/^S^ + bS^ (8.18)
and the net entropy change as calculated from Eq. 8.18 must be greater than zero (irre-
versible process), or in the limit be equal to zero (completely reversible process, internally
and externally). This type of calculation is illustrated in the following example.
EXAMPLE 8,3 Suppose that 1 kg of saturated water vapor at 100°C is condensed to a saturated liquid at
100°C in a constant-pressure process by heat transfer to the surrounding air, which is at
25°C. What is the net increase in entropy of the water plus surroundings?
Solution
For the control mass (water), from the steam tables, we obtain
AS cm = _ mjj6 = -i x 6.0480 = -6.0480 kJ/K
Concerning the surroundings, we have
fios—d ^ = mh& = 1 X 2257 -° = 2257 W
AS net = AS cnL + AS surt = -6.0480 + 7.5700 = 1.5220 kJ/K
This increase in entropy is in accordance with the principle of the increase of entropy
and tells us, as does our experience, that this process can take place.
It is interesting to note how this heat transfer from the water to the surroundings
might have taken place reversibly. Suppose that an engine operating on the Carnot cycle
received heat from the water and rejected heat to the surroundings, as shown in Fig.
8.16. The decrease in the entropy of the water is equal to the increase in the entropy of
the surroundings.
AS cjn . = -6.0480 kJ/K
A5 iun = 6.0480 kJ/K
0*™^ = T^ AS = 298.15(6.0480) = 1803.2 kj
W = Q H - Q L = 2257 - 1803.2 = 453.8 kj
Since this is a reversible cycle, the engine could be reversed and operated as a heat
pump. For this cycle the work input to the heat pump would be 453.8 kJ.
Principle of the Increase of Entropy H 271
FIGURE 8.16
Reversible heat transfer
with the surroundings.
M i I in
H 2 :
T= 373.2 K
Reversible
engine
Ql
Surroundings
T = 298.2 K
W
1
373.2 K
2
4
3
298.2 K
Example 8.3E
Suppose that 1 Ibm of saturated water vapor at 212 F is condensed to a saturated liquid
at 212 F in a constant-pressure process by heat transfer to the surrounding air, which is
at 80 F. What is the net increase in entropy of the water plus surroundings?
Solution
For the control mass (water), from the steam tables,
AS^tea = s /g - - 1.4446 Btn/lbm R
Considering the surroundings, we have
2to surroundings = h fg = 970.3 Btu/lbm
^su* = y = ^Ql = 1 .7980 Btu/lbm R
AS, mm + AS^ = -1.4446 + 1.7980 - 0.3534 Btu/lbm R
This increase in entropy is in accordance with the principle of the increase of entropy
and tells us, as does our experience, that this process can take place.
It is interesting to note how this heat transfer from the water to the surroundings
might have taken place reversibly. Suppose that an engine operating on the Camot cycle
received heat from the water and rejected heat to the surroundings, as shown in Fig. 8. 16E.
FIGURE 8.16E
Reversible heat transfer
with the surroundings.
H 2
T=672R
Reversibie
engine
O
Surroundings
T a = 540 R
1
672 R
2
4
3
540 R
272 H Chapter Eight Entropy
The decrease in the entropy of the water is equal to the increase in the entropy of the
surroundings.
AS C ,,, - 1.4446 Btu/lbm R
&S smc - 1.4446 Btu/lbmR
Q lm ^ ■ ; T (l l\S - 540(1.4446) - 780.1 Btu/lbm
W ~ Q.h Ql 970.3 ~ 78(U "** 190.2 Btu/lbm
Since this is a reversible cycle, the engine could be reversed and operated as a heat
pump. For this cycle the work input to the heat pump would be 190.2 Btu/lbm.
8.9 Entropy Change of a Solid or Liquid
In Section 5.6 we considered the calculation of the internal energy and enthalpy changes
with temperature for solids and liquids and found that, in general, it is possible to express
both in terms of the specific heat, in the simple manner of Eq. 5.17, and in most instances
in the integrated form of Eq. 5.18. We can now use this result and the thermodynamic
property relation, Eq. 8.7, to calculate the entropy change for a solid or liquid. Note that
for such a phase the specific volume term in Eq. 8.7 is very small, so that substituting Eq.
5.17 yields
ds^^dT (8.19)
Now, as was mentioned in Section 5.6, for many processes involving a solid or liquid, we
may assume that the specific heat remains constant, in which case Eq. 8.19 can be inte-
grated. The result is
5 2 -5, = Cln^ (8.20)
If the specific heat is not constant, then commonly C is known as a function of T, in which
case Eq. 8.19 can also be integrated to find the entropy change. Equation 8.20 illustrates
what happens in a reversible (ds m = 0) adiabatic (dq = 0) process, which therefore is
isentropic. In this process, the approximation of constant u leads to constant temperature,
which explains why pumping liquid does not change the temperature.
EXAMPLE 8.4 One kilogram of liquid water is heated from 20°C to 90°C. Calculate the entropy change,
assuming constant specific heat, and compare the result with that found when using the
steam tables.
Control mass:
Initial and final states:
Model:
Water.
Known.
Constant specific heat, value at room temperature.
Entropy Change of an Ideal Gas H 273
Solution '
For constant specific heat, from Eq. 8,20,
s 2 - si = 4.184 In (^j^J ^ 0.8958 kJ/kg K
Comparing this result with that obtained by using the steam tables, we have
s 2 ~si= s fWC - 5 /20 = c = 1.1925 - 0.2966
= 0.8959 kJ/kgK
8,10 Entropy Change of an Ideal Gas
Two very useful equations for computing the entropy change of an ideal gas can be devel-
oped from Eq. 8.7 by substituting Eqs, 5.20 and 5.24;
For an ideal gas
Therefore,
Similarly,
For an ideal gas
Therefore,
Tds = du + Pdv
du = C„o dT and ~ = §
ds^C u0 f + ^ (8.21)
s 2 - Sl = f^Cojr + R^ (8.22)
Tds = dh-vdP
dh = C^dT and | = |
ds=C^-R^ (8.23)
s 2 ~ Sl = ^C^-Rln^ (8.24)
To integrate Eqs. 8.22 and 8.24, we must know the temperature dependence of the
specific heats. However, if we recall that their difference is always constant as expressed
by Eq. 5.27, we realize that we need to examine the temperature dependence of only one
of the specific heats.
274 H CHAPTER EIGHT ENTROPY
As in Section 5,7, let us consider the specific heat C p0 . Again, there are three possi-
bilities to examine, the simplest of which is the assumption of constant specific heat. In
this instance it is possible to integrate Eq. 8.24 directly to
s t - Sl = C^]n^~R\n^ (8.25)
Similarly, integrating Eq. 8.22 for constant specific heat, we have
s 2 -s i = C uQ \n^+R]n^ (8.26)
The second possibility for the specific heat is to use an analytical equation for
C p0 as a function of temperature, for example, one of those listed in Table A.6. The
third possibility is to integrate the results of the calculations of statistical thermody-
namics from reference temperature T to any other temperature T and define the stan-
dard entropy
4- ( T ^-dT (8.27)
This function can then be tabulated in the single-entry (temperature) ideal-gas table, as for
air in Table A.7 or for other gases in Table A.8. The entropy change between any two
states 1 and 2 is then given by
*2 = In (8.28)
As with the energy functions discussed in Section 5.7, the ideal-gas tables, Tables
A.7 and A. 8, would give the most accurate results, and the equations listed in Table A. 6
would give a close empirical approximation. Constant specific heat would be less accu-
rate, except for monatomic gases and for other gases below room temperature. Again, it
should be remembered that all these results are part of the ideal-gas model, which may or
may not be appropriate in any particular problem.
EXAMPLE 8.5 Consider Example 5.6, in which oxygen is heated from 300 to 1 500 K. Assume that dur-
ing this process the pressure dropped from 200 to 150 kPa. Calculate the change in en-
tropy per kilogram.
Solution
The most accurate answer for the entropy change, assuming ideal-gas behavior, would
be from the ideal-gas tables, Table A.8. This result is, using Eq. 8.28,
s 2 ~ Ji = (8.0649 - 6.4168) - 0.2598 In f |™
= 1.7228 kJ/kgK
ENTROPY CHANGE OF AN IDEAL GAS U 275
The empirical equation from Table A.6 should give a good approximation to this result.
Integrating Eq. 8,24, we have
which is within 1.0% of the previous value. For constant specific heat, using the value at
300 K from Table A .5, we have
which is too low by 9.5%. If, however, we assume that the specific heat is constant at its
value at 900 K, the average temperature, as in Example 5.6, then
which is high by 4.9%.
Calculate the change in entropy per kilogram as air is heated from 300 to 600 K while
pressure drops from 400 to 300 kPa. Assume:
1. Constant specific heat.
2. Variable specific heat.
Solution
1. From Table A.5 for air at 300 K,
1.7058 kJ/kgK
= 1.5586 kJ/kgK
C p0 = 1.004 kJ/kgK
Therefore, using Eq, 8.25, we have
s 2 ~s^ 1.004 In (0j - 0.287 In {^j = 0.7785 kJ/kg K
276 M Chapter eight entropy
2. From Table A.7,
s° n = 6.8693 kJ/kg K, s° T2 = 7.5764 kJ/kg K
Using Eq. 8.28 gives
- Sl = 7.5764 - 6.8693 - 0.287 in ^ = 0.7897 kJ/kgK
EXAMPLE 8.6E Calculate the change in entropy per pound as air is heated from 540 R to 1200 R while
pressure drops from 50 Ibf/in. 2 to 40 lbf/in. 2 . Assume:
1. Constant specific heat.
2. Variable specific heat.
Solution
1. From Table F.4 for air at 80 F,
- 0.24 Btu/lbm R
Therefore, using Eq. 8.25, we have
- Sl = 0.24 In - ^ In (^j = 0.2068 Btu/fbm R
j 2 - a, = 0.7963 - 0.6008 - in ^ = 0.2108 Btu/lbmR
Let us now consider the case of an ideal gas undergoing an isentropic process, a sit-
uation that is analyzed frequently. We conclude that Eq. 8.24 with the left side of the
equation equal to zero then expresses the relation between the pressure and temperature at
the initial and final states, with the specific relation depending on the nature of the specific
heat as a function of T. As was discussed following Eq. 8.24, there are three possibilities
to examine. Of these, the most accurate is the third, that is, the ideal-gas Tables A.7 or
A.8 and Eq. 8.28, with the integrated temperature function s° T denned by Eq. 8.27. The
following Example illustrates the procedure.
2. From Table F.5
s° n = 0.6008 Btu/lbm R & = 0-7963 Btu/ibm R
Using Eq. 8.28 gives
Entropy Change of an Ideal Gas 277
EXAMPLE 8.7 One kilogram of air is contained in a cylinder fitted with a piston at a pressure of 400
kPa and a temperature of 600 K. The air is expanded to 150 kPa in a reversible, adia-
batic process. Calculate the work done by the air.
Control mass: Air.
Initial state: P u Tjj state 1 fixed.
Final state: P 2 .
Process: Reversible and adiabatic.
Model: Ideal gas and air tables, Table A.7.
Analysis
From the first law we have
i< 2 ■ tt t + w
The second law gives us
^2 = ^
Solution
From Table A. 7,
FromEq. 8.28 )
From Table A.7,
Therefore,
u x = 435.10 kJ/kg, s° n = 7.5764 kJ/kg K
s 2 ~s x = = *In§
= - 7.5764) - 0.287 In ^jg
4i = 7.2949 kJ/kgK
r 2 = 457K, u 2 = 328.14 kJ/kg
w = 435.10 - 328.14 = 106.96 kJ/kg
The first of the three possibilities, constant specific heat, is also worth analyzing as a
special case. In this instance, the result is Eq. 8.25 with the left side equal to zero, or
278 1 Chapter Eight Entropy
This expression can also be written as
or
However,
R _ Cp<s C vQ _ k- 1
(8.29)
(8.30)
where k, the ratio of the specific heats, is defined as
k = ^ (8.31)
wo
Equation (8.29) is now conveniently written as
From this expression and the ideal-gas equation of state, it also follows that
(8.32)
and
(8.33)
(8.34)
From this last expression, we note that for this process
Pv k ~ constant (8.35)
This is a special case of a polytropic process in which the polytropic exponent n is equal
to the specific heat ratio k.
8.11 THE REVERSIBLE POLYTROPIC
PROCESS FOR AN IDEAL GAS
When a gas undergoes a reversible process in which there is heat transfer, the process fre-
quently takes place in such a manner that a plot of log P versus log V is a straight line, as
shown in Fig. 8.17. For such a process PV* is a constant.
A process having this relation between pressure and volume is called a polytropic
process An example is the expansion of the combustion gases in the cylinder of a water-
cooled reciprocating engine. If the pressure and volume are measured during the expan-
sion stroke of a polytropic process, as might be done with an engine indicator, and the
Tiie Reversible Polytropic Process for an Ideal Gas fl 279
log/"
FIGURE 8.17
Example of a polytropic
process.
slope =
tog v
logarithms of the pressure and volume are plotted, the result would be similar to the
straight line in Fig. 8.17. From this figure it follows that
^InP
d\nV "
rflnP + nd hi V=
If n is a constant (which implies a straight line on the log P versus log rplot), this equa-
tion can be integrated to give the following relation:
PV = constant - P X V* X = P 2 V n 2 (8.36)
From this equation the following relations can be written for a polytropic process:
t 2 (p t \>«-iy* ( M"" 1
For a control mass consisting of an ideal gas, the work done at the moving boundary dur-
ing a reversible polytropic process can be derived from the relations
,W> = \ PdV and PV = constant
X W 2 ~ j PdV^ constant j
dV
i r
P 2 V 2 ~P X V, _ mR{T 2 ~T 1 )
1 — n 1 — n
(8.38)
for any value of n except n = 1.
The polytropic processes for various values of n are shown in Fig. 8.18 on P-u and
T-s diagrams. The values of n for some familiar processes are
Isobaric process: n — 0, P= constant
Isothermal process: n = 1, T= constant
Isentropic process: n = k, s = constant
Isochoric process: n — » ) y = constant
280 CHAPTER EIGHT ENTROPY
P
FIGURE 8.18 Polytropic process on P-u and T~s diagrams.
EXAMPLE 8.8 In a reversible process, nitrogen is compressed in a cylinder from 100 kPa and 20°C to
500 kPa. During this compression process, the relation between pressure and volume is
pjA 3 = constant. Calculate the work and heat transfer per kilogram, and show this
process on P~Vand T S diagrams.
Control mass: Nitrogen.
Initial state: P x , 7\; state 1 known.
Final state: P 2 .
Process: Reversible, polytropic with exponent n < k.
Diagram: Fig. 8.19
Model: Ideal gas, constant specific heat — value at 300 K.
Analysis
We need to find the boundary movement work. From Eq. 8.38, we have
J i 1 - n 1 - n
•Area = work
Area = heat
transfer
FIGURE 8.19
Diagram for Example 8,8.
(b)
Tiie Reversible PoLYTROPrc Process for an Ideal Gas M 281
The first law is
i?2 = "2 - «i + iw 2 = c^zi - ro + iW 2
Solution
From Eq. 8.37
r, IpJ I iooJ
1.4498
T 2 = 293.2 X 1.4498 = 425 K
Then
5(r 2 - T x ) 0.2968(425 - 293.2)
lW2 = - — ( ; _ L3) = - 13 °- 4 ^
and from the first law,
x q 2 = C vQ (T 2 - T x ) + ,w 2
- 0.745(425 - 293.2) - 130.4 = -32.2 kj/kg
The reversible isothermal process for an ideal gas is of particular interest. In this
process
PV = constant = P X V X = P 2 V 2 (8.39)
The work done at the boundary of a simple compressible mass during a reversible isother-
mal process can be found by integrating the equation
X W % = j"' ' PdV
The integration is
X W 2 = f 2 P^F- constant f 2 ~r = ^i I^tt = W In tt ( 8 - 40 )
or
X W 2 = mRT\s& = mRT In (8.41)
Because there is no change in internal energy or enthalpy in an isothermal process,
the heat transfer is equal to the work (neglecting changes in kinetic and potential energy).
Therefore, we could have derived Eq. 8.40 by calculating the heat transfer.
For example, using Eq. 8.7, we have
282 W ChapterEight Entropy
But du = and Pv = constant = = P 2 v 2 , such that
f 2 , v 2
,q 2 = J Prfu =P,Uiln^
which yields the same result as Eq. 8.40.
8.12 ENTROPY AS A RATE EQUATION
The second law of thermodynamics was used to write the balance of entropy in Eq. 8.11
for a variation and in Eq. 8.14 for a finite change. In some cases the equation is needed m
a rate form so that a given process can be tracked in time. The rate form is also the basis
for the development of the entropy balance equation in the general control volume analy-
sis for an unsteady situation.
Take the incremental change in S from Eq. 8.11 and divide by St. We get
dS = 1 ?f? , (8i4 2)
St T 8t St
For a given control volume we may have more than one source of heat transfer, each at a
certain surface temperature (semidistributed situation). Since we did not have to consider
the temperature at which the heat transfer crossed the control surface for the energy equa-
tion all the terms were added into a net heat transfer in a rate form m Eq. 5.3 1. Using this
and'a dot to indicate a rate, the final form for the entropy equation in the limit is
+ v C8 - 43)
expressing the rate of entropy change as due to the flux of entropy into the control mass
from heat transfer and an increase due to irreversible processes inside the control mass. It
only reversible processes take place inside the control volume, the rate of change of en-
tropy is determined by the rate of heat transfer divided by the temperature terms alone.
EXAMPLE 8.9 Consider an electric space heater that converts 1 kW of electric power into a heat flux of
1 kW delivered at 600 K from the hot wire surface. Let us look at the process of the energy
conversion from electricity to heat transfer and find the rate of total entropy generation.
Control mass: The electric heater wire.
State: Constant wire temperature 600 K.
Analysis
The first and the second laws of thermodynamics in rate form become
^=^ = o = ^un-a>«
dt dt '
dt
Key Concepts and Formulas H 283
Notice that we neglected kinetic and potential energy changes in going from a rate of E
to a rate of U. Then the left-hand side is zero since it is steady state and the right-hand
side of the energy equation is electric work in minus the heat transfer out. For the en-
tropy equation the left-hand side is zero because of steady state and the right-hand side
has a flux of entropy out due to heat transfer, and entropy is generated in the wire.
Solution
We now get the entropy generation as
The inequality of Clausius and the property entropy (s) are modern statements of the sec-
ond law. The final statement of the second law is the entropy balance equation that in-
cludes generation of entropy. All the results that were derived from the classical
formulation of the second law in Chapter 7 can be re-derived with the entropy balance
equation applied to the cyclic devices. For all reversible processes, entropy generation is
zero and all real (irreversible) processes have a positive entropy generation. How large the
entropy generation is depends on the actual process.
Thermodynamic property relations for s are derived from consideration of a re-
versible process and leads to Gibbs relations. Changes in the property s are covered
through general tables, approximations for liquids and solids, as well as ideal gases.
Changes of entropy in various processes are examined in general together with special
cases of polytropic processes. Just as a reversible specific boundary work is the area
below the process curve in a P-v diagram, the reversible heat transfer is the area below
the process curve in a T-s diagram.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Know that Clausius inequality is an alternative statement of the second law.
• Know the relation between the entropy and the reversible heat transfer.
• Locate states in the tables involving entropy.
• Understand how a Carnot cycle looks in a T-s diagram.
• Know how different simple process curves look in a T-s diagram.
• Understand how to apply the entropy balance equation for a control mass,
• Recognize processes that generate entropy and where the entropy is made.
• Evaluate changes in s for liquids, solids, and ideal gases.
• Know the various property relations for a polytropic process in an ideal gas.
• Know the application of the unsteady entropy equation and what a flux of s is.
'gen
Q JT= 1/600 - 0.001 67 kW/K
Summary
Key Concepts
and formulas
Clausius inequality
Entropy
284 M Chapter Eight Entropy
Rate equation for entropy = 2 + S,
'gen
Entropy equation m(s 2 — ■Sj) — j -jr + 1^;
l u 2 gen> l u 2 gsn
1 J
Total entropy change AS net = AiS cm + AS iUtr = AS^ s
Lost work JF lost = fTdS^ n
Actual boundary work X W 2 = fP dV — W loa
Gibbs relations Tds = du + P dv
Tds = dh ~vdP
Solids, Liquids v = constant, dv =
Change ins J 2 ~ J i ~ j" = J* ^ ^ U1
Ideal Gas
Standard entropy
Change in s
Ratio of specific heats
Polytropic processes
Specific work
dT
(Function of T)
p
s 2 ~ 5i = 4 ~ s n ~ * In -W (Using Table A.7 or A.8)
T P
s 2 ~ sj = C o ln^ ~ J? In ^ (For constant C p , C„)
1 i "i
T 2 v 2
s 2 — s x — Q, In + i? In — (For constant C p> C v )
k — Cp^i C^o
Pv" — constant; PV" ~ constant
1
1 — ti
(P 2 v 2 - PjVd
R
(T 2 ~ T,)
1 -71
The work is moving boundary work w = j P dv
n 1
71= 1
Homework Problems
285
Identifiable processes
n — Oj P = constant;
n = 1; T = constant;
n = k; s — constant;
rt = co- y = constant:
Isobaric
Isothermal
Isentropic ■
Isochoric or isometric
Concept-Study Guide Problems
8.1 Does Clausius say anything about the sign for $ g?
8.2 When a substance has completed a cycle, y, i/, A,
and s are unchanged. Did anytliing happen? Explain.
8.3 Assume a heat engine with a given Q H , Can you
say anything about Q L if the engine is reversible? if
it is irreversible?
8.4 How can you change s of a substance going
through a reversible process?
8.5 Does the statement of Clausius require a constant T
for the heat transfer as in a Carnot cycle?
8.6 A reversible process adds heat to a substance. If T
is varying, does that influence the change in s?
8.7 Water at 100 kPa, 150°C, receives 75 kJ/kg in a re-
versible process by heat transfer. Which process
changes s the most: constant T, constant y, or con-
stant PI
8.8 A substance has heat transfer out. Can you say any-
thing about changes in s if the process is re-
versible? if it is irreversible?
8.9 A substance is compressed adiabatically so P and T
go up. Does that change s7
8.10 Saturated water vapor at 200 kPa is compressed to
600 kPa in a reversible adiabatic process. Find the
new u and T.
8.11 A computer chip dissipates 2 kJ of electric work
over time and rejects that as heat transfer from its
50°C surface to 25°C air. How much entropy is
generated in the chip? How much, if any, is gener-
ated outside the chip?
8.12 A car uses an average power of 25 hp for a one-
hour round trip. With a thermal efficiency of 35%
Homework problems
Inequality of Clausius
8.21 Consider the steam power plant in Example 6.9
and assume an average T in the line between 1 and
2. Show that this cycle satisfies the inequality of
Clausius.
how much fuel energy was used? What happened
to all the energy? What change in entropy took
place if we assume ambient at 20°C?
8.13 A liquid is compressed in a reversible adiabatic
process. What is the change in 71
8.14 Two 5-kg blocks of steel, one at 250°C, the other at
25°C, come in thermal contact. Find the final temper-
ature and the total entropy generation in the process.
8.15 One kg of air at 300 K is mixed with one kg of air
at 400 K in a process at a constant 1 00 kPa and Q ~
0. Find the finai T and the entropy generation in the
process.
8.16 One kg of air at 100 kPa is mixed with one kg of
air at 200 kPa, both at 300 K, in a rigid insulated
tank. Find the final state (P, T) and the entropy
generation in the process.
8.17 An ideal gas goes through a constant T reversible
heat addition process. How do the properties (v, u }
h,s,P) change (up, down, or constant)?
8.18 Carbon dioxide is compressed to a smaller volume in
a polytropic process with n — 1,2. How do the prop-
erties (it, h, s, P, T) change (up, down, or constant)?
8.19 Hot combustion air at 1500 K expands in a poly-
tropic process to a volume six times as large with
n = 1.5. Find the specific boundary work and the
specific heat transfer.
8.20 A window receives 200 W of heat transfer at the in-
side surface of 20°C and transmits the 200 W from
its outside surface at 2°C continuing to ambient air
at — 5°C. Find the flux of entropy at all three sur-
faces and the window's rate of entropy generation.
8.22 Assume the heat engine in Problem 7.25 has a high
temperature of 1200 K and a low temperature of
400 K. What does the inequality of Clausius say
about each of the four cases?
8.23 Let the steam power plant in Problem 7.26 have
700°C in the boiler and 40°C during the heat
286 B CHAPTER EIGHT ENTROPY
rejection in the condenser. Does that satisfy the in-
equality of Clausius? Repeat the question for the
cycle operated in reverse as a refrigerator,
8.24 A heat engine receives 6 kW from a 250°C source
and rejects heat at 30°C. Examine each of three
cases with respect to the inequality of Clausius.
a. IK = 6 kW b. W = kW c. Carnot cycle
8.25 Examine the heat engine given in Problem 7.50 to
see if it satisfies the inequality of Clausius.
Entropy of a Pure Substance
8.26 Find the entropy for the following water states and
indicate each state on a Ts diagram relative to the
two-phase region.
a. 250°C, v = 0.02 mVkg
b. 250°C, 2000 kPa
c. -2°C, 100 kPa
d. 20°C, 100 kPa
e. 20°C, lOOOOkPa
8.27 Find the missing properties and give the phase of
the substance.
a. H 2 5 = 7.70kJ/kgK, ft = ?r=?* = ?
P = 25 kPa
b. H 2 u = 3400 kJ/kg, T = ? x = ? s = ?
p - 10 MPa
c. R-12 r=0°C, *=?x=7
P = 200 kPa
d. R-134a r=-10°C ) i> = ?s = ?
x = 0.45
e. NH 3 r=20 o C, » = ?* = ?
s = 5.50kJ/kgK
8.28 Saturated liquid water at 20°C is compressed to a
higher pressure with constant temperature. Find the
changes in u and s when the final pressure is
a. 500 kPa b. 2000 kPa c. 20 000 kPa
8.29 Saturated vapor water at 150°C is expanded to a
lower pressure with constant temperature. Find the
changes in u and s when the final pressure is
a. 100 kPa b.SOkPa c.lOkPa
8.30 Determine the missing property among P, T, s, and
x for the following states:
a. Ammonia 25°C, if = 0.10 nrVkg
b. Ammonia 1000 kPa,s = 5.2 kJ/kgK
c. R-134a
d. R-134a
e. R-22
Reversible Processes
5°C,s = 1.7 kJ/kgK
50°C,s = 1-9 kJ/kg K
100 kPa, v = 0.3 nrVkg
8.31 Consider a Camot-cycle heat engine with water as
the working fluid. The heat transfer to the water oc-
curs at 300°C, during which process the water
changes from saturated liquid to saturated vapor.
The heat is rejected from the water at 40°C. Show
the cycle on a Ts diagram and find the quality of
the water at the beginning and the end of the heat
rejection process. Determine the net work output
per kg water and the cycle thermal efficiency.
8.32 In a Carnot engine with ammonia as the working
fluid the high temperature is T H = 60°C, and as Q H
is received the ammonia changes from saturated
liquid to saturated vapor. The ammonia pressure at
the low temperature is P low = 190 kPa. Find T L> the
cycle thermal efficiency, the heat added per kg, and
the entropy, s, at the beginning of the heat rejection
process.
8.33 Water is used as the working fluid in a Camot-
cycle heat engine, where it changes from saturated
liquid to saturated vapor at 200°C as heat is added.
Heat is rejected in a constant-pressure process (also
constant T) at 20 kPa. The heat engine powers a
Camot-cycle refrigerator that operates between
-15°C and +20°C, shown in Fig. P8.33. Find the
heat added to the water per kg water. How much
heat should be added to the water in the heat en-
gine so the refrigerator can remove 1 kJ from the
cold space?
/ -15° C \ FIGURE P8.33
8,34 Consider a Camot-cycle heat pump with R-22 as
the working fluid. Heat is rejected from the R-22 at
homework problems M 287
40 o C, during which process the R-22 changes from
saturated vapor to saturated liquid. The heat is
transferred to the R-22 at 0°C.
a. Show the cycle on a T-s diagram.
b. Find the quality of the R-22 at the beginning
and end of the isothermal heat addition process
at 0°C.
c. Determine the coefficient of performance for the
cycle.
8.35 Do Problem 8.34 using refrigerant R-134a instead
ofR-22.
8.36 Water at 200 kPa with x = 1.0 is compressed in a
piston cylinder to 1 MPa and 250°C in a reversible
process. Find the sign for the work and the sign for
the heat transfer.
8.37 Water at 200 kPa with.r = 1.0 is compressed in a
piston cylinder to 1 MPa and 350°C in a reversible
process. Find the sign for the work and the sign for
the heat transfer.
8.38 Ammonia at 1 MPa and 50°C is expanded in a pis-
ton cylinder to 500 kPa and 20°C in a reversible
process. Find the sign for both the work and the
heat transfer.
8.39 One kilogram of ammonia in a piston cylinder at
50°C and 1000 kPa is expanded in a reversible
isothermal process to 100 kPa. Find the work and
heat transfer for this process.
8.40 One kilogram of ammonia in a piston cylinder at
50°C and 1000 kPa is expanded in a reversible iso-
baric process to 140°C, shown in Fig. 8.40. Find
the work and heat transfer for this process.
■F~ constant
FIGURE P8.40
8.41 One kilogram of ammonia in a piston cylinder at
50°C and 1000 kPa is expanded in a reversible adi-
abatic process to 100 kPa, shown in Fig. 8.41, Find
the work and heat transfer for this process.
-4 p
FIGURE P8.41
8.42 A cylinder fitted with a piston contains ammonia at
50°C and 20% quality with a volume of 1 L. The
ammonia expands slowly, and during this process
heat is transferred to maintain a constant tempera-
ture. The process continues until all the liquid is
gone. Determine the work and heat transfer for this
process.
8.43 An insulated cylinder fitted with a piston contains
0.1 kg of water at 100°C with 90% quality. The
piston is moved, compressing the water until it
reaches a pressure of 1 .2 MPa. How much work is
required in the process?
8.44 Compression and heat transfer brings R-134a in a
piston cylinder from 500 kPa and 50°C to saturated
vapor in an isothermal process. Find the specific
heat transfer and the specific work.
8.45 One kilogram of water at 300°C expands against a
piston in a cylinder until it reaches ambient pres-
sure, 100 kPa, at which point the water has a- qual-
ity of 90.2%. It may be assumed that the expansion
is reversible and adiabatic. What was the initial
pressure in the cylinder and how much work is
done by the water?
8.46 Water in a piston/cylinder device at 400°C and
2000 kPa is expanded in a reversible adiabatic
process. The specific work is measured to be
415.72 kj/kg out. Find the final P and Tand show
the P-v and the T-s diagrams for the process.
8.47 A piston cylinder with 2 kg ammonia at 50°C and
100 kPa is compressed to 1000 kPa. The process
happens so slowly that the temperature is constant.
Find the heat transfer and the work for the process
assuming it to be reversible.
8.48 A piston cylinder with R-134a at -20°C and 100
kPa is compressed to 500 kPa in a reversible adia-
batic process. Find the final temperature and the
specific work.
8.49 A closed tank, with V = 1 L, containing 5 kg of
water initially at 25°C is heated to 175°C in a
288 6 Chapter Eight entropy
reversible process. Find the heat transfer to the
water and its change in entropy.
8.50 A cylinder containing R-134a at 10°C and 150 kPa
has an initial volume of 20 L. A piston compresses
the R-134a in a reversible, isothermal process until
it reaches the saturated vapor state. Calculate the
required work and heat transfer to accomplish this
process.
8.51 A heavily insulated cylinder fitted with a friction-
less piston, as shown in Fig. P8.51, contains am-
monia at 5°C and 92.9% quality, at which point the
volume is 200 L. The external force on the piston
is now increased slowly, compressing the ammonia
until its temperature reaches 50°C. How much
work is done by the ammonia during this process?
V/S///////////////////////////////^
ft/////////////////////////////////'
FIGURE P8.51
8.52 A piston/cylinder device with 2 kg water at 1000
kPa and 250°C is cooled with a constant loading on
the piston. This isobaric process ends when the
water has reached a state of saturated liquid. Find
the work and heat transfer and sketch the process
in both a P-v and a T-s diagram.
$.53 Water at 1000 kPa and 250°C is brought to saturated
vapor in a piston/cylinder assembly with an isother-
mal process. Find the specific work and heat trans-
fer. Estimate the specific work from the area in the
P-v diagram and compare it to the correct value,
8.54 Water at 1000 kPa and 250°C is brought to satu-
rated vapor in a rigid container, shown in Fig.
P8.54. Find the final rand the specific heat transfer
in this isometric process.
8.55 Estimate the specific heat transfer from the area in
the T-s diagram and compare it to the correct value
for the states and process in Problem 8.54.
8.56 Water at 1000 kPa and 250°C is brought to satu-
rated vapor in a piston/cylinder setup with an iso-
baric process. Find the specific work and heat
transfer. Estimate the specific heat transfer from
the area in the T-s diagram and compare it to the
correct value.
8.57 A heavily insulated cylinder/piston contains am-
monia at 1200 kPa, 60°C. The piston is moved,
expanding the ammonia in a reversible process
until the temperature is — 20°C. During the process
600 kJ of work is given out by the ammonia. What
was the initial volume of the cylinder?
8.58 Water at 1000 kPa and 250°C is brought to satu-
rated vapor in a piston/cylinder device with an adi-
abatic process. Find the final T and the specific
work. Estimate the specific work from the area in
the P-v diagram and compare it to the correct
value.
A rigid, insulated vessel contains superheated
vapor steam at 3 MPa, 400 o C. A valve on the ves-
sel is opened, allowing steam to escape as shown in
Fig. P8.59. The overall process is irreversible, but
the steam remaining inside the vessel goes through
a reversible adiabatic expansion. Determine the
fraction of steam that has escaped when the final
state inside is saturated vapor.
8.59
FIGURE P8.54
FIGURE P8.59
8.60 A piston/cylinder setup contains 2 kg of water at
200°C and 10 MPa. The piston is slowly moved to
expand the water in an isothermal process to a
pressure of 200 kPa. Heat transfer takes place with
an ambient surrounding at 200°C, and the whole
process may be assumed reversible. Sketch the
Homework Problej^s H 289
process in a P-V diagram and calculate both the
heat transfer and the total work.
Entropy Generation
8.61 One kg of water at 500°C and 1 kg of saturated
water vapor, both at 200 kPa, are mixed in a
constant-pressure and adiabatic process. Find the
final temperature and the entropy generation for the
process.
8.62 The unrestrained expansion of the reactor water in
Problem 5.48 has a final state in the two-phase re-
gion. Find the entropy generated in the process.
8.63 A mass- and atmosphere-loaded piston/cylinder
device contains 2 kg of water at 5 MPa and 100°C.
Heat is added from a reservoir at 700°C to the
water until it reaches 700°C. Find the work, heat
transfer, and total entropy production for the sys-
tem and surroundings.
8.64 Ammonia is contained in a rigid sealed tank of un-
known quality at 0°C. When heated in boiling
water to 100°C, its pressure reaches 1200 kPa. Find
the initial quality, the heat transfer to the ammonia,
and the total entropy generation.
8.65 An insulated cylinder/piston arrangement contains
R-134a at 1 MPa and 50°C, with a volume of
100 L. The R-134a expands, moving the piston
until the pressure in the cylinder has dropped to
100 kPa. It is claimed that the R-134a does 190 kj
of work against the piston during the process. Is
that possible?
8.66 A piece of hot metal should be cooled rapidly
(quenched) to 25°C, which requires removal of
1000 kj from the metal. There are three possible
ways to remove this energy: (1) Submerge the
metal into a bath of liquid water and ice, thus melt-
ing the ice. (2) Let saturated liquid R-22 at -20°C
absorb the energy so that it becomes saturated
vapor. (3) Absorb the energy by vaporizing liquid
nitrogen at 101.3 kPa pressure.
a. Calculate the change of entropy of the cooling
media for each of the three cases.
b. Discuss the significance of the results.
8.67 A piston/cylinder setup has 2.5 kg of ammonia at
50 kPa and -20°C. Now it is heated to 50°C at
constant pressure through the bottom of the cylinder
from external hot gas at 200°C. Find the heat trans-
fer to the ammonia and the total entropy generation.
8.68 A cylinder fitted with a movable piston contains
water at 3 MPa with 50% quality, at which point
the volume is 20 L. The water now expands to 1.2
MPa as a result of receiving 600 kJ of heat from a
large source at 300°C. It is claimed that the water
does 124 kj of work during this process. Is this
possible?
8.69 A piston/cylinder device loaded so it gives constant
pressure has 0.75 kg of saturated vapor water at
200 kPa. It is now cooled so that the volume be-
comes half the initial volume by heat transfer to the
ambient surroundings at 20°C. Find the work, heat
transfer, and total entropy generation.
8.70 A piston/cylinder setup contains 1 kg of water at
150 kPa and 20°C. The piston is loaded so that
pressure is linear in volume. Heat is added from a
600°C source until the water is at 1 MPa and
500°C. Find the heat transfer and total change hi
entropy.
8.71 A piston/cylinder has ammonia at 2000 kPa, 80°C
with a volume of 0. 1 m 3 . The piston is loaded with
a linear spring, and the outside ambient is at 20°C,
shown in Fig. P8.71. The ammonia now cools
down to 20°C at which point it has a quality of
10%. Find the work, heat transfer, and total en-
tropy generation in the process.
20' c
NH 3 "
11
FIGURE P8.71
8.72 A cylinder/piston assembly contains water at 200
kPa and 200°C with a volume of 20 L. The piston
is moved slowly, compressing the water to a pres-
sure of 800 kPa. The loading on the piston is such
that the product PV is a constant. Assuming that
the room temperature is 20°C, show that this
process does not violate the second law.
8.73 One kilogram of ammonia (NH 3 ) is contained in
a spring-loaded piston/cylinder, Fig. P8.73, as
290 H CHAPTER. EIGHT ENTROPY
r
100° c
FIGURE P8.73
saturated liquid at -20°C. Heat is added from a
reservoir at 100°C until a final condition of 800
kPa and 70°C is reached. Find the work, heat trans-
fer, and entropy generation, assuming the process
is internally reversible.
8.74 A piston/cylinder device keeping a constant pres-
sure of 500 kPa has 1 kg of water at 20°C and 1 kg
of water at 100°C separated by a membrane, shown
in Fig. P8.74. The membrane is broken and the
water comes to a uniform state with no external
heat transfer. Find the final temperature and the en-
tropy generation.
inside constant temperature of 5°C. Assume the
milk to have the properties of liquid water and find
the entropy generated in the cooling process.
8.78 A foundry form box with 25 kg of 200°C hot sand
is dumped into a bucket with 50 L of water at
15°C. Assuming no heat transfer with the sur-
roundings and no boiling away of liquid water, cal-
culate the net entropy change for the process.
8.79 A 5-kg steel container is cured at 500°C. An amount
of liquid water at 15°C, 100 kPa, is added to the
container so that a final uniform temperature of the
steel and the water becomes 75°C. Neglect any
water that might evaporate during the process and
any air in the container. How much water should be
added, and how much entropy was generated?
8.80 A pan in an autoshop contains 5 L of engine oil at
20°C and 100 kPa. Now 2 L of hot 100°C oil is
mixed into the pan. Neglect any work term and find
the final temperature and the entropy generation.
8.81 Find the total work the heat engine can give out as it
receives energy from the rock bed as described in
Problem 7.61 (see Fig. P8.81). Hint. Write the en-
tropy balance equation for the control volume that is
the combination of the rockbed and the heat engine.
■F- constant
FIGURE P8.74
Entropy of a Liquid or Solid
8.75 A piston/cylinder setup has constant pressure- of
2000 kPa with water at 20°C. It is now heated up to
100°C. Find the heat transfer and the entropy change
using the steam tables. Repeat the calculation using
constant heat capacity and incompressibility.
8.76 A large slab of concrete, 5 m X 8 m X 0.3 m, is
used as a thermal storage mass in a solar-heated
house. If the slab cools overnight from 23°C to
18°C in an 18°C house, what is the net entropy
change associated with this process?
8.77 A 4-L jug of milk at 25°C is placed in your refrig-
erator where it is cooled down to the refrigerator's
FIGURE P8.81
8.82 Two kg of liquid lead initially at 500°C are poured
into a form. It then cools at constant pressure down
to room temperature of 20°C as heat is transferred
to the room. The melting point of lead is 327°C,
and the enthalpy change between the phases, h i/s is
24.6 kJ/kg. The specific heats are in Tables A.3 and
A.4. Calculate the net entropy change for this
process.
8.83 A 12-kg steel container has 0.2 kg of superheated
water vapor at 1000 kPa and 200°C. The total mass
is now cooled to the ambient temperature of 30°C.
Homework Problems □ 291
How much heat transfer occurs, and what is the
total entropy generation?
8.84 A 5-kg aluminum radiator holds 2 kg of liquid
R-134a at — 10°C. The setup is brought indoors and
heated with 220 kJ from a heat source at 100°C.
Find the total entropy generation for the process
assuming the R-134a remains a liquid.
8.85 A piston/cylinder of total 1 kg steel contains 0.5 kg
ammonia at 1600 kPa with both masses at 120°C.
Some stops are placed so that a minimum volume
is 0.02 m 3 , shown in Fig. P8.85. Now the whole
system is cooled down to 30°C by heat transfer to
the ambient at 20°C, and during the process the
steel keeps the same temperature as the ammonia.
Find the work, heat transfer, and total entropy gen-
eration in the process.
1
II
ZD.-',
NH 3
20° C
FIGURE P8.85
8.86 A hollow steel sphere with a 0.5 -m inside diameter
and a 2 -mm thick wall contains water at 2 MPa,
250°C. The system (steel plus water) cools to the
ambient temperature, 30°C. Calculate the net en-
tropy change of the system and surroundings for
this process.
Entropy of Ideal Gases
8.87 A mass of 1 kg of air contained in a cylinder at
1.5 MPa and 1000 K expands in. a reversible
isothermal process to a volume 10 times larger.
Calculate the heat transfer during the process and
the change of entropy of the air.
8.88 A piston/cylinder setup containing air at 100 kPa
and 400 K is compressed to a final pressure of
1000 kPa. Consider two different processes: (1) a
reversible adiabatic process and (2) a reversible
isothermal process. Show both processes in a P~v
diagram and a T-s diagram. Find the final temper-
ature and the specific work for both processes.
8.89 Consider a Carnot-cycle heat pump having 1 kg
of nitrogen gas in a cylinder/piston arrangement.
This heat pump operates between reservoirs at
300 K and 400 K. At the beginning of the low-
temperature heat addition, the pressure is 1 MPa.
During this process the volume triples. Analyze
each of the four processes in the cycle and
determine
a. the pressure, volume, and temperature at each
point.
b. the work and heat transfer for each process.
8.90 Consider a small air pistol with a cylinder volume
of 1 cm 3 at 250 kPa and 27°C. The bullet acts as a
piston initially held by a trigger, shown in Fig.
P8.90. The bullet is released so that the air ex-
pands in an adiabatic process. If the pressure
should be 100 kPa as the bullet leaves the cylin-
der, find the final volume and the work done by
the air.
FIGURE P8.90
8.91 Oxygen gas in a piston/cylinder assembly at 300 K
and 100 kPa with volume 0.1 m 3 is compressed in
a reversible adiabatic process to a final tempera-
ture of 700 K. Find the final pressure and volume
using Table A. 5.
8.92 Oxygen gas in a piston/cylinder device at 300 K
and 100 kPa with volume 0.1 m 3 is compressed in
a reversible adiabatic process to a final tempera-
ture of 700 K. Find the final pressure and volume
using Table A.8.
8.93 A hand-held pump for a bicycle has a volume of
25 cm 3 when fully extended. You now press the
plunger (piston) in while holding your thumb over
the exit hole so that an air pressure of 300 kPa is
obtained. The outside atmosphere is at P and T .
FIGURE P8.93
Consider two cases: (1) It is done quickly (-1 s)
and (2) it is done very slowly (~i h).
a. State assumptions about the process for each
case.
b. Find the final volume and temperature for both
cases.
8.94 An insulated cylinder/piston setup contains car-
bon dioxide gas at 120 kPa and 400 K. The gas is
compressed to 2.5 MPa in a reversible adiabatic
process. Calculate the final temperature and the
work per unit mass, assuming
a. Variable specific heat (Table A.8).
b. Constant specific heat (value from Table A.5).
c. Constant specific heat (value at an intermediate
temperature from Table A. 6).
8.95 A piston/cylinder assembly shown in Fig. P8.95,
contains air at 1380 K and 15 MPa, with V x = 10
cm 3 and^ cyi = 5 cm 2 . The piston is released, and
just before the piston exits the end of the cylinder,
the pressure inside is 200 kPa. If the cylinder is
insulated, what is its length? How much work is
done by the air inside?
f ///////////////////////////^^
FIGURE P8.95
8,96 Two rigid tanks shown in Fig. P8.96 each con-
tain 10 kg of N 2 gas at 1000 K and 500 kPa.
They are now thermally connected to a re-
versible heat pump, which heats one and cools
the other with no heat transfer to the surround-
ings. When one tank is heated to 1500 K the
process stops. Find the final (P, T) in both tanks
and the work input to the heat pump, assuming
constant heat capacities.
FIGURE P8.96
8.97 A spring-loaded piston/cylinder setup contains
1.5 kg of air at 27°C and 160 kPa. It is now
heated in a process where pressure is linear in
volume, P = A +- BV, to twice the initial vol-
ume where it reaches 900 K. Find the work, heat
transfer, and total entropy generation assuming
a source at 900 K.
8.98 A rigid storage tank of 1.5 m 3 contains 1 kg of
argon at 30°C. Heat is then transferred to the
argon from a furnace operating at 1300°C until
the specific entropy of the argon has increased by
0.343 kJ/kg K. Find the total heat transfer and the
entropy generated in the process.
8.99 A rigid tank contains 2 kg of air at 200 kPa and
ambient temperature, 20°C. An electric current
now passes through a resistor inside the tank.
After a total of 100 kJ of electrical work has
crossed the boundary, the air temperature inside is
80°C. Is this possible?
8.100 Argon in a light bulb is at 90 kPa and heated from
20°C to 60°C with electrical power. Do not con-
sider any radiation, or the glass mass. Find the
total entropy generation per unit mass of argon.
8.101 We wish to obtain a supply of cold helium gas by
applying the following technique. Helium con-
tained in a cylinder at ambient conditions, 100
kPa and 20°C, is compressed in a reversible
isothermal process to 600 kPa, after which the gas
is expanded back to 100 kPa in a reversible adia-
batic process.
a. Show the process on a T-s diagram.
b. Calculate the final temperature and the net
work per kilogram of helium.
8.102 A 1 m 3 insulated, rigid tank contains air at 800
kPa and 25°C. A valve on the tank is opened, and
the pressure inside quickly drops to 150 kPa, at
which point the valve is closed. Assuming that the
air remaining inside has undergone a reversible
HOMEWORK PROBLEMS S 293
adiabatic expansion, calculate the mass with-
drawn during the process.
8.103 Nitrogen at 200°C and 300 kPa is in a piston/
cylinder device of volume 5 L, with the piston
locked with a pin. The forces on the piston re-
quire a pressure inside of 200 kPa to balance it
without the pin. The pin is removed and the pis-
ton quickly comes to its equilibrium position
without any heat transfer. Find the final P, T, and
Tand the entropy generation due to this partly un-
restrained expansion.
8.104 A rigid container with a volume of 200 L is di-
vided into two equal volumes by a partition,
shown in Fig. P8.104. Both sides contain nitro-
gen; one side is at 2 MPa and 200°C, while the
other is at 200 kPa and 100°C. The partition rup-
tures, and the nitrogen comes to a uniform state at
70°C. Assume the temperature of the surround-
ings to be 20°C. Determine the work done and the
net entropy change for the process.
A .
B
N 2 V
H 2
FIGURE P3.104
8.105 Nitrogen at 600 kPa and 127°C is in a 0.5 m 3 in-
sulated tank connected to a pipe with a valve to a
second insulated initially empty tank with a vol-
ume of 0.5 m 3 , shown in Fig. P8.105. The valve is
FIGURE P8.105
opened, and the nitrogen fills both tanks at a uni-
form state. Find the final pressure and tempera-
ture and the entropy generation this process
causes. Why is the process irreversible?
Polytropic Processes
8.106 Neon at 400 kPa and 20°C is brought to 100°C in
a polytropic process with n = 1.4. Give the sign
for the heat transfer and work terms and explain.
8.107 A mass of 1 kg of air contained in a cylinder at
1 .5 MPa and 1000 K expands in a reversible adia-
batic process to 100 kPa. Calculate the final tem-
perature and the work done during the process,
using
a. Constant specific heat (value from Table A.5).
b. The ideal gas tables (Table A.7).
8.108 An ideal gas having a constant specific heat un-
dergoes a reversible polytropic expansion with
exponent n = 1.4. If the gas is carbon dioxide,
will the heat transfer for this process be positive,
negative, or zero?
8.109 A cylinder/piston setup contains 1 kg of methane
gas at 100 kPa and 20°C. The gas is compressed
reversibly to a pressure of 800 kPa. Calculate the
work required if the process is
a. Adiabatic
b. Isothermal
c. Polytropic, with exponent b =1.15
8.110 Helium in a piston/cylinder assembly at 20°C and
100 kPa is brought to 400 K in a reversible poly-
tropic process with exponent n = 1.25. You may
assume helium is an ideal gas with constant spe-
cific heat. Find the final pressure and both the spe-
cific heat transfer and specific work.
8.111 The power stroke in an internal combustion en-
gine can be approximated with a polytropic ex-
pansion. Consider air in a cylinder volume of
0.2 L at 7 MPa and 1 800 K, shown in Fig. P 8.1 1 1.
FIGURE P8.111
It now expands in a reversible polytropic process
with exponent n = 1.5, through a volume ratio of
8:1. Show this process on P-v and T-s diagrams,
and calculate the work and heat transfer for the
process.
8.112 A piston/cylinder contains air at 300 K., 100 kPa.
It is now compressed in a reversible adiabatic
process to a volume seven times as small. Use
constant heat capacity and find the final pressure
and temperature, the specific work, and specific
heat transfer for the process.
8.113 A cylinder/piston device contains carbon dioxide
at 1 MPa and 300°C with a volume of 200 L. The
total external force acting on the piston is propor-
tional to K 3 . This system is allowed to cool to
room temperature, 20°C. What is the total entropy
generation for the process?
8.114 A device brings 2 kg of ammonia from 150 kPa
and -20°C to 400 kPa and 80°C in a polytropic
process. Find the polytropic exponent, n, the
work, and the heat transfer. Find the total entropy
generated assuming a source at 100°C.
8.115 A cylinder/piston setup contains 100 L of air at
1 10 kPa and 25°C. The air is compressed in a re-
versible polytropic process to a final state of 800
kPa and 200°C. Assume the heat transfer is with
the ambient surroundings at 25°C and determine
the polytropic exponent n and the final volume
of the air. Find the work done by the air, heat trans-
fer, and total entropy generation for the process.
8.116 A mass of 2 kg of ethane gas at 500^ kPa and
100°C undergoes a reversible polytropic expan-
sion with exponent n = 1.3 to a final temperature
of the ambient surroundings, 20°C. Calculate the
total entropy generation for the process if the heat
is exchanged with the ambient surroundings.
8.117 A piston/cylinder contains air at 300 K, 100 kPa.
A reversible polytropic process with n = 1.3
brings the air to 500 K. Any heat transfer if it
comes in is from a 325°C reservoir, and if it goes
out it is to the ambient at 300 K. Sketch the
process in a P-v and a T-s diagram. Find the spe-
cific work and specific heat transfer in the
process. Find the specific entropy generation (ex-
ternal to the air) in the process.
8.118 A cylinder/piston device contains saturated vapor
R-22 at 10°C; the volume is 10 L. The R-22 is
compressed to 2 MPa at 60°C in a reversible (in-
ternally) polytropic process. If all the heat transfer
during the process is with the ambient surround-
ings at 10°C, calculate the net entropy change.
8.119 A cylinder/piston setup contains air at ambient
conditions, 100 kPa and 20°C, with a volume
of 0.3 m 3 . The air is compressed to 800 kPa in
a reversible polytropic process with exponent
n = 1.2, after which it is expanded back to
100 kPa in a reversible adiabatic process.
a. Show the two processes in P-v and T-s dia-
grams.
b. Determine the final temperature and the net
work.
Rates or Fluxes of Entropy
8.120 A reversible heat pump uses 1 k\V of power input
to heat a 25°C room, drawing energy from the
outside at 15°C. Assuming every process irre-
versible, what are the total rates of entropy into
the heat pump from the outside and from the heat
pump to the room?
8.121 An amount of power, say 1000 kW, comes from a
furnace at 800°C going into water vapor at 400°C.
From the water the power goes to solid metal at
200°C and then into some air at 70°C. For each
location calculate the flux of s as (QIT). What,
makes the flux larger and larger?
8.122 Room air at 23°C is heated by a 2000 W space
heater with a surface filament temperature of
700 K, shown in Fig. P8.122. The room at steady
state loses heat to the outside, which is at 7°C.
Find the rate(s) of entropy generation and specify
where it is made.
23° C
Wall
7°C
FIGURE PS.122
HOMEWORK PROBLEMS H 295
8.123 A small halogen light bulb receives an electrical
power of 50 W. The small filament is at 1000 K
and gives out 20% of the power as light and the
rest as heat transfer to the gas, which is at 500 K;
the glass is at 400 K. All the power is absorbed by
the room walls at 25°C. Find the rate of genera-
tion of entropy in the filament, in the entire bulb
including glass, and in the entire room including
the bulb.
8.124 A fanner runs a heat pump using 2 kW of power
input. It keeps a chicken hatchery at a constant
30°C, while the room loses 10 kW to the colder
outside ambient at 10°C. What is the rate of en-
tropy generated in the heat pump? What is the
rate of entropy generated in the heat loss process?
8.125 The automatic transmission in a car receives
25 kW shaft work and gives out 24 kW to the
drive shaft. The balance is dissipated in the hy-
draulic fluid and metal casing, all at 45°C, which
in rum transmits it to the outer atmosphere at
20°C. What is the rate of entropy generation inside
the transmission unit? What is it outside the unit?
Review Problems
8.126 An insulated cylinder/piston arrangement has an
initial volume of 0.15 m 3 and contains steam at 400
kPa and 200°C. The steam is expanded adiabati-
cally, and the work output is measured very care-
fully to be 30 kJ. It is claimed that the final state of
the water is in the two-phase (liquid and vapor) re-
gion. What is your evaluation of the claim?
8.127 A closed tank, V — 10 L, containing 5 kg of water
initially at 25°C, is heated to 175°C by a heat
pump that is receiving heat from the surroundings
at 25°C. Assume that this process is reversible.
Find the heat transfer to the water and the work .
input to the heat pump.
8.128 Two tanks contain steam, and they are both con-
nected to a piston cylinder, as shown in Fig.
PS. 128. Initially, the piston is at the bottom, and
the mass of the piston is such that a pressure of
1.4 MPa below it will be able to lift it. Steam in A
has a mass of 4 kg at 7 MPa and 700°C, and B has
2 kg at 3 MPa and 350°C. The two valves are
opened, and the water comes to a uniform state.
Find the final temperature and the total entropy
generation, assuming no heat transfer.
'////////////■/////////////.
FIGURE P8.128
8.129 A piston/cylinder with constant loading of piston
contains 1 L water at 400 kPa, quality 15%. It has
some stops mounted so the maximum possible
volume is 1 1 L. A reversible heat pump extracting
heat from the ambient at 300 K, 100 kPa, heats
the water to 300°C. Find the total work and heat
transfer for the water and the work input to the
heat pump.
8.130 Water in a piston/cylinder shown in Fig. P8. 130 is
at 1 MPa, 500°C. There are two stops: a lower one
at which = 1 m 3 and an upper one at V^ x =
3 m 3 . The piston is loaded with a mass and out-
side atmosphere such that it floats when the pres-
sure is 500 kPa. This setup is now cooled to
100°C by rejecting heat to the surroundings at
20°C. Find the total entropy generated in the
process.
SI
fi>0
FIGURE P8.130
8.131 A cylinder fitted with a frictionless piston con-
tains water, as shown in Fig. P8.131. A constant
296 ■ CHAPTER EIGHT ENTROPY
Constant hydraulic
pressure
(F)
1 (•
J
Ql
FIGURE P8.131
hydraulic pressure on the back face of the piston
maintains a cylinder pressure of 10 MPa. Initially,
the water is at 700°C, and the volume is 100 L.
The water is now cooled and condensed to satu-
rated liquid. The heat released during this process
is the Q supply to a cyclic heat engine that in turn
rejects heat to the ambient at 30°C. If the overall
process is reversible, what is the net work output
of the heat engine?
8.132 A cylinder/piston contains 3 kg of water at 500
kPa, 600°C. The piston has a cross-sectional area
of 1 m 2 and is restrained by a linear spring with
spring constant 10 kN/m. The setup is allowed to
cool down to room temperature due to heat trans-
fer to the room at 20°C. Calculate the total (water
and surroundings) change in entropy for the
process.
8.133 An insulated cylinder fitted with a frictionless
piston contains saturated vapor R-12 at ambient
temperature, 20°C. The initial volume is 10 L.
The R-12 is now expanded to a temperature of
-30°C. The insulation is then removed from the
cylinder, allowing it to warm at constant pres-
sure to ambient temperature. Calculate the net
work and the net entropy change for the overall
process.
8 134 A piston/cylinder assembly contains 2 kg of
liquid water at 20°C, 100 kPa, and it is now
heated to 300°C by a source at 500°C. A pres-
sure of 1000 kPa will lift the piston off the
tower stops, shown in Fig. P8.134. Find the
final volume, work, heat transfer, and total en-
tropy generation.
FIGURE P8.134
8 135 An uninsulated cylinder fitted with a piston con-
tains air at 500 kPa, 200°C, at which point the
volume is 10 L. The external force on the piston
is now varied in such a manner that the air ex-
pands to 150 kPa, 25 L volume. It is claimed that
in this process the air produces 70% of the work
that would have resulted from a reversible, adia-
batic expansion from the same initial pressure and
temperature to the same final pressure. Room
temperature is 20°C
a. What is the amount of work claimed?
b. Is this claim possible?
8.136 A cylinder fitted with a piston contains 0.5 kg of
R-134a at 60°C, with a quality of 50 percent. The
R-134a now expands in an internally reversible
polytropic process to ambient temperature 20°C,
at which point the quality is 100%. Any heat -
transfer is with a constant-temperature source,
which is at 60°C. Find the polytropic exponent n
and show that this process satisfies the second law
of thermodynamics.
8 137 A cylinder with a linear spring-loaded piston con-
tains carbon dioxide gas at 2 MPa with a volume
of 50 L. The device is of aluminum and has a
mass of 4 kg. Everything (Al and gas) is initially
at 200°C. By heat transfer the whole system cools
to the ambient temperature of 25°C, at which
point the gas pressure is 1.5 MPa. Find the total
entropy generation for the process.
8.138 A vertical cylinder/piston contains R-22 at
-20°C, 70% quality, and the volume is 50 L,
shown 'in Fig. P8.138. This cylinder is brought
into a 20°C room, and an electric current of 10 A
is passed through a resistor inside the cylinder.
The voltage drop across the resistor is 12 V. It is
claimed that after 30 min the temperature inside
the cylinder is 40°C Is this possible?
ENGLISH UNIT PROBLEMS M 297
J" 8,139 A gas in a rigid vessel is at ambient temperature
11 • and at a pressure, P h slightly higher than ambient
pressure, P . A valve on the vessel is opened, so
gas escapes and the pressure drops quickly to am-
bient pressure. The valve is closed, and after a long
1 D + time the remaining gas returns to ambient tempera-
S~®~ ture at which point the pressure is P 2 . Develop an
expression that allows a determination of the ratio
FIGURE P8.138 of specific heats, k, in terms of the pressures.
English Unit Problems
Concept Problems
8.140E Water at 20 psia, 240 F, receives 40 Btu/lbm in
a reversible process by heat transfer. Which
process changes s the most: constant T, constant
v, or constant P?
8. 14 IE Saturated water vapor at 20 psia is compressed
to 60 psia in a reversible adiabatic process. Find
the change in v and T.
8.142E A computer chip dissipates 2 Btu of electric
work over time and rejects that as heat transfer
from its 125 F surface to 70 F air. How much
entropy is generated in the chip? How much, if
any, is generated outside the chip?
8.143E Two 10-lbm blocks of steel, one at 400 F and
the other at 70 F, come in thermal contact. Find
the final temperature and the total entropy gen-
eration in the process.
8.144E One lbm of air at 540 R is mixed with one lbm
air at 720 R in a process at a constant 15 psia
and Q = 0, Find the final T and the entropy gen-
eration in the process.
8.145E One lbm of air at 15 psia is mixed with one lbm
air at 30 psia, both at 540 R, in a rigid insulated
tank. Find the final state (P, T) and the entropy
generation in the process.
8.146E A window receives 600 Btu/h of heat transfer at
the inside surface of 70 F and transmits the 600
Btu/h from its outside surface at 36 F, continu-
ing to ambient air at 23 F. Find the flux of en-
tropy at all three surfaces and the window's rate
of entropy generation.
English Unit Problems
8.147E Consider the steam power plant in Problem
7.100E and show that this cycle satisfies the in-
equality of Clausius.
8.148E Find the missing properties and give the phase
of the substance.
a.
H 2
j =
1.75 Btu/lbm R,
h =
? T
P =
4 Ibf/in. 2
x —
?
b.
H 2
u —
1350 Btu/lbm,
T =
?JC
P =
1500 Ibf/in. 2
s —
?
c.
R-22
T =
30 F,
s ~
?.v
P =
60 ibf/in 2
d.
R-134a
T =
10 F,
V =
?J
x —
0.45
e.
NH 3
T =
60 F,
u —
?x
s =
1.35 Btu/lbm R
8.149E In a Carnot engine with water as the working
fluid, the high temperature is 450 F, and as Q H is
received, the water changes from saturated liq-
uid to saturated vapor. The water pressure at the
low temperature is 14.7 lbflin. 2 . Find T L , cycle
thermal efficiency, heat added per pound-mass,
and entropy, s, at the beginning of the heat re-
jection process,
8.150E Consider a Carnot-cycle heat pump with R-22
as the working fluid. Heat is rejected from the
R-22 at 100 F, during which process the R-22
changes from saturated vapor to saturated liq-
uid. The heat is transferred to the R-22 at 30 F.
a. Show the cycle on a T-s diagram.
b. Find the quality of the R-22 at the beginning
and end of the isothermal heat addition
process at 30 F.
c. Determine the coefficient of performance for
the cycle.
8.151E Do Problem 8.150 using refrigerant R-134a in-
stead of R-22.
8 152E Water at 30 lbf/in. 2 , x = 1.0 is compressed in a
piston/cylinder to 140 lbf/in. 2 1 600 F in a re-
versible process. Find the sign for the work and
the sign for the heat transfer.
8 153E Two pound-mass of ammonia in a piston/cylin-
der at 120 F, 150 lbf/in. 2 is expanded in a re-
versible adiabatic process to 15 lbf/in. 2 . Find the
work and heat transfer for this process.
8 154E A cylinder fitted with a piston contains ammo-
nia at 120 F, 20% quality with a volume of
60 in. 3 . The ammonia expands slowly, and dur-
ing this process heat is transferred to maintain a
constant temperature. The process continues
until all the liquid is gone. Determine the work
and heat transfer for this process.
8.155E One pound-mass of water at 600 F expands
against a piston in a cylinder until it reaches am-
bient pressure, 14.7 lbf/in. 2 , at which point the
water has a quality of 90%. It may be assumed
that the expansion is reversible and adiabatic.
a. What was the initial pressure in the cylinder?
b. How much work is done by the water?
8 156E A closed tank, V = 0.35 ft 3 , containing 10 Ibm
of water initially at 77 F is heated to 350 F by a
heat pump that is receiving heat from the sur-
roundings at 77 F. Assume that this process is
reversible. Find the heat transfer to the water
and the work input to the heat pump.
8 157E A cylinder containing R-134a at 60 F, 30 lbf/in. 2
has an initial volume of 1 ft 3 . A piston com-
presses the R-134a in a reversible, isothermal
process until it reaches the saturated vapor state.
Calculate the required work and heat transfer to
accomplish this process,
8 158E A rigid, insulated vessel contains superheated
vapor steam at 450 lbt7in. 2 , 700 F. A valve on
the vessel is opened, allowing steam to escape.
It may be assumed that the steam remaining in-
side the vessel goes through a reversible adia-
batic expansion. Determine the fraction of steam
that has escaped, when the final state inside is
saturated vapor.
8.159E An insulated cylinder/piston contains R-134a at
150 lbf/in. 2 , 120 F, with a volume of 3.5 ft 3 . The
R-134a expands, moving the piston until the
pressure in the cylinder has dropped to 15
lbf/in. 2 . It is claimed that the R-134a does 180
Btu of work against the piston during the
process. Is that possible?
8.160E A mass- and atmosphere-loaded piston/cylinder
contains 4 lbm of water at 500 lbf/in. 2 , 200 F.
Heat is added from a reservoir at 1200 F to the
water until it reaches 1200 F. Find the work,
heat transfer, and total entropy production for
the system and surroundings.
8.161E A 1 -gallon jug of milk at 75 F is placed in your
refrigerator where it is cooled down to the re-
frigerators inside temperature of 40 F. Assume
the milk has the properties of liquid water and
find the entropy generated in the cooling
process.
8.162E A cylinder/piston contains water at 30 lbffiru ,
400 F with a volume of 1 ft 3 . The piston is
moved slowly, compressing the water to a pres-
sure of 120 lbf/in. 2 . The loading on the piston is
such that the product PV is a constant. Assum-
ing that the room temperature is 70 F, show that
this process does not violate the second law.
8.163E One pound-mass of ammonia (NH 3 ) is con-
tained in a linear spring-loaded piston/cylinder
as saturated liquid at F. Heat is added from a
reservoir at 225 F until a final condition of 125
lbf/in. 2 , 160 F, is reached. Find the work, heat
transfer, and entropy generation, assuming the
process is internally reversible.
8.164E A foundry form box with 50 lbm of 400 F hot
sand is dumped into a bucket with 2 ft 3 water
at 60 F. Assuming no heat transfer with the
surroundings and no boiling away of liquid
water, calculate the net entropy change for the
process.
8.165E Four pounds of liquid lead at 900 F are poured
into a form. It then cools at constant pressure
down to room temperature at 68 F as heat is
transferred to the room. The melting point of
lead is 620 F, and the enthalpy change between
the phases fyis 10.6Btu/lbm. The specific heats
are in Tables F.2 and F.3. Calculate the net en-
tropy change for this process.
English Unit Problems H 299
8.166E A hollow steel sphere with a 2-ft inside diameter
and a 0.1-in. thick wall contains water at 300
lbf/in. 2 , 500 F, The system (steel plus water)
cools to the ambient temperature, 90 F. Calcu-
late the net entropy change of the system and
surroundings for this process.
8.1 67E Oxygen gas in a piston/cylinder at 500 R and
1 atm with a volume of 1 ft 3 is compressed in a
reversible adiabatic process to a final tempera-
ture of 1000 R. Find the final pressure and vol-
ume using constant heat capacity from Table F.4.
8.168E Oxygen gas in a piston/cylinder at 500 R and
1 atm with a volume of 1 ft 3 is compressed in a
reversible adiabatic process to a final tempera-
ture of 1000 R. Find the final pressure and vol-
ume using Table F.6.
8.1 69E A handheld pump for a bicycle has a volume of
2 in. 3 when fully extended. You now press the
plunger (piston) in while holding your thumb
over the exit hole so an air pressure of 45 lbf/in. 2
is obtained. The outside atmosphere is at P 0> T .
Consider two cases: (1) it is done quickly
(~- 1 s), and (2) it is done very slowly (— 1 h).
a. State assumptions about the process for each
case.
b. Find the final volume and temperature for
both cases.
8.170E A piston/cylinder contains air at 2500 R, 2200
lbf/in. 2 ,with V x = 1 in. 3 , A cyl = 1 in. 2 , as shown
in Fig. P8.95. The piston is released, and just
before the piston exits the end of the cylinder
the pressure inside is 30 lbf/in. 2 . If the cylinder
is insulated, what is its length? How much work
is done by the air inside?
8.171E A 25-ft 3 insulated, rigid tank contains air at 110
lbf/in. 2 , 75 F. A valve on the tank is opened, and
the pressure inside quickly drops to 15 lbf/in. 2 , '
at which point the valve is closed. Assuming
that the air remaining inside has undergone a re-
versible adiabatic expansion, calculate the mass
withdrawn during the process.
8.172E A rigid container with volume 7 ft 3 is divided
into two equal volumes by a partition. Both
sides contain nitrogen, one side is at 300 lbf/in, 2 ,
400 F, and the other at 30 lbf/in. 2 , 200 F. The par-
tition ruptures, and the nitrogen comes to a uni-
form state at 160 F. Assuming the temperature of
the surroundings is 68 F, determine the work
done and the net entropy change for the process.
8.173E Nitrogen at 90 lbf/in. 2 , 260 F, is in a 20 ft 3 insu-
lated tank connected to a pipe with a valve to a
second insulated, initially empty tank of volume
20 ft 3 . The valve is opened, and the nitrogen fills
both tanks. Find the final pressure and tempera-
ture and the entropy generation this process
causes. Why is the process irreversible?
8.174E Helium in a piston/cylinder at 70 F, 15 lbf/in 2 ,
is brought to 720 R in a reversible polytropic
process with exponent n = 1.25. You may as-
sume helium is an ideal gas with constant spe-
cific heat. Find the final pressure and both the
specific heat transfer and specific work.
8.175E A cylinder/piston contains air at ambient con-
ditions, 14.7 lbf/in. 2 and 70 F, with a volume
of 10 ft 3 . The air is compressed to 100 lbf/in. 2
in a reversible polytropic process with expo-
nent, n = 1 .2, after which it is expanded back
to 14.7 lbf/in. 2 in a reversible adiabatic
process.
a. Show the two processes in P-v and T~s
diagrams.
b. Determine the final temperature and net work.
c. What is the potential refrigeration capacity
(in British thermal units) of the air at the final
state?
8.176E A cylinder/piston contains carbon dioxide at
150 lbf/in. 2 , 600 F, with a volume of 7 ft 3 . The
total external force acting on the piston is pro-
portional to V s . This system is allowed to cool to
room temperature, 70 F. What is the total en-
tropy generation for the process?
8.177E A cylinder/piston contains 4 ft 3 of air at 16
lbf/in. 2 , 77 F. The air is compressed in a re-
versible polytropic process to a final state of 120
lbf/in. 2 , 400 F. Assume the heat transfer is with
the ambient at 77 F and determine the polytropic
exponent n and the final volume of the air. Find
the work done by the air, heat transfer, and total
entropy generation for the process.
8.178E A reversible heat pump uses 1 kW of power
input to heat a 78 F room, drawing energy from
the outside at 60 F. Assume every process is re-
300 H Chapter Eight entropy
versible, what are the total rates of entropy into
the heat pump from the outside and from the
heat pump to the room?
8.179E A farmer runs a heat pump using 2.5 hp of
power input. It keeps a chicken hatchery at a
constant 86 F, while the room loses 20 Btu/s to
the colder outside ambient at 50 F. What is the
rate of entropy generated in the heat pump?
What is the rate of entropy generated in the heat
loss process?
8.1 8 0E A cylinder/piston contains 5 Ibm of water at 80
lbf/in. 2 , 1000 F. The piston has a cross-sectional
area of 1 ft 2 and is restrained by a linear spring
with spring constant 60 lbf/in. The setup is al-
lowed to cool down to room temperature due to
heat transfer to the room at 70 F. Calculate the
total (water and surroundings) change in en-
tropy for the process.
8.181E Water in a piston/cylinder is at 150 lbf/in. 2 ,
900 F, as shown in Fig. P8.130. There are two
stops: a lower one at which V min — 35 ft 3 and an
■upper one at F ra „ = 105 ft 3 . The piston is loaded
with a mass and outside atmosphere such that it
floats when the pressure is 75 lbf/in. 2 . This setup
is now cooled to 210 F by rejecting heat to the
surroundings at 70 F. Find the total entropy gen-
erated in the process.
8.182E A cylinder with a linear spring-loaded piston
contains carbon dioxide gas at 300 lbf/in. 2 with
a volume of 2 ft 3 . The device is of aluminum
and has a mass of 8 Ibm. Everything (Al and
gas) is initially at 400 F. By heat transfer the
whole system cools to the ambient temperature
of 77 F, at which point the gas pressure is 220
lbf/in. 2 . Find the total entropy generation for the
process.
Computer, design, and Open-Ended problems
8.183 Write a computer program to solve Problem 8,78
using constant specific heat for both the sand and
the liquid water. Let the amount and the initial
temperatures be input variables.
8.184 Write a program to solve Problem 8.81 with the
thermal storage rock bed in Problem 7.68. Let the
size and temperatures be input variables so that
the heat engine work output can be studied as a
function of the system parameters.
8.185 Write a program to solve the following problem.
One of the gases listed in Table A.6 undergoes a
reversible adiabatic process in a cylinder from P u
T v to P 2 . We wish to calculate the final temperature
and the work for the process by three methods:
a. Integrating the specific heat equation.
b. Assuming constant specific heat at tempera-
ture, T x .
c. Assuming constant specific heat at the average
temperature (by iteration).
8.186 Write a program to solve Problem 8.87. Let the
initial state and the expansion ratio be input
variables.
8.187 Write a program to solve a problem similar to
Problem 8.107, but instead of the ideal gas tables
use the formula for the specific heat as a function
of temperature in Table A.6.
8.188 Write a program to study a general polytropic
process in an ideal gas with constant specific heat.
Take Problem 8.106 as an example.
8.189 Write a program to solve the general case of
Problem 8.111, in which the initial state and the
expansion ratio are input variables.
8.190 A piston/cylinder maintaining constant pressure
contains 0.5 kg of water at room temperature
20°C and 100 kPa. An electric heater of 500 W
heats the water up to 500°C. Assume no heat
losses to the ambient and plot the temperature and
total accumulated entropy production as a func-
tion of time. Investigate the first part of the
process, namely, bringing the water to the boiling
point, by measuring it in your kitchen and know-
ing the rate of power added,
8.191 Air in a piston/cylinder is used as a small air-
spring that should support a steady load of 200 N.
Assume that the load can vary with ±10% over a
period of 1 s and that the displacement should be
limited to ±0.01 m. For some choice of sizes
show the spring displacement, x, as a function of
load and compare that to an elastic linear coil
spring designed for the same conditions.
8.192 Consider a piston/cylinder arrangement with am-
monia at — 10°C, 50 kPa that is compressed to
computer, design and open-ended problems M 301
200 kPa. Examine the effect of heat transfer
to/from the ambient at 15°C on the process and
the required work. Some limiting processes are a
reversible adiabatic compression giving an exit
temperature of about 90°C and as mentioned in
the text an isothermal compression. Evaluate the
work and heat transfer for both cases and for
cases in between assuming a polytropic process.
Which processes are actually possible and how
would they proceed?
Second-Law Analysis for a
CONTROL VOLUME
In the preceding two chapters we discussed the second law of thermodynamics and the
thermodynamic property entropy. As was done with the first-law analysis, we now con-
sider the more general application of these concepts, the control volume analysis, and a
number of cases of special interest. We will also discuss usual definitions of thermody-
namic efficiencies.
9.1 The Second Law of thermodynamics
for a control volume
The second law of thermodynamics can be applied to a control volume by a procedure
similar to that used in Section 6.1, where the first law was developed for a control vol-
ume. We start with the second law expressed as a change of the entropy for a control
mass in a rate form from Eq. 8.43,
to which we now will add the contributions from the mass flow rates in and out of the
control volume, A simple example of such a situation is illustrated in Fig. 9.1. The flow
of mass does carry an amount of entropy, s, per unit mass flowing, but it does not give
rise to any other contributions. As a process may take place in the flow, entropy can be
generated, but this is attributed to the space it belongs to (i.e., either inside or outside of
the control volume).
The balance of entropy as an equation then states that the rate of change in total en-
tropy inside the control volume is equal to the net sum of fluxes across the control surface
plus the generation rate. That is,
rate of change = + in - out + generation
^ = 2 »A " 2 "A + 2 + 4™ ' (9-2)
These fluxes are mass flow rates carrying a level of entropy and the rate of heat transfer
that takes place at a certain temperature (the temperature right at the control surface). The
302
The second Law of tiiermodynamics for a control volume M 303
FIGURE 9.1 The
entropy balance for a
control volume on a rate
form.
V;€i
i —
fl Q
accumulation and generation terms cover the total control volume and are expressed in
the lumped (integral form) so that
S c v . — j ps dV = m c v s = m A s A + m B s B + m^c +
^gen ~ j P^gen dV = S giILji + S gtnB + + ■ ■ •
(9.3)
If the control volume has several different accumulation units with different fluid states
and processes occurring in them, we may have to sum the various contributions over the
different domains. If the heat transfer is distributed over the control surface, then an inte-
gral has to be done over the total surface area using the local temperature and rate of heat
transfer per unit area, QiA, as
Z T } T
(QIA)
dA
(9.4)
These distributed cases typically require a much more detailed analysis, which is beyond
the scope of the current presentation of the second law.
The generation term(s) in Eq. 9.2 from a summation of individual positive intemal-
irreversibility entropy-generation terms in Eq. 9.3 is necessarily positive (or zero), such
that an inequality is often written as
dS r
dt
- > 2 MM - 2
m e s t
(9.5)
Now the equality applies to internally reversible processes and the inequality to internally
irreversible processes. The form of the second law in Eq. 9.2 or 9.5 is general, such that
any particular case results in a form that is a subset (simplification) of this form. Exam-
ples of various classes of problems are illustrated in the following sections.
— ^ If there is no mass flow into or out of the control volume, it simplifies to a control
THERMQNET maSS and the e< * uation for the total entropy reverts back to Eq. 8.43. Since that version of
— the second law has been covered in Chapter 8 here we will consider the remaining cases
that were done for the first law of thermodynamics in Chapter 6.
304 m CHAPTER Nl»E SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
9.2 The Steady-State process
and the Transient process
We now consider in turn the application of the second-law control volume equation, Eq.
9.2 or 9.5, to the two control volume model processes developed in Chapter 6.
Steady-State Process
For the steady-state process, which has been defined in Section 6.3, we conclude that
there is no change with time of the entropy per unit mass at any point within the control
volume, and therefore the first term of Eq. 9.2 equals zero. That is,
= o (9-6)
dt
so that, for the steady-state process,
2 m e s e ~ 2 MM - 2 %^ + 5 gea (9-7)
c.v.
in which the various mass flows, heat transfer and entropy generation rates, and states are
all constant with time.
If in a steady-state process there is only one area over which mass enters the control
volume at a uniform rate and only one area over which mass leaves the control volume at
a uniform rate, we can write
= 2% + 4c (9 ' 8)
c.v. J
and dividing the mass flow rate out gives
S e ^ s i + 2l f + S S™
Since s s , a is always greater than or equal to zero, for an adiabatic process it follows that
= Si + s gin > S{ ( 9 *9)
where the equality holds for a reversible adiabatic process.
EXAMPLE 9.1 Steam enters a steam turbine at a pressure of 1 MPa, a temperature of 300 C, and a ve-
locity of 50 m/s. The steam leaves the turbine at a pressure of 150 kPa and a velocity ot
200 m/s. Determine the work per kilogram of steam flowing through the turbine, assum-
ing the process to be reversible arid adiabatic.
Control volume: Turbine.
Sketch: Fig. 9.2.
Inlet state: Fixed (Fig. 9.2).
Exit state: P e) V e known.
Process: Steady state, reversible and adiabatic.
Model: Steam tables.
The Steady-State Process and the Transient Process B 305
P t = 1 MPa
i r t - = 300°C
P e = 150 kPa
^ = 200 m/s
T
FIGURE 9.2 Sketch
for Example 9.1.
Analysis
The continuity equation gives us
m e = m i = m
From the first law we have
V 2
and the second law is
Solution
From the steam tables, we get
ft, = 3051.2 kJ/kg, s;= 7.1228 kJ/kgK
The two properties known in the final state are pressure and entropy:
P e = 0.15 MPa, ff e = j, = 7.1228 kJ/kg K
The quality and enthalpy of the steam leaving the turbine can be determined as follows:
s e =7A22S=s f + -x e Sfi= 1.4335 -Kr e 5.7897
.v f = 0.9827
K = h f + x e h /g = 467.1 4- 0.9827(2226.5)
- 2655.0 kJ/kg
Therefore, the work per kilogram of steam for this isentropic process may be found
using the equation for the first law:
EXAMPLE 9.2 Consider the reversible adiabatic flow of steam through a nozzle. Steam enters the nozzle
at 1 MPa and 300°C, with a velocity of 30 m/s. The pressure of the steam at the nozzle
w = 3051.2 +
50 X 50
2 X 1000
- 2655.0 -
200 X 200
2 X 1000
- 377.5 kJ/kg
306 B CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
FIGURE 9.3 Sketch
for Example 9.2.
p. = 1 MPa
7)=300°C
V- = 30 mis
T
P e = 0.3 MPa
S„ = S;
exit is 0.3 MPa. Determine the exit velocity of the steam from the nozzle, assuming a re-
versible, adiabatic, steady-state process.
Control volume: Nozzle.
Sketch: Fig. 9.3.
Inlet state: Fixed (Fig. 9.3).
Exit State: P e known.
Process: Steady state, reversible, and adiabatic.
Model: Steam tables.
Analysis
Because this is a steady-state process in which the work, heat transfer, and changes in
potential energy are zero, we can write
Continuity equation: m e = m, -m
First law: A, + - K + "J"
Second law: s e = Sj
Solution
From the steam tables, we have
ft, = 3051.2 kJ/kg, 5 ,. = 7.1228 kJ/kgK
The two properties that we know in the final state are entropy and pressure:
^ = s , = 7.1228kJ/kgK, P e = 0.3MPa
Therefore,
T e - i59.1°C, K = 2780.2 kJ/kg
Substituting into the equation for the first law, we have
= 3051.2 - 2780.2 + = 271.5 kJ/kg
V - V2000 X 271.5 = 737 m/s
the Steady-State Process and the Transient Process M 307
EXAMPLE 9.2E Consider the reversible adiabatic flow of steam through a nozzle. Steam enters the noz-
zle at 100 lbf/in. 2 , 500 F, with a velocity of 100 ft/s. The pressure of the steam at the
nozzle exit is 40 lbf/in. 2 . Determine the exit velocity of the steam from the nozzle, as-
suming a reversible adiabatic, steady-state process.
Control volume; Nozzle.
Sketch: Fig. 9.3E.
Inlet state: Fixed (Fig. 9.3E).
Exit state: P e known.
Process: Steady state, reversible, and adiabatic.
Model: Steam tables.
Analysis
Because this is a steady-state process in which the work, the heat transfer, and changes
in potential energy are zero, we can write
Continuity equation: m e = m { = m
V 2 V 2
First law: h t + y = h e + -j-
Second law; s e = s f
Solution
From the steam tables, we have
h( = 1279.1 Btu/lbm s t = 1.7085 Btu/lbm R
The two properties that we know in the final state are entropy and pressure.
s e = Si = 1.7085 Btu/lbm R, P £ = 40 tbf/hi. 3
Therefore,
T e — 314.2 F /z e = 1193.9 Btu/lbm
FIGURE 9.3E Sketch
for Example 9.2E.
Substituting into the equation for the first law, we have
V? V?
2 ' e 2
- 1279.1 - 1193.9 +
100 X 100
2 X 32.17 X 778
V = V2 X 32.17 X 778 X 85.4 = 2070 fl/s
85.4 Btu/lbm
EXAMPLE 9.3 An inventor reports having a refrigeration compressor that receives saturated R-134a
vapor at -2CPC and delivers the vapor at 1 MPa and 40°C. The compression process is
adiabatic. Does the process described violate the second law?
Control volume:
Inlet state:
Exit state:
Process:
Model:
Compressor.
Fixed (saturated vapor at 7)).
Fixed (P e , T e known).
Steady state, adiabatic.
R-134a tables.
Analysis
Because this is a steady-state adiabatic process, we can write the second law as
Solution
From the R-134a tables, we read
s = 1.7148 kJ/kgK,
s,= 1.7395 kJ/kgK
Therefore, s e < s b whereas for this process the second law requires that s e ^ s. The
process described involves a violation of the second law and thus would be impossible.
EXAMPLE 9.4 An air compressor in a gas station, see Fig. 9.4, takes in a flow of ambient air at 100 kPa,
290 K, and compresses it to 1000 kPa in a reversible adiabatic process. We want to
know the specific work required and the exit air temperature.
Solution
C.V. air compressor, steady state, single flow through it, and we assume adiabatic fi - 0.
Continuity Eq. 6.11
Energy Eq. 6.12
Entropy Eq. 9.8:
Process
mhf = mh € + W a
ihs; + S gtu = ms e
J gen
The Steady-State Process and the Transient Process □ 309
FIGURE 9,4 Diagram
for Example 9.4.
Use constant specific heat from Table A.5, Cpo = 1.004 kJ/kg K > k= 1.4.
Entropy equation gives constant s, which gives the relation in Eq. 8.32:
EXAMPLE 9.4E
r. = 290
Pi
loopy- 2857
100 j
559.9 K
The energy equation per unit mass gives the work term
w c = h - ft e = C n (T t - T e ) = 1.004(290 - 559.9) = -271 kJ/kg
An air compressor in a gas station, see Fig. 9.4E, takes in a flow of ambient air at 14.7
lbf/in. 2 , 520 R, and compresses it to 147 lbf/in. 2 in a reversible adiabatic process. We
want to know the specific work required and the exit air temperature.
310 M CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
Solution
C.V. air compressor, steady state, single flow through it, and we assume adiabatic Q ~ 0.
Continuity Eq. 6.11
Energy Eq. 6.12
Entropy Eq. 9.8:
Process
m } = m e = m,
mhi = mhe + W Ci
ms { + S gen = ms e
Reversible S gia =
Use constant specific heat from Table F.4, = 0,24 Btu/lbm R, * = 1.4. The entropy
equation gives constant s> which gives the relation in Eq. 8.32:
T e = 520 {j^jj m7 = 1003.9 R
The energy equation per unit mass gives the work term
w c - h t -K= - T e ) = 0.24(520 - 1003.9) = -116.1 Btu/lbm
EXAMPLE 9.5 A de-superheater works by injecting liquid water into a flow of superheated steam. With
2 kg/s at 300 kPa, 200°C, steam flowing in, what mass flow rate of liquid water at 20°C
should be added to generate saturated vapor at 300 kPa? We also want to know the rate
of entropy generation in the process.
Solution
C.V. De-superheater, see Fig. 9.5, no external heat transfer, and no work.
FIGURE 9.5 Sketch
and diagram, for Example
9.5.
Continuity Eq. 6.9
Energy Eq. 6.10
Entropy Eq. 9.7
Process
/»! + m 2 = mi,
mihi + m 2 h 2 = m 3 ft 3 = 0»i + in ^ h 3
m x s x + m 2 s 2 = S glia = m^Ss ^
P = constant, W= 0, and Q =
De-superheater
300 kPa
The Steady-State process and the transient process H 311
AH the states are specified (approximate state 2 with saturated liquid 20°C)
B.1.3:^ 2865.54 M ^ = 7.3115^; A 3 = 2725.3 j| , 3 = 6.9918^
B.1.2:A2 = 83.94^, s 2 = 0.2966 kJ
kg' 2 kgK
Now we can solve for the flow rate m 2 from the energy equation, having eliminated m 3
by the continuity equation
. h x ~ h _ 9 2865.54 - 2725.3 _ ft lftfi9 w_
" h = " h h^h 2 ~ 2 2725.3 ^ 83.94 ~ °- 1062 kg/s
?ii 3 = m x + »i 2 = 2.1062 kg/s
Generation is from the entropy equation
5 gen = m 3 s 3 - ?Vi " *"2^2
£ gen = 2.1062 X 6.9918 - 2 X 7.3115 - 0.1062 X 0.2966 = 0.072 kW/K
Transient Process
For the transient process, which was described in Section 6.5, the second law for a control
volume, Eq. 9.2, can be written in the following form:
If this is integrated over the time interval t s we have
rt j
Therefore, for this period of time /, we can write the second law for the transient
process as
(m 2 s 2 - 7»iS])cv = 2 ~ 2 + f 2 %^ + Asm C 9 - 11 )
-'O c.v. J
312 H chapter nine second-Law Analysis for a Control volume
Since in this process the temperature is uniform throughout the control volume at any in-
stant of time, the integral on the right reduces to
and therefore the second law for the transient process can be written
EXAMPLE 9.6 Assume an air tank has 40 L of 100 kPa air at ambient temperature 17 C. The adiabatic
and reversible compressor is started so that it charges the tank up to a pressure of 10OO
kPa and then it shuts off. We want to know how hot the air in the tank gets and the total
amount of work required to fill the tank.
Solution
C.V. compressor and atr tank in Fig. 9.6.
Continuity Eq. 6,15
Energy Eq. 6.16:
Entropy Eq. 9. 12
Process
5i S; n
m 2 ~ mi = '"in
7»2« 2 - mi"l = 1^2 ~ 1^2 + m inMn
Adiabatic Process ideal ]S lm = 0,
=> m 2 s 2 = m l s l + Mii^in = (mi + m-Js x = =^>s 2 = s l
Constant s => Eq. 8.28 s\ 2 = s° Tl + R HP^P t )
s° n - 6.83521 + 0.287 In (10) = 7.49605 kJ/kg K
Interpolate in Table A.7 => T 2 = 555.7 K, u 2 = 401.49 kJ/kg
m , = p x y x IRT x = 100 X 0.04/(0.287 X 290) = 0.04806 kg
m 2 = P 2 V 2 IRT 2 - 1000 X 0.04/(0.287 X 555.7) = 0.2508 kg
=> m fa = 0.2027 kg
1^2 = mfaftjn + m x u x - m 2 u 2
= m in (290.43) + mi (207.19) - m 2 (401.49) = -31.9 kJ
Remark. The high final' temperature makes the assumption of zero heat transfer poor.
The charging process does not happen rapidly so there will be a heat transfer loss. We
need to know this to make a better approximation about the real process.
The Reversible Steady-State Process M 313
9.3 The Reversible Steady-State process
An expression can be derived for the work in a reversible, adiabatic, steady-state process
that is of great help in understanding its significant variables. We have noted that when a
steady-state process involves a single flow of fluid into and out of the control volume, the
first law, Eq. 6.13, can be written.
v 2 v 2
q + h i + ^- J t- gli = ft e + y + 8 Z e + w
and the second law, Eq. 9.8, is
m(s e - J/) = 2 %T + V
c.v, -*
Let us now consider two types of flow, a reversible adiabatic process and a re-
versible isothermal process.
If the process is reversible and adiabatic, the second-law equation reduces to
s e = Si
It follows from the property relation
Tds = dh-vdP
that
-*< = />
dP (9.13)
Substituting these relations into Eq. 6.13 and noting that q = 0, we have for the re-
versible, adiabatic process
V? - V 2
w = (ft, - K) + * + g(z f - z e )
fe V? - V 2
= _ J „ dP + -L^JL + g( z. - z e ) (9.14)
If, instead, the process is reversible and isothermal, the second law reduces to
or
S e ~ St) - -jjr- ~ q
and the property relation can be integrated to give
T(s e ~ sd = {h e ~ A/) - j'v dP (9.16)
Substituting Eqs. 9.15 and 9.16 into the first law, Eq. 6.13, gives us the same ex-
pression as for the reversible adiabatic process, Eq. 9.14. We further note that any other
reversible process can be constructed, in the limit, from a series of alternate adiabatic and
isothermal processes. Thus, we may conclude that Eq. 9.14 is valid for any reversible,
steady-state process without the restriction that it be either adiabatic or isothermal.
314 M CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
This expression has a wide range of application. If we consider a reversible steady-
state process in which the work is zero (such as flow through a nozzle) and the fluid is in-
compressible (u = constant), Eq. 9.14 can be integrated to give
v{P e - Pj) + + g(Z e - 50 = (9.17)
This equation, known as the Bernoulli equation (after Daniel Bernoulli), is very important
in fluid mechanics.
Equation 9.14 is also frequently applied to the large class of flow processes involving
work (such as turbines and compressors) in which changes in kinetic and potential energies
of the working fluid are small. The model process for these machines is then a reversible,
steady-state process with no change in kinetic or potential energy (and commonly, though
not necessarily, adiabatic as well). For this process Eq. 9. 14 reduces to the form
w=-j £ vdP (9-18)
From this result, we conclude that the shaft work associated with this type of
process is given by the area shown in the diagram of Fig. 9.7. It is important to note that
this result applies to a very special case—the area Jf v dP is not the same as the area
p £ v — anc i i s applicable only in entirely different circumstances. We also note that the
shaft work associated with this type of process is closely related to the specific volume of
the fluid during the process. To amplify this point further, consider the simple steam
power plant show in Fig. 9.8. Suppose that this is an ideal power plant with no pressure
FIGURE 9.8 Simple
steam power plant.
The Reversible Steady-State Process B 315
drop in the piping, the boiler, or the condenser. Thus, the pressure increase in the pump is
equal to the pressure decrease in the turbine. Neglecting kinetic and potential energy
changes, the work done in each of these processes is given by Eq. 9. 1 8. Since the pump
handles liquid, which has a very small specific volume compared to that of the vapor that
flows through the turbine, the power input to the pump is much less than the power output
of the turbine. The difference is the net power output of the power plant.
This same line of reasoning can be qualitatively applied to actual devices that involve
steady-state processes, even though the processes are not exactly reversible and adiabatic.
EXAMPLE 9.7 Calculate the work per kilogram to pump water isentropically from 100 kPa and 30°C to
5 MPa.
Control volume: Pump.
Inlet state: P u T t known; state fixed.
Exit state: P e known.
Process: Steady-state, isentropic.
Model: Steam tables.
Analysis
Since the process is steady, state, reversible, and adiabatic, and because changes in ki-
netic and potential energies can be neglected, we have
First law: h { = h e + w
Second law: s e — s t =
Solution
Since P e and s e are known, state e is fixed and therefore h e is known and w can be found
from the first law. However, the process is reversible and steady state, with negligible
changes in kinetic and potential energies, so that Eq. 9. 1 8 is also valid. Furthermore, since a
liquid is being pumped, the specific volume will change very little during the process.
From the steam tables, v { = 0.001 004 mVkg. Assuming that the specific volume
remains constant and using Eq. 9. 1 8, we have
dP = v{P 2 - P x ) = 0.001 004(5000 - 100) = 4.92 kJ/kg
As a final application of Eq. 9.14, we recall the reversible polytropic process for an
ideal gas, discussed in Section 8.11 for a control mass process. For the steady-state
process with no change in kinetic and potential energies, we have the relations
dP and Pv" - constant = C
= -\\dP=-c[
dP
pVn
316 M CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
If the process is isothermal, then n = 1 and the integral becomes
w =- j' v dP = -constant j' ^ = In ^ (9.20)
Note that the P-v and T-s diagrams of Fig. 8.17 are applicable to represent the slope of
polytropic processes in this case as well.
These evaluations of the integral
1
vdP
i
may also be used in conjunction with Eq. 9.14 for instances in which kinetic and potential
energy changes are not negligibly small.
9.4 Principle of the Increase of entropy
The principle of the increase of entropy for a control mass analysis was discussed in Sec-
tion 8.8. The same general conclusion is reached for a control volume analysis. To
demonstrate this, consider a control volume, Fig. 9.9, that exchanges both mass and heat
with its surroundings. At the point in the surroundings where the heat transfer occurs, the
temperature is T . From Eq. 9.5, the second law for this process is
dS t
dt
We recall that the first term represents the rate of change of entropy within the con-
trol volume, and the next terms the net entropy flow out of the control volume resulting
from the mass flow. Therefore, for the surroundings, we can write
dt
(9.2!)
Adding Eqs. 9.5 and 9.21, we have
dt
dS t ,
dt
dt
.. . _ "^1 Qc.Y. _ Qc.\.
(9.22)
FIGURE 9,9 Entropy
change for a control
volume plus surroundings.
Control volume
Surroundings
Efficiency M 317
Because £ cv . > when T > Tand £? c . v . < when T < T } it follows that
-S a = -# + -5 = = 2^*0 (9.23)
which can be termed the general statement of the principle of the increase of entropy.
When we use Eq. 9.23 to check any particular process for a possible violation of the
second law, it will be in connection with one of our mode! processes. For example, in a
steady-state process, as we consider the two terms in Eq. 9.23, we realize that, in accor-
dance with Eq. 9.6, the first term is zero. As a result, all the entropy change that is due to
irreversibilities in this type of process is observed in the surroundings. This term may then
be evaluated using Eq. 9.21. In contrast, for the transient process, there are both control
volume and surroundings terms to evaluate. Each term is integrated over the time t of the
process, as was done in Section 9.2. Thus, Eq. 9.23 is integrated to
A5 Det - A5 c , Vi + AS^ (9.24)
in which the control volume term is
AS™. = (rn 2 s 2 - miJi) tv . (9.25)
The term for the surroundings is, after applying Eq. 9.21 to the surroundings and integrating,
A^surr = r CV ' + 2 *» t s t ~ 2 W (9.26)
9.5 EFFICIENCY
In Chapter 7 we noted that the second law of thermodynamics led to the concept of. ther-
mal efficiency for a heat engine cycle, namely
^ Qh
where W ait is the net work of the cycle and Q H is the heat transfer from the high-tempera-
ture body.
In this chapter we have extended our consideration of the second law to control vol-
ume processes, which leads us now to consider the efficiency of a process. For example,
we might be interested in the efficiency of a turbine in a steam power plant or the com-
pressor in a gas turbine engine.
In general, we can say that to determine the efficiency of a machine in which a
process takes place, we compare the actual performance of the machine under given con-
ditions to the performance that would have been achieved in an ideal process. It is in the
definition of this ideal process that the second law becomes a major consideration. For ex-
ample, a steam turbine is intended to be an adiabatic machine. The only heat transfer is
the unavoidable heat transfer that takes place between the given turbine and the surround-
ings. We also note that for a given steam turbine operating in a steady-state manner, the
state of the steam entering the turbine and the exhaust pressure are fixed. Therefore, the
ideal process is a reversible adiabatic process, which is an isentropic process, between
the inlet state and the turbine exhaust pressure. In other words, the variables P is 7), and P e
are the design variables — the first two because the working fluid has been prepared in
318 a CHAPTERNlNE second-law ANALYSIS for a control volume
prior processes to be at these conditions at the turbine inlet, while the exit pressure is fixed
by the environment into which the turbine exhausts. Thus, the ideal turbine process would
go from state i to state e $i as shown in Fig. 9.10, whereas the real turbine process is irre-
versible, with the exhaust at a larger entropy at the real exit state e. Figure 9.10 shows typ-
ical states for a steam turbine, where state e s is in the two-phase region, and state e may be
as well, or may be in the superheated vapor region, depending on the extent of irre-
versibility of the real process. Denoting the work done in the real process / to e as w, and
that done in the ideal, isentrophic process from the same P„ T x to the same P e as w„ we de-
fine the efficiency of the turbine as
= w_ = h ( ~ h e (9,27)
The same definition applies to a gas turbine, where all states are in the gaseous phase.
Typical turbine efficiencies are 0.70-O.88, with large turbines usually having higher effi-
ciencies than small ones.
EXAMPLE 9.8 A steam turbine receives steam at a pressure of 1 MPa and a temperature of 300 C. The
steam leaves the turbine at a pressure of 15 kPa. The work output of the turbine is mea-
sured and is found to be 600 kJ/kg of steam flowing through the turbine. Determine the
efficiency of the turbine.
Control volume: Turbine.
Inlet state: P h T t known; state fixed.
Exit state: P e known.
Process: Steady-state.
Model: Steam tables.
Analysis
The efficiency of the turbine is given by Eq. 9.27:
turbine W,
EFFICIENCY M 319
Thus, to determine the turbine efficiency, we calculate the work that would be
done in an isentropic process between the given inlet state and final pressure. For this
isentropic process, we have
Continuity equation: mi = m e — in
First law: h t = h e> + w s _
Second law: s { = s e
Solution
From the steam tables, we get
h t = 3051.2 kJ/kg, s t = 7.1228 kJ/kg K
Therefore, atP e = 15kPa,
J« = s t = 7-1228 = 0.7548 + x^ 7.2536
x u = 0.8779
h„ = 225.9 + 0.8779(2373.1) - 2309.3 kJ/kg
From the first law for the isentropic process,
w, = h l -k a = 3051.2 - 2309.3 = 741.9 kJ/kg
But, since
w a = 600 kJ/kg
we find that
In connection with this example, it should be noted that to find the actual state e of the
steam exiting the turbine, we need to analyze the real process taking place. For the real
process
rhi = m e = m
A, = h s + w a
S e >Si
Therefore, from the first law for the real process, we have
K = 3051.2 - 600 = 2451.2 kJ/kg
2451.2 = 225.9 +^2373.1
x e = 0.9377
It is important to keep in mind that the turbine efficiency is defined in terms of an
ideal, isentropic process from P t and 7} to P e , even when one or more of these variables is
unknown. This is illustrated in the following example.
320 a
and 830 K. The turbine efficiency
is estimated to be 85%. What is the turbine inlet pressure?
Control volume: Turbine.
Inlet state: T, known.
Exit state: P e , T e known; state fixed.
Process: Steady state.
Model: Air tables, Table A.7.
Analysis
The efficiency, which is 85%, is given by Eq. 9.27,
__ w_
^turbine \V S
The first law for the real, irreversible process is '
1^ - h e + u<
For the ideal, isentropic process from P u T, to P„ the first law is
lU ~ h ejs +
and the second law is, from Eq. 8.28,
(Note that this equation is only for the ideal, isentropic process and not for the real
process, for which s e ~ s t > 0.)
Solution
From the air tables, Table A.7, at 1600 K, we get
h { = 1757.3 kJ/kg, si = 8.6905 kJ/kg K
From the air tables at 830 K (the actual turbine exit temperature),
ft ( = 855.3 kJ/kg
Therefore, from the first law for the real process.
w = 1757.3 - 855.3 = 902.0 kJ/kg
Using the definition of turbine efficiency,
w = 902.0/0.85 = 1061.2 kJ/kg
Efficiency M 321
From the first law for the isentropic process,
h es = 1757.3 - 1061.2 - 696.1 kJ/kg
so that, from the air tables,
T„ = 683.7 K, s% = 7.7148 kf/kg K
and the turbine inlet pressure is determined from
100
or
- 7.7148 - 8.6905 - 0.287 In
P t = 2995 kPa
As was discussed in Section 6.4, unless specifically noted to the contrary, we nor-
mally assume compressors or pumps to be adiabatic. In this case the fluid enters the com-
pressor at P- c and T h the condition at which it exists, and exits at the desired value of P ei
the reason for building the compressor. Thus, the ideal process between the given inlet
state i and the exit pressure would be an isentropic process between state / and state e st as
shown in Fig. 9.1 1 with a work input of w s . The real process, however, is irreversible, and
the fluid exits at the real state e with a larger entropy, and a larger amount of work input w
is required. The compressor (or pump, in the case of a liquid) efficiency is defined as
Typical compressor efficiencies are in the range of 0.70-0.88 with large compres-
sors usually having higher efficiencies than small ones.
If an effort is made to cool a gas during compression by using a water jacket or fins,
the ideal process is considered a reversible isothermal process, the work input for which is
w T , compared to the larger required work w for the real compressor. The efficiency of the
cooled compressor is then
"^cooled comp ~ (9.29)
322 M Chapter Nine Second-Law Analysis for a Control Volume
EXAMPLE 9.10 Air enters an automotive superhcharger at 100 kPa and 300 K and is compressed to 150
kPa. The efficiency is 70%. What is the required work input per kg of air? What is the
exit temperature?
Control volume: Supercharger (compressor).
Inlet state: P h T t known; state fixed.
Exit state: P e known.
Process: Steady-state.
Model: Ideal gas, 300 K specific heat, Table A. 5.
Analysis
The efficiency, which is 70%, is given by Eq. 9,28,
The first law for the real, irreversible process is
^ = K + w = CpoiT; - T e )
For the ideal, isentropic process from P f , T f to P e> the first law is
hi = h es + w„ w s = C p0 (Ti - T es )
and the second law is, from Eq. 8.32
Solution
Using and k from Table A.5, from the second law, we get
150
0.286
r « = 300 \jSJ = 336,9 K
From the first law for the isentropic process, we have
ft a = 1.004(300 ™ 336.9) = "37.1 kJ/kg
so that, from the efficiency, the real work input is
w = -37.1/0.70 = -53.0kJ/kg
and from the first law for the real process, the temperature at the supercharger exit is
T e = 300 -^|| = 352.8 K
Our final example is that of nozzle efficiency. As discussed in Section 6.4, the pur-
pose of a nozzle is to produce a high-velocity fluid stream, or in terms of energy, a large
kinetic energy, at the expense of the fluid pressure. The design variables are the same as
for a turbine, P„ T h and P e . A nozzle is usually assumed to be adiabatic, such that the
Some General Comments Regarding Entropy H 323
ideal process is an isentropic process from state / to state e s , as shown in Fig. 9.12, with
the production of velocity V^. The real process is irreversible, with the exit state e having
a larger entropy, and a smaller exit velocity V e . The nozzle efficiency is defined in terms
of the corresponding kinetic energies,
Vl/2
TW = ~ (9.30)
Nozzles are simple devices with no moving parts. As a result, nozzle efficiency may be
very high, typically 0.90-0.97.
In summary, to determine the efficiency of a device that carries out a process (rather
than a cycle), we compare the actual performance to what would be achieved in a related,
but well-defined ideal process.
9,6 Some General comments
Regarding Entropy
It is quite possible at this point that a student may have a good grasp of the material that
has been covered and yet may have only a vague understanding of the significance of en-
tropy. In fact, the question "What is entropy?" is frequently raised by students with the
implication that no one really knows! This section has been included in an attempt to give
insight into the qualitative and philosophical aspects of the concept of entropy, and to il-
lustrate the broad application of entropy to many different disciplines.
First, we recall that the concept of energy rises from the first law of thermodynam-
ics and the concept of entropy from the second law of thermodynamics. Actually, it is just
as difficult to answer the question "What is energy?" as it is to answer the question "What
is entropy?" However, since we regularly use the term energy and are able to relate this
term to phenomena that we observe every day, the word energy has a definite meaning to
us and thus serves as an effective vehicle for thought and communication. The word en-
tropy could serve in the same capacity. If, when we observed a highly irreversible process
(such as cooling coffee by placing an ice cube in it), we said, "That surely increases the
entropy," we would soon be as familiar with the word entropy as we are with the word en-
ergy. In many cases when we speak about a higher efficiency we are actually speaking
about accomplishing a given objective with a smaller total increase in entropy.
324 m CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
A second point to be made regarding entropy is that in statistical thermodynamics,
the property entropy is denned in terms of probability. Although this topic will not be ex-
amined in detail in this text, a few brief remarks regarding entropy and probability may
prove helpful. From this point of view, the net increase in entropy that occurs during an
irreversible process can be associated with a change of state from a less probable state to a
more probable state. For instance, to use a previous example, one is more likely to find
gas on both sides of ruptured membrane in Fig. 7.15 than to find a gas on one side and a
vacuum on the other. Thus, when the membrane ruptures, the direction of the process is
from a less probable state to a more probable state and associated with this process is an
increase in entropy. Similarly, the more probable state is that a cup of coffee will be at the
same temperature as its surroundings than at a higher (or lower) temperature. Therefore,
as the coffee cools as the result of a transferring of heat to the surroundings, there is a
change from a less probable to a more probable state, and associated with this is an in-
crease in entropy.
SUMMARY ' The secon d law of thermodynamics is extended to a general control volume with mass
flow rates in or out for steady and transient processes. The vast majority of common de-
vices and complete systems can be treated as nearly steady-state operation even if they
have slower transients as in a car engine or jet engine. Simplification of the entropy equa-
tion arises when applied to steady-state and single-flow devices like a turbine, nozzle,
compressor, or pump. The second law and Gibbs property relation are used to develop a
general expression for reversible shaft work in a single flow that is useful m understand-
ing the importance of the specific volume (or density) that influences the magnitude of the
work. For a flow with no shaft work, consideration of the reversible process also leads to
the derivation of the energy equation for an incompressible fluid as the Bernoulli equa-
tion This covers the flows of liquids such as water or hydraulic fluid as well as airflow at
low speeds, which can be considered incompressible for velocities less than a third of the
speed of sound.
Many actual devices operate with some irreversibility in the processes that occur, so
we also have entropy generation in the flow processes and the total entropy is always in-
creasing. The characterization of performance of actual devices can be done with a com-
parison to a corresponding ideal device, giving efficiency as the ratio of two energy terms
(work or kinetic energy). ,
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Apply the second law to more general control volumes.
• Analyze steady-state, single-flow devices such as turbines, nozzles, compressors,
and pumps, both reversible and irreversible.
• Know how to extend the second law to transient processes.
• Analyze complete systems as a whole or divide them into individual devices.
• Apply the second law to multiple-flow devices such as heat exchangers, mixing
chambers, and turbines with several inlets and outlets.
• Recognize when you have an incompressible flow where you can apply the
Bernoulli equation or the expression for reversible shaft work.
• Know when you can apply the Bernoulli equation and when you cannot.
• Know how to evaluate the shaft work for a polytropic process.
CONCEPT-STUDY GUIDE PROBLEMS M 325
* Know how to apply the analysis to an actual device using an efficiency and identify
the closest ideal approximation to the actual device.
* Know the difference between a cycle efficiency and a device efficiency.
* Have a sense of entropy as a measure of disorder or chaos.
Hey concepts d f , _ — _ — _ —
AND FORMULAS equation for entropy rate of change = + in - out + generation
Steady state single flow s e = s f + f ~ + s ea
j { i
Reversible shaft work w = - j v dP + ~ Vf - ^ \J 2 e + gZ t - gZ e
Reversible heat transfer q = jjds = h e - h, ~ j'v dP (from Gibbs relation)
Bernoulli equation r v(P, - i>) + I V? - I V* + gZ . ~ g z e ^0 ( v = constant)
Polytropic process work w = (P e v e - Pp t ) = (T e - 7)) n # 1
w = -P t v t In ^ = -£7) In ^ = i?r, In ^ „ = 1
The work is shaft work w = - f waEP and for ideal gas
■' j
Isentropic efficiencies 7? turbil!e - w T Jw Ts (Turbine work is out)
^compressor = w cJ w Cac (Compressor work is in)
— WpJw Pac (Pump work is in)
^nozzle ~ A ^ V^/A i Vf (Kinetic energy is out)
Concept-Study guide Problems
9.1 In a steady state single flow s is either constant or it
increases. Is that true?
9.2 Which process will make the previous statement
true?
9.3 A reversible adiabatic flow of liquid water in a
pump has increasing P. How about 77
9.4 A reversible adiabatic flow of air in a compressor
has increasing P. How about 77
9.5 An irreversible adiabatic flow of liquid water in a
pump has higher P. How about 77
9.6 A compressor receives R-134a at - 10°C, 200 kPa,
with an exit of 1200 kPa, 50°C. What can you say
about the process?
9.7 An air compressor has a significant heat transfer
out. See Example 9.4 for how high X becomes if
there is no heat transfer. Is that good, or should the
compressor be insulated?
9.8 A large condenser in a steam power plant dumps
15 MW at 45°C with an ambient at 25°C. What is
the entropy generation rate?
9.9 Air at 1000 kPa, 300 K, is throttled to 500 kPa.
What is the specific entropy generation?
9.10 Friction in a pipe flow causes a slight pressure de-
crease and a slight temperature increase. How does
that affect entropy?
9.11 A flow of water at some velocity out of a nozzle is
used to wash a car. The water then falls to the
326 M CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
ground. What happens to the water state in terms of
Viands?
9.12 The shaft work in a pump to increase the pressure
is small compared to the shaft work in an air com-
pressor for the same pressure increase. Why?
9.13 If the pressure in a flow is constant, can you have
shaft work?
9.14 A pump has a 2 kW motor. How much liquid water
at 15°C can I pump to 250 kPa from 100 kPa?
9.15 Liquid water is sprayed into the hot gases before
they enter the turbine section of a large gasturbine
power plant. It is claimed that the larger mass flow
rate produces more work. Is that the reason?
9.16 A polytropic flow process with n = might be
which device?
9.17 A steam turbine inlet is at 1200 kPa, 500°C. The exit
is at 200 kPa. What is the lowest possible exit tem-
perature? Which efficiency does that correspond to?
9.18 A steam turbine inlet is at 1200 kPa, 500°C. The exit
is at 200 kPa. What is the highest possible exit tem-
perature? Which efficiency does that correspond to?
9.19 A steam turbine inlet is at 1200 kPa, 500°C. The
exit is at 200 kPa, 275°C. What is the isentropic
efficiency?
9.20 The exit velocity of a nozzle is 500 m/s. If i7 noz2k =
0.88, what is the ideal exit velocity?
Homework problems
Steady-State Reversible Processes
Single Flow
9.21 A first stage in a turbine receives steam at 10 MPa
and 800°C, with an exit pressure of 800 kPa. As-
sume the stage is adiabatic and neglect kinetic ener-
gies. Find the exit temperature and the specific work.
9.22 Steam enters a turbine at 3 MPa and 450°C, ex-
pands in a reversible adiabatic process, and ex-
hausts at 10 kPa, Changes in kinetic and potential
energies between the inlet and the exit of the fur-
bine are small. The power output of the turbine is
800 kW. What is the mass flow rate of steam
through the turbine?
9.23 A reversible adiabatic compressor receives 0.05
kg/s saturated vapor R-22 at 200 kPa and has an
exit pressure of 800 kPa. Neglect kinetic energies
and find the exit temperature and the minimum
power needed to drive the unit.
9.24 In a heat pump that uses R-134a as the working
fluid, the R-134a enters the compressor at 150
kPa and -10°C at a rate of 0.1 kg/s. In the com-
pressor the R-134a is compressed in an adiabatic
process to 1 MPa. Calculate the power input re-
quired to the compressor, assuming the process to
be reversible.
9.25 A boiler section boils 3 kg/s saturated liquid water at
2000 kPa to saturated vapor in a reversible constant-
pressure process. Assume you do not know that
there is no work. Prove that there is no shaftwork
using the first and second laws of thermodynamics.
9.26 Consider the design of a nozzle in which nitrogen
gas flowing in a pipe at 500 kPa and 200°C at a ve-
locity of 10 m/s is to be expanded to produce a ve-
locity of 300 m/s. Determine the exit pressure and
cross-sectional area of the nozzle if the mass flow
rate is 0.15 kg/s and the expansion is reversible and
adiabatic.
9.27 Atmospheric air at -45°C and 60 kPa enters the
front diffuser of a jet engine, shown in Fig. P9.27,
with a velocity of 900 km/h and frontal area of
1 m 2 . After leaving the adiabatic diffuser, the ve-
locity is 20 m/s. Find the diffuser exit temperature
and the maximum pressure possible.
FIGURE P9.27
9.28 A compressor receives air at 290 K and 100 kPa
and a shaft work of 5,5 kW from a gasoline engine.
It should deliver a mass flow rate of 0,01 kg/s air to
a pipeline. Find the maximum possible exit pres-
sure of the compressor.
9.29 A compressor is surrounded by cold R-134a so it
works as an isothermal compressor. The inlet state
is 0°C, 100 kPa, and the exit state is saturated
vapor. Find the specific heat transfer and specific
work.
Homework Problems M 327
9.30 A difiuser is a steady-state device in which a fluid
flowing at high velocity is decelerated such that the
pressure increases in the process. Air at 120 kPa
and 30°C enters a diffuser with a velocity of 200
m/s and exits with a velocity of 20 m/s. Assuming
the process is reversible and adiabatic, what are the
exit pressure and temperature of the air?
9.31 The exit nozzle in a jet engine receives air at 1200 K
and 150 kPa with negligible kinetic energy. The
exit pressure is 80 kPa, and the process is re-
versible and adiabatic. Use constant heat capacity
at 300 K to find the exit velocity.
9.32 Do the previous problem using the air tables in
Table A.7.
9.33 An expander receives 0.5 kg/s air at 2000 kPa, 300 K
with an exit state of 400 kPa, 300 K. Assume the
process is reversible and isothermal. Find the rates
of heat transfer and work neglecting kinetic and
potential energy changes.
9.34 Air enters a turbine at 800 kPa and 1200 K and ex-
pands in a reversible adiabatic process to 100 kPa.
Calculate the exit temperature and the work output
per kilogram of air, using
a. The ideal gas tables (Table A.7).
b. Constant specific heat (value at 300 K from
Table A.5).
9.35 A flow of 2 kg/s saturated vapor R-22 at 500 kPa is
heated at constant pressure to 60°C. The heat is
supplied by a heat pump that receives heat from the
ambient at 300 K and work input shown in Fig.
P9.35. Assume everything is reversible and find the
rate of work input.
+-*»R-22
Qh
o
Ql
300 K
FIGURE P9.3S
FIGURE P9.36
at 600 K and 100 kPa. Heat transfer of 800 kW is
added from a 1000 K reservoir, 100 kW is rejected
at 350 Kj and some heat transfer takes place at
500 K. Find the heat transferred at 500 K and the
rate of work produced.
Multiple Devices and Cycles
9.37 Air at 100 kPa and 17°C is compressed to 400 kPa,
after which it is expanded through a nozzle back to
the atmosphere. The compressor and the nozzle are
both reversible and adiabatic, and kinetic energy in
and out of the compressor can be neglected. Find
the compressor work and its exit temperature, and
find the nozzle exit velocity.
9.38 A small turbine delivers 150 kW and is supplied
with steam at 700°C and 2 MPa. The exhaust
passes through a heat exchanger where the pressure
is 10 kPa and exits as saturated liquid. The turbine
is reversible and adiabatic. Find the specific turbine
work and the heat transfer in the heat exchanger.
9.39 One technique for operating a steam turbine in
part-load power output is to throttle the steam to a
lower pressure before it enters the turbine, as
shown in Fig. P9.39. The steamline conditions are
2 MPa and 400°C, and the turbine exhaust pressure
Steam line
Throttling
© valve
9.36 A reversible steady-state device receives a flow of
1 kg/s air at 400 K and 450 kPa, and the air leaves FIGURE P9.39
328 B CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
is fixed at 10 kPa. Assume the expansion inside the
turbine to be reversible and adiabatic.
a. Determine the full-load specific work output of
the turbine.
b. Find the pressure the steam must be throttled to
for 80% of full- load output.
c. Show both processes in a T-s diagram.
9.40 Two flows of air are both at 200 kPa; one has 1
kg/s at 400 K, and the other has 2 kg/s at 290 K.
The two lines exchange energy through a number
of ideal heat engines, taking energy from the hot
line and rejecting it to the colder line. The two
flows then leave at the same temperature. Assume
the whole setup is reversible and find the exit
temperature and the total power out of the heat
engines.
9.41 A certain industrial process requires a steady sup-
ply of saturated vapor steam at 200 kPa, at a rate of
0.5 kg/s. Also required is a steady supply of com-
pressed air at 500 kPa, at a rate of 0.1 kg/s. Both
are to be supplied by the process shown in Fig.
P9.41. Steam is expanded in a turbine to supply the
power needed to drive the air compressor, and the
exhaust steam exits the turbine at the desired state.
Air into the compressor is at the ambient condi-
tions, 100 kPa and 20°C. Give the required steam
inlet pressure and temperature, assuming that both
the turbine and the compressor are reversible and
adiabatic.
w T =~\v c
Process
steam supply
FIGURE P9.41
Compressed
air supply
p 4 = F, = 20 MPa
p 2 = P 3 = 20 kPa
r, =700^
73 = 40*0
bine and the pump processes are reversible and
adiabatic. Neglecting any changes in kinetic and
potential energies, calculate
a. The specific turbine work output and the turbine
exit state.
b. The pump work input and enthalpy at the pump
exit state.
c. The thermal efficiency of the cycle.
9.43 A turbocharger boosts the inlet air pressure to an
automobile engine. It consists of an exhaust gas-,
driven turbine directly connected to an air com-
pressor, as shown in Fig. P9.43. For a certain
engine load, the conditions are given in the figure.
Assume that both the turbine and the compressor
are reversible and adiabatic, having also the same
Engine
j power
out
9.42 Consider a steam turbine power plant operating
near critical pressure, as shown in Fig. P9.42. As a
first approximation, it may be assumed that the tur-
Inlet air
F 1 = 100 kPa
= 30°C
m = 0.1 kg/s
FIGURE P9.43
Exhaust
P 4 = 100 kPa
mass flow rate. Calculate the turbine exit tempera-
ture and power output. Find also the compressor
exit pressure and temperature.
9.44 A two-stage compressor having an interstage
cooler takes in air, 300 K and 100 kPa, and com-
presses it to 2 MPa, as shown in Fig. P9.44. The
cooler then cools the air to 340 K, after which it
enters the second stage, which has an exit pressure
of 15.74 MPa. Both stages are adiabatic and re-
versible. Find the specific heat transfer in the inter-
cooler and the total specific work. Compare this to
the work required with no intercooler.
FIGURE P9.44
9.45 A heat-powered portable air compressor consists of
three components: (a) an adiabatic compressor; (b) a
constant-pressure heater (heat supplied from an out-
side source); and (c) an adiabatic turbine (see Fig.
P9.45). Ambient air enters the compressor at 100
kPa and 300 K and is compressed to 600 kPa. All of
the power from the turbine goes into the compressor,
and the turbine exhaust is the supply of compressed
air. If this pressure is required to be 200 kPa, what
must the temperature be at the exit of the heater?
Homework problems 3 329
9.46 A certain industrial process requires a steady 0.5
■kg/s supply of compressed air at 500 kPa, at a max-
imum temperature of 30°C, as shown in Fig. P9.46.
This air is to be supplied by installing a compressor
and aftercooler. Local ambient conditions are 100
kPa and 20°C. Using a reversible compressor, de-
termine the power required to drive the compressor
and the rate of heat rejection in the aftercooler.
Ambient air
FIGURE P9.46
Steady-State Irreversible Processes
9.47 Analyze the steam turbine described in Problem
6.78. Is it possible?
9.48 Carbon dioxide at 300 K and 200 kPa flows through
a steady device where it is heated to 500 K by a
600 K reservoir in a constant-pressure process. Find
the specific work, specific heat transfer, and specific
entropy generation.
9.49 Consider the steam turbine in Example 6.6. Is this
a reversible process?
9.50 The throttle process described in Example 6.5 is an
irreversible process. Find the entropy generation
per kg of ammonia in the throttling process.
9.51 A geothermal supply of hot water at 500 kPa and
150°C is fed to an insulated flash evaporator at the
rate of 1.5 kg/s, shown in Fig. P9.51. A stream of
330 B
saturated liquid at 200 RPa is drained from the bot-
tom of the chamber, and a stream of saturated
vapor at 200 kPa is drawn from the top and fed to a
turbine. Find the rate of entropy generation m the
flash evaporator.
9 52 Two flowstreams of water, one of saturated vapor
at 6 MPa, and the other at 0.6 MPa and 600 C,
mix adiabatically in a steady flow to produce a sin-
gle flow out at 0.6 MPa and 400X. Find the total
entropy generation for this process.
9 53 A condenser in a power plant receives 5 kg/s steam
at 1 5 kPa with a quality of 90% and rejects the heat
to cooling water with an average temperature of
\TC. Find the power given to the cooling water m
this constant-pressure process, shown in Fig.
P9.53, and the total rate of entropy generation
when saturated liquid exits the condenser.
Steam
Cooling
water
40° C
FIGURE P9.54
9 55 A heat exchanger that follows a compressor re-
ceives 0. 1 kg/s air at 1 000 kPa and 500 K and cools
it in a constant-pressure process to 320 K. The heat
is absorbed by ambient air at 300 K. Find the total
rate of entropy generation.
9 56 Air at 327°C and 400 kPa with a volume flow 1 m 3 /s
runs through an adiabatic turbine with exhaust pres-
sure of 100 kPa. Neglect kinetic energies and use -
constant specific heats. Find the lowest and highest
possible exit temperature. For each case find also
the rate of work and the rate of entropy generation.
9.57 In a heat-driven refrigerator with ammonia as the
working fluid, a turbine with inlet conditions of 2.0
MPa and 70°C is used to drive a compressor with
inlet saturated vapor at -20°C. The exhausts, both
at 1.2 MPa, are then mixed together, shown m Fig.
P9.57. The ratio of the mass flow rate to the turbine
to the total exit flow was measured to be 0.62. Can
this be true?
FIGURE P9.53
9 54 A mixing chamber receives 5 kg/min of ammonia
as saturated liquid at -20°C from one line and am-
monia at 40°C and 250 kPa from another Ime
through a valve. The chamber also receives 325
kJ/min of energy as heat transferred from a 40°C
reservoir, shown in Fig. P9.54. This should pro-
duce saturated ammonia vapor at -20°C m the exit
line. What is the mass flow rate in the second line,
and* what is the total entropy generation in the
process?
FIGURE F9.57
9 58 Two flows of air are both at 200 kPa; one has 1
kg/s at 400 K, and the other has 2 kg/s at 290 K.
The two flows are mixed together in an insulated
box to produce a single exit flow at 200 kPa. Find
the exit temperature and the total rate of entropy
generation.
Homework Problems H 331
9.59 One type of feedwater heater for preheating the
water before entering a boiler operates on the prin-
ciple of mixing the water with steam that has been
bled from the turbine. For the states as shown in
Fig. P9.59, calculate the rate of net entropy in-
crease for the process, assuming the process to be
steady flow and adiabatic.
7 2 = 200°C
Feedwater
heater
®
P 3 = 1 MPa
r 3 = 160°C
m 3 = 4 kg/s
FIGURE P9.59
9.60 A supply of 5 kg/s ammonia at 500 kPa and 20°C
is needed. Two sources are available: One is satu-
rated liquid at 20°C ) and the other is at 500 kPa and
140°C. Flows from the two sources are fed through
valves to an insulated mixing chamber, which then
produces the desired output state. Find the two
source mass flow rates and the total rate of entropy
generation by this setup.
9.61 A counterflowing heat exchanger has one line with
2 kg/s air at 125 kPa and 1000 K entering, and the
air is leaving at 100 kPa and 400 K. The other line
has 0.5 kg/s water coming in at 200 kPa and 20°C
and leaving at 200 fcPa, What is the exit temperature
of the water and the total rate of entropy generation?
Air
am
am
CD
1H0O
FIGURE P9.61
9.62 A coflowing (same direction) as shown in Fig.
P9.62, heat exchanger has one line with 0.25 kg/s
FIGURE P9.62
oxygen at 17°C and 200 kPa entering, and the other
line has 0.6 kg/s nitrogen at 150 kPa and 500 K en-
tering. The heat exchanger is long enough so that
the two flows exit at the same temperature. Use
constant heat capacities and find the exit tempera-
ture and the total rate of entropy generation.
Transient Processes
9.63 Calculate the specific entropy generated in the fill-
ing process given in Example 6.1 1.
9.64 Calculate the total entropy generated in the filling
process given in Example 6.12.
9.65 An initially empty 0.1 m 3 canister is filled with
R-12 from a line flowing saturated liquid at -5°C.
This is done quickly such that the process is adia-
batic. Find the final mass, and determine liquid and
vapor volumes, if any, in the canister. Is the
process reversible?
9.66 Aim 3 rigid tank contains 100 kg of R-22 at ambi-
ent temperature, 15°C. A valve on top of the tank is
opened, and saturated vapor is throttled to ambient
pressure, 100 kPa, and flows to a collector system,
shown in Fig. P9.66. During the process, the tem-
perature inside the tank remains at 15°C. The valve
is closed when no more liquid remains inside. Cal-
culate the heat transfer to the tank and the total en-
tropy generation in the process.
FIGURE P9.66
332
9.67 Air in a tank is at 300 kPa and 400 K with a vol-
ume of 2 m 3 . A valve on the tank is opened to let
some air escape to the ambient surroundings to
leave a final pressure inside of 200 kPa. Find the
final temperature and mass assuming a reversible
adiabatic process for the air remaining inside the
tank.
9.68 An empty canister of 0.002 m 3 is filled with
R- 1 34a from a line flowing saturated liquid R- 1 34a
at 0°C. The filling is done quickly so it is adiabatic.
Find the final mass in the canister and the total en-
tropy generation.
9 69 An old abandoned salt mine, 100 000 m 3 in vol-
ume, contains air at 290 K and 100 kPa. The mine
is used for energy storage so the local power plant
pumps it up to 2.1 MPa using outside air at 290 K
and 100 kPa. Assume the pump is ideal and the
process is adiabatic. Find the final mass and tem-
perature of the air and the required pump work.
9.70 Air in a tank is at 300 kPa and 400 K with a vol-
ume of 2 m 3 . A valve on the tank is opened to let
some air escape to the ambient surroundings to
leave a final pressure inside of 200 kPa. At the
same time the tank is heated so the air remaining
has a constant temperature. What is the mass aver-
age value (Table A.7 reference) of the s leaving,
assuming this is an internally reversible process?
9.71 An insulated 2 m 3 tank is to be charged with
R-134a from a line flowing the refrigerant at 3
MPa. The tank is initially evacuated, and the valve
is closed when the pressure inside the tank reaches
3 MPa. The line is supplied by an insulated com-
pressor that takes in R-134a at 5°C, with a quality
of 96.5%, and compresses it to 3 MPa in a re-
versible process. Calculate the total work input to
the compressor to charge the tank.
9 72 A 0.2 m 3 initially empty container is filled with
water from a line at 500 kPa and 200°C until there
is no more flow. Assume the process is adiabatic
and find the final mass, final temperature, and total
entropy generation.
9.73 Air from a line at 12 MPa and 15°C flows into a
500 L rigid tank that initially contained air at ambi-
ent conditions, 100 kPa and 15°C. The process oc-
curs rapidly and is essentially adiabatic. The valve
is closed when the pressure inside reaches some
value, F 2 . The tank eventually cools to room tem-
perature, at which time the pressure inside is
5 MPa. What is the pressure P 2 ? What is the net
entropy change for the overall process?
9.74 An initially empty canister with a volume of 0.2 m 3
is filled with carbon dioxide from a line at 1000
kPa and 500 K. Assume the process is adiabatic
and the flow continues until it stops by itself. Use
constant heat capacity to solve for the final mass
and temperature of the carbon dioxide in the canis-
ter and the total entropy generation by the process.
9.75 A cook filled a pressure cooker with 3 kg water at
20°C and a small amount of air and forgot about it.
The pressure cooker has a vent valve so if P > 200
kPa, steam escapes to maintain the pressure at 200
kPa. How much entropy was generated in the throt-
tling of the steam through the vent to 100 kPa
when half the original mass has escaped?
Reversible Shaft Work, Bernoulli Equation
9.76 A large storage tank contains saturated liquid nitro-
gen at ambient pressure, 100 kPa; it is to be
pumped to 500 kPa and fed to a pipeline at the rate
of 0.5 kg/s. How much power input is required for
the pump, assuming it to be reversible?
9.77 Liquid water at ambient conditions, 100 kPa and
25°C, enters a pump at the rate of 0.5 kg/s. Power
input' to the pump is 3 kW. Assuming the pump
process to be reversible, determine the pump exit
pressure and temperature.
9.78 A small dam has a 0.5-m-diameter pipe carrying
liquid water at 150 kPa and 20°C with a flow rate of
2000 kg/s. The pipe runs to the bottom of the dam
15 m lower into a turbine with pipe diameter 0.35
m, shown in Fig. P9.78. Assume no friction or heat
transfer in the pipe and find the pressure of the tur-
bine inlet. If the turbine exhausts to 100 kPa with
negligible kinetic energy, what is the rate of work?
/i 2 = /i 3 = m
FIGURE P9.78
Homework Problems H 333
9.79 A firefighter on a ladder 25 m above ground should
be able to spray water an additional 10 m up with
the hose nozzle of exit diameter 2.5 cm. Assume a
water pump on the ground and a reversible flow
(hose, nozzle included) and find the minimum re-
quired power.
9.80 A small pump is driven by a 2 kW motor with liq-
uid water at 150 kPa and 10°C entering. Find the
maximum water flow rate you can get with an exit
pressure of 1 MPa and negligible kinetic energies.
The exit flow goes through a small hole in a spray
nozzle out to the atmosphere at 100 kPa, shown in
Fig. P9.80. Find the spray velocity.
Nozzle
FIGURE P9.80
9.81 A garden water hose has liquid water at 200 kPa
and 15°C. How high a velocity can be generated in
a small ideal nozzle? If you direct the water spray
straight up how high will it go?
9.82 Saturated R-134a at -10°C is pumped/compressed
to a pressure of 1 .0 MPa at the rate of 0.5 kg/s in a
reversible adiabatic process. Calculate the power
required and the exit temperature for the two cases
of inlet state of the R-134a:
a. Quality of 100% b. Quality of 0%.
9.83 A small water pump on ground level has an inlet
pipe down into a well at a depth H with the water
at 100 kPa and 15°C. The pump delivers water at
400 kPa to a building. The absolute pressure of the
water must be at least twice the saturation pressure
to avoid cavitation. What is the maximum depth
this setup w r ill allow?
9.84 A small pump takes in water at 20°C and 100 kPa
and pumps it to 2.5 MPa at a flow rate of 100
kg/min. Find the required pump power input.
9.85 A pump/compressor pumps a substance from 100
' kPa and 10°C to 1 MPa in a reversible adiabatic
process. The exit pipe has a small crack, so that a
small amount leaks to the atmosphere at 100 kPa.
If the substance is (a) water, (b) R-12, find the tem-
perature after compression and the temperature of
the leak flow as it enters the atmosphere, neglect-
ing kinetic energies.
9.86 Atmospheric air at 100 kPa and 17°C blows at 60
km/h toward the side of a building. Assuming the
air is nearly incompressible, find the pressure and
the temperature at the stagnation (zero-velocity)
point on the wall.
9.87 You drive on the highway at 1 20 km/h on a day
with 17°C, 100 kPa atmosphere. When you put
your hand out of the window flat against the wind
you feel the force from the ah* stagnating (i.e., it
comes to relative zero velocity on your skin). As-
sume that the air is nearly incompressible and find
the air temperature and pressure right on your
hand.
9.88 An airflow at 100 kPa, 290 K, and 200 m/s is di-
rected toward a wall. At the wall the flow stagnates
(comes to zero velocity) without any heat transfer,
as shown in Fig. P9. 88. Find the stagnation pressure
(a) assuming incompressible flow, (b) assuming an
adiabatic compression. Hint'. T comes from the en-
ergy equation.
5
r
Zero
"velocity
FIGURE P9.88
9.89 Calculate the air temperahire and pressure at the
stagnation point right in front of a meteorite enter-
ing the atmosphere (— 50°C, 50 kPa) with a veloc-
ity of 2000 m/s. Do this assuming air is
incompressible at the given state and repeat for air
being a compressible substance going through an
adiabatic compression.
334 ffl Chapter Nine second-Law analysis for a Control Volume
9.90 Helium gas enters a steady-flow expander at 800
kPa and 300°C and exits at 120 kPa. The mass flow
rate is 0.2 kg/s, and the expansion process can be
considered as a reversible polytropic process with
exponent n = 1.3. Calculate the power output of
the expander.
9.91 Air at 100 kPa and 300 K flows through a device at
steady state with the exit at 1000 K during which it
went through a polytropic process with n = 1.3.
Find the exit pressure, the specific work, and heat
transfer.
9.92 A 4 kg/s flow of ammonia goes through a device in
a polytropic process with an inlet state of 150 kPa,
-20°C and an exit state of 400 kPa, 80°C. Find the
polytropic exponent n> the specific work, and the
specific heat transfer.
9.93 Carbon dioxide flows through a device entering at
300 K and 200 kPa and leaving at 500 K. The
process is steady-state polytropic with n = 3.8, and
heat transfer comes from a 600 K source. Find the
specific work, specific heat transfer, and specific
entropy generation due to this process.
9.94 An expansion in a gas turbine can be approxi-
mated with a polytropic process with exponent
n = 1.25. The inlet air is at 1200 K, 800 kPa, and
the exit pressure is 125 kPa with a mass flow rate
of 0.75 kg/s. Find the turbine heat transfer and
power output.
Device Efficiency
9.95 Find the isentropic efficiency of the R-134a com-
pressor in Example 6.10, assuming the ideal com-
pressor is adiabatic.
9.96 A compressor is used to bring saturated water
vapor at 1 MPa up to 17.5 MPa, where the actual
exit temperature is 650X. Find the isentropic
compressor efficiency and the entropy generation.
9.97 Liquid water enters a pump at 15°C and 100 kPa
and exits at a pressure of 5 MPa. If the isentropic
efficiency of the pump is 75%, determine the en-
thalpy (steam table reference) of the water at the
pump exit.
9.98 A centrifugal compressor takes in ambient air at
100 kPa and 15°C and discharges it at 450 kPa.
The compressor has an isentropic efficiency of
80%. What is your best estimate for the discharge
temperature?
9.99 An emergency drain pump, shown in Fig. P9.99,
should be able to pump 0.1 mVs of liquid water at
15°C, 10 m vertically up delivering it with a ve-
locity of 20 m/s. It is estimated that the pump,
pipe, and nozzle have a combined isentropic effi-
ciency expressed for the pump as 60%. How
much power is needed to drive the pump?
Nozzle
10m
Drain pump
FIGURE P9.99
9.100 A pump receives water at 100 kPa and 15°C and
has a power input of 1.5 kW. The pump has an
isentropic efficiency of 75%, and it should flow
1.2 kg/s delivered at 30 m/s exit velocity. How
high an exit pressure can the pump produce?
9.101 A small air turbine with an isentropic efficiency
of 80% should produce 270 kJ/kg of work. The
inlet temperature is 1000 K, and the turbine ex-
hausts to the atmosphere. Find the required inlet
pressure and the exhaust temperature.
9.102 Repeat Problem 9.42 assuming the turbine and the
pump each have an isentropic efficiency of 85%.
9.103 Repeat Problem 9.41 assuming the steam turbine
and the air compressor each have an isentropic ef-
ficiency of 80%.
9.104 Steam enters a turbine at 300°C 5 600 kPa, and ex-
hausts as saturated vapor at 20 kPa. What is the
isentropic efficiency?
9.105 A turbine receives air at 1500 K and 1000 kPa
and expands it to 100 kPa. The turbine has an
isentropic efficiency of 85%. Find the actual tur-
bine exit air temperature and the specific entropy
increase in the actual turbine.
9.106 The small turbine in Problem 9.38 was ideal. As-
sume instead that the isentropic turbine efficiency
is 88%. Find the actual specific turbine work and
the entropy generated in the turbine.
HOMEWORK PROBLEMS B 335
9.107
9.108
9.109
9.110
9.111
9.112
9.113
9.114
9.115
9.116
Air enters an insulated turbine at 50°C and exits
the turbine at -30°C and 100 kPa. The isentropic
turbine efficiency is 70%, and the inlet volumetric
flow rate is 20 L/s. What is the turbine inlet pres-
sure and the turbine power output?
Carbon dioxide, C0 2) enters an adiabatic com-
pressor at 100 kPa and 300 K and exits at 1000
kPa and 520 K. Find the compressor efficiency
and the entropy generation for the. process.
Air enters an insulated compressor at ambient
conditions, 100 kPa and 20°C, at the rate of 0.1
kg/s and exits at 200°C. The isentropic efficiency
of the compressor is 70%. Assume the ideal and
actual compressor have the same exit pressure.
What is the exit pressure? How much power is re-
quired to drive the unit?
Assume an actual compressor has the same exit
pressure and specific heat transfer as the ideal
isothermal compressor in Problem 9.29 with
an isothermal efficiency of 80%. Find the spe-
cific work and exit temperature for the actual
compressor.
A water-cooled air compressor takes air in at
20°C and 90 kPa and compresses it to 500 kPa.
The isothermal efficiency is 80%, and the actual
compressor has the same heat transfer as the ideal
one. Find the specific compressor work and the
exit temperature.
A nozzle in a high-pressure liquid water sprayer
has an area of 0.5 cm 2 . It receives water at 250
kPa, 20°C, and the exit pressure is 100 kPa. Ne-
glect the inlet kinetic energy and assume a noz-
zle isentropic efficiency of 85%. Find the ideal
nozzle exit velocity and the actual nozzle mass
flow rate.
A nozzle should produce a flow of air with 200
m/s at 20°C and 100 kPa. It is estimated that the '
nozzle has an isentropic efficiency of 92%. What
nozzle inlet pressure and temperature are required
assuming the inlet kinetic energy is negligible?
Redo Problem 9.79 if the water pump has an isen-
tropic efficiency of 85% including hose and nozzle.
Find the isentropic efficiency of the nozzle in
Example 6.4.
Air flows into an insulated nozzle at 1 MPa and
1200 K with 15 m/s and a mass flow rate of 2
kg/s. It expands to 650 kPa, and the exit tempera-
ture is 1 100 K. Find the exit velocity and the noz-
zle efficiency.
Review Problems
9.117 A coflowing heat exchanger has one line with 2
kg/s saturated water vapor at 100 kPa entering. The
other line is 1 kg/s air at 200 kPa, 1200 K. The heat
exchanger is very long so the two flows exit at the
same temperature. Find the exit temperature by trial
and error. Calculate the rate of entropy generation.
9.118 A vortex tube has an air inlet flow at 20°C, 200
kPa, and two exit flows of 100 kPa: one at 0°C
and the other at 40°C. The tube, shown in Fig.
P9. 1 1 8, has no external heat transfer and no work,
and all the flows are steady and have negligible
kinetic energy. Find the fraction of the inlet flow
that comes out at 0°C. Is this setup possible?
FIGURE P9.118
9.119 An initially empty spring-loaded piston/cylinder
requires 100 kPa to float the piston. A compressor
with a line and valve now charges the cylinder
with water to a final pressure of 1,4 MPa at which
point the volume is 0.6 m 3 , state 2. The inlet con-
dition to the reversible adiabatic compressor is sat-
urated vapor at 100 kPa. After charging, the valve
is closed, and the water eventually cools to room
temperature, 20°C, state 3. Find the final mass of
water, the piston work from 1 to 2, the required
compressor work, and the final pressure, P 3 .
9.120 In a heat-powered refrigerator, a turbine is used to
drive the compressor using the same working
fluid. Consider the combination shown in Fig.
P9.120 where the turbine produces just enough
power to drive the compressor and the two exit
flows are mixed together. List any assumptions
336 B CHAFTERNINB SECOND-LAW Analysis FOR A CONTROL VOLUME
j> 2 = p 4 =j» 6 =1.0MPa
To condenser
FIGURE P9.120
made and find the ratio of mass flow rates m^mi
and T 5 (x s if in two-phase region) if the turbine
and the compressor are reversible and adiabatic.
9.121 A stream of ammonia enters a steady flow device
at 100 kPa and 50°C, at the rate of 1 kg/s. Two
streams exit the device at equal mass flow rates;
one is at 200 kPa and 50°C, and the other is a sat-
urated liquid at 10°C. It is claimed that the device
operates in a room at 25°C on an electrical power
input of 250 fcW. Is this possible?
9.122 A frictionless piston/cylinder is loaded with a lin-
ear spring, spring constant 100 kN/m, and the pis-
ton cross-sectional area is 0.1 m 2 . The cylinder
initial volume of 20 L contains air at 200 kPa and
ambient temperature, 10°C. The cylinder has a set
of stops that prevents its volume from exceeding
50 L. A valve connects to a line flowing air at 800
kPa, 50°C, as shown in Fig. P9.122. The valve is
Air line
u
FIGURE P9.122
now opened, allowing air to flow in until the
cylinder pressure reaches 800 kPa, at which point
the temperature inside the cylinder is 80°C. The
valve is then closed and the process ends.
a. Is the piston at the stops at the final state?
b. Taking the inside of the cylinder as a control
volume, calculate the heat transfer during the
process.
c. Calculate the net entropy change for this
process.
9.123 An insulated piston/cylinder contains R-22 at
20°C, 85% quality, at a cylinder volume of 50 L.
A valve at the closed end of the cylinder is con-
nected to a line flowing R-22 at 2 MPa, 60°G. The
valve is now opened, allowing R-22 to flow in,
and at the same time the external force on the pis-
ton is decreased, and the piston moves. When the
valve is closed, the cylinder contents are at 800
kPa, 20°C J and a positive work of 50 kJ has been
done against the external force. What is the final
volume of the cylinder? Does this process violate
the second law of thermodynamics?
9.124 Air enters an insulated turbine at 50°C and exits
the turbine at -30°C, 100 kPa. The isentropic tur-
bine efficiency is 70%, and the inlet volumetric
flow rate is 20 L/s. What is the turbine inlet pres-
sure and the turbine power output? ,
9.125 A certain industrial process requires a steady 0.5
kg/s supply of compressed air at 500 kPa, at a
maximum temperature of 30°C, as shown in Fig.
P9.46. This air is to be supplied by installing a
Homework problems ffl 337
compressor and aftercooler. Local ambient condi-
tions are 100 kPa, 20°C. Using an isentropic com-
pressor efficiency of 80%, determine the power
required to drive the compressor and the rate of
heat rejection in the aftercooler.
9.126 Consider the scheme shown in Fig. P9. 126 for pro-
ducing fresh water from salt water. The conditions
are as shown in the figure. Assume that the proper-
ties of salt water are the same as for pure water,
and that the pump is reversible and adiabatic.
a. Determine the ratio (th^m^ the fraction of salt
water purified.
b. Determine the input quantities, w p and q H .
c. Make a second law analysis of the overall
system.
Heat source
T H => 200°
© r 4 = i5u°c
Liquid t 1
seawater
7t = 15°C
^1 = 100 kPa
FIGURE P9.126
P B = 100 kPa
sat. liquid
saltwater
(concentrated)
T 7 = 35°C
pure liquid
H 2 out
percharger (compressor) has an isentropic effi-
ciency of 75%, and uses 20 kW of power input.
Assume that the ideal and actual compressor have
the same exit pressure. Find the ideal specific
work and verify that the exit pressure is 175 kPa.
Find the percent increase in air density entering
the engine due to the supercharger and the en-
tropy generation.
9.128 A jet-ejector pump, shown schematically in Fig.
P9.128, is a device in which a low-pressure (sec-
ondary) fluid is compressed by entrainment in a
high-velocity (primary) fluid stream. The com-
pression results from the deceleration in a dif-
fuser. For purposes of analysis, this can be
considered as equivalent to the turbine-compres-
sor unit shown in Fig. P9.120 with the states 1, 3,
and 5 corresponding to those in Fig. P9.128. Con-
sider a stream jet-pump with state 1 as saturated
vapor at 35 kPa; state 3 is 300 kPa, 150°C; and
the discharge pressure, P 5 , is 100 kPa.
a. Calculate the ideal mass flow ratio, m u m 3 .
b. The efficiency of a jet pump is defined as
n
fjet pump
for the same inlet conditions and discharge pres-
sure. Determine the discharge temperature of the
jet pump if its efficiency is 10%.
High-pressure
primary fluid
D T ^
^ Low-velocity
discharge
Diffuser
Secondary
fluid
FIGURE P9.128
9.127 Supercharging of an engine is used to increase the
inlet air density so that more fuel can be added,
the result of which is an increased power output.
Assume that ambient air, 100 kPa and 27°C, en-
ters the supercharger at a rate of 250 L/s. The su-
9.129 A rigid steel bottle, with V = 0.25 m 3 , contains air
at 100 kPa and 300 K. The bottle is now charged
with air from a line at 260 K and 6 MPa to a bot-
tle pressure of 5 MPa, state 2, and the valve is
closed. Assume that the process is adiabatic and
338 H CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
that the charge always is uniform. In storage, the
bottle slowly returns to room temperature at 300
K, state 3. Find the final mass, the temperature T 2 ,
the final pressure P 3 , the heat transfer x Q 3l and the
total entropy generation.
9.130 A horizontal, insulated cylinder has a frictionless
piston held against stops by an external force of
500 kN, as shown in Fig. P9.130. The piston
cross-sectional area is 0.5 m 2 , and the initial vol-
ume is 0.25 m 3 . Argon gas in the cylinder is at
200 kPa and 100°C. A valve is now opened to a
line flowing argon at 1.2 MPa and 200°C, and gas
flows in until the cylinder pressure just balances
the external force, at which point the valve is
closed. Use constant head capacity to verify that
the final temperature is 645 K and find the total
entropy generation.
Ar line
Ar
5
FIGURE P9.130
9.131 A rigid 1,0 m 3 tank contains water initially at
120°C, with 50% liquid and 50% vapor, by vol-
ume. A pressure-relief valve on the top of the tank
is set to 1.0 MPa (the tank pressure cannot exceed
1 .0 MPa— water will be discharged instead). Heat
is now transferred to the tank from a 200°C heat
source until the tank contains saturated vapor at
1 .0 MPa. Calculate the heat transfer to the tank
and show that this process does not violate the
second law.
9.132 A certain industrial process requires a steady 0,5
kg/s of air at 200 m/s, at the condition of 150 kPa,
300 K, as shown in Fig. P9.132. This air is to be
the exhaust from a specially designed turbine
whose inlet pressure is 400 kPa. The turbine
process may be assumed to be reversible and
polytropic, with polytropic exponent n = 1.20.
a. What is the turbine inlet temperature?
b. What are the power output and heat transfer
rate for the turbine?
c. Calculate the rate of net entropy increase, if
the heat transfer comes from a source at a tem-
perature 100°C higher than the turbine inlet
temperature.
To process
m, P 2 , T 2 , V 2
FIGURE F9.132
9,133 Assume both the compressor and the nozzle in
Problem 9.37 have an isentropic efficiency of
90%, the rest being unchanged. Find the actual
compressor work and its exit temperature and find
the actual nozzle exit velocity.
English unit problems
Concept Problems
9.134E A compressor receives R-134a at 20 F, 30 psia,
with an exit of 200 psia, * = 1. What can you
say about the process?
9.135E A large condenser in a steam power plant dumps
15 000 Btu/s at 1 15 F with an ambient at 77 F.
What is the entropy generation rate?
9.13 6E Air at 150 psia, 540 R, is throttled to 75 psia.
What is the specific entropy generation?
9.137E A pump has a 2 kW motor. How much liquid
water at 60 F can I pump to 35 psia from 14.7
psia?
9.138E A steam turbine inlet is at 200 psia, 900 F. The
exit is at 40 psia. What is the lowest possible
exit temperature? Which efficiency does that
correspond to?
9.139E A steam turbine inlet is at 200 psia, 900 F. The
exit is at 40 psia. What is the highest possible
exit temperature? Which efficiency does that
correspond to?
English Unit Problems H 339
9.140E A steam turbine inlet is at 200 psia, 900 F. The
exit is at 40 psia, 600 F. What is the isentropic
efficiency?
9.141E The exit velocity of a nozzle is 1500 ft/s. If
i^ozzie = 0-88, what is the ideal exit velocity?
9.142E Steam enters a turbine at 450 lbf/in. 2 , 900 F,
expands in a reversible adiabatic process, and
exhausts at 130 F. Changes in kinetic and poten-
tial energies between the inlet and the exit of the
turbine are small. The power output of the tur-
bine is 800 Btu/s. What is the mass flow rate of
steam through the turbine?
9.143E In a heat pump that uses R-134a as the working
fluid, the R-134a enters the compressor at 30
lbf/in. 2 , 20 F, at a rate of 0.1 Ibm/s. In the com-
pressor the R-134a is compressed in an adia-
batic process to 150 lbf/in. 2 . Calculate the power
input required to the compressor, assuming the
process to be reversible.
9.144E A diffuser is a steady-state, steady-flow device
in which a fluid flowing at high velocity is de-
celerated such that the pressure increases in the
process. Air at 18 lbf/in. 2 , 90 F, enters a diffuser
with velocity 600 ft/s and exits with a velocity
of 60 ft/s. Assuming the process is reversible
and adiabatic, what are the exit pressure and
temperature of the air?
9.145E The exit nozzle in a jet engine receives air at
2100 R, 20 psia, with negligible kinetic energy.
The exit pressure is 10 psia, and the process is
reversible and adiabatic. Use constant heat ca-
pacity at 77 F to find the exit velocity.
9.146E Air at 1 atm, 60 F, is compressed to 4 atm, after
which it is expanded through a nozzle back to
the atmosphere. The compressor and the nozzle
are both reversible and adiabatic, and kinetic en-
ergy in/out of the compressor can be neglected.
Find the compressor work and its exit tempera-
ture, and find the nozzle exit velocity.
9.147E An expander receives 1 Ibm/s air at 300 psia,
540 R, with an exit state of 60 psia, 540 R. As-
sume the process is reversible and isothermal.
Find the rates of heat transfer and work, neglect-
ing kinetic and potential energy changes.
9.148E A flow of 4 Ibm/s saturated vapor R-22 at 100
psia is heated at constant pressure to 140 F. The
heat is supplied by a heat pump that receives
heat from the ambient at 540 R and work input
as shown in Fig. P9.35. Assume everything is
reversible and find the rate of work input.
9.149E One technique for operating a steam turbine in
part-load power output is to throttle the steam to
a lower pressure before it enters the turbine, as
shown in Fig. P9.39. The steamline conditions
are 200 lbf/in. 2 , 600 F, and the turbine exhaust
pressure is fixed at 1 lbf/in. 2 . Assuming the ex-
pansion inside the turbine to be reversible and
adiabatic,
a. Determine the full-load specific work output
of the turbine.
b. Determine the pressure the steam must be
throttled to for 80% of full-load output.
c. Show both processes in a T-s diagram.
9.150E Analyze the steam turbine described in Problem
6.1 6 IE. Is it possible?
9.151E Two flowstreams of water, one at 100 lbf/in. 2 ,
saturated vapor, and the other at 100 lbf/in. 2 ,
1000 F, mix adiabatically in a steady flow
process to produce a single flow out at 100
lbf/in. 2 , 600 F. Find the total entropy generation
for this process.
9.152E A mixing chamber receives 10 lbm/min ammo-
nia as saturated liquid at F from one line and
ammonia at 100 F, 40 lbf/in. 2 from another line
through a valve. The chamber also receives 340
Btu/min energy as heat transferred from a 1 00-F
reservoir. This should produce saturated ammo-
nia vapor at F in the exit line. What is the
mass flow rate at state 2, and what is the total
entropy generation in the process?
9.153E A condenser in a power plant receives 10 Ibm/s
steam at 130 F, quality 90%, and rejects the heat
to cooling water with an average temperature of
62 F. Find the power given to the cooling water
in this constant pressure process and the total
rate of entropy generation when condenser exit
is saturated liquid.
9.154E Air at 540 F, 60 lbf/in. 2 , with a volume flow
40 ftVs runs through an adiabatic turbine with
exhaust pressure of 15 lbf/in. 2 . Neglect kinetic
energies and use constant specific heats. Find
the lowest and highest possible exit temperature.
For each case find also the rate of work and the
rate of entropy generation.
340 S chapterNine second-law analysis for a Control Volume
9.155E A supply of 10 Ibm/s ammonia at 80 lbf/in. 2 ,
80 F, is needed. Two sources are available: one
is saturated liquid at 80 F, and the other is at 80
lbf/in. 2 , 260 F. Flows from the two sources are
fed tlirough valves to an insulated SSSF mixing
chamber, which then produces the desired out-
put state. Find the two source mass flow rates
and the total rate of entropy generation by this
setup.
9.156E An old abandoned saltmine, 3.5 X 10 6 ft 3 in vol-
ume, contains air at 520 R, 14.7 lbf/in. 2 . The
mine is used for energy storage so the local
power plant pumps it up to 310 lbf/in. 2 using
outside air at 520 R, 14.7 lbf/in. 2 . Assume the
pump is ideal and the process is adiabatic. Find
the final mass and temperature of the air and the
required pump work. Overnight, the air in the
mine cools down to 720 R. Find the final pres-
sure and heat transfer.
9.157E Air from a line at 1800 lbf/in. 2 , 60 F, flows into a
20-ft 3 rigid tank that initially contained air at
ambient conditions, 14.7 lbf/in. 2 , 60 F. The
process occurs rapidly and is essentially adia-
batic. The valve is closed when the pressure in-
side reaches some value, P 2 . The tank eventually
cools to room temperature, at which time the
pressure inside is 750 lbf/in. 2 . What is the pres-
sure P 2 ? What is me net entr °Py change for the
overall process?
9.158E Liquid water at ambient conditions, 14.7 lbf/in. 2 ,
75 F, enters a pump at the rate of 1 lbm/s. Power
input to the pump is 3 Btu/s. Assuming the
pump process to be reversible, determine the
pump exit pressure and temperature.
9.159E A fireman on a ladder 80 ft above ground should
be able to spray water an additional 30 ft up
with the hose nozzle of exit diameter 1 in. As-
sume a water pump on the ground and a re-
versible flow (hose, nozzle included) and find
the minimum required power.
9.160E Saturated R-134a at 10 F is pumped/compressed
to a pressure of 150 lbf/in 2 at the rate of 1.0
lbm/s in a reversible adiabatic SSSF process. Cal-
culate the power required and the exit tempera-
ture for the two cases of inlet state of the R-134a:
a. Quality of 100%
b. Quality of 0%
9.161E A small pump takes in water at 70 F, 14.7
lbf/in. 2 , and pumps it to 250 lbf/in. 2 at a flow
rate of 200 lbm/min. Find the required pump
power input.
9.162E An expansion in a gas turbine can be approxi-
mated with a polytropic process with exponent
n = 1.25. The inlet air is at 2100 R, 120 psia,
and the exit pressure is 1 8 psia with a mass flow
rate of 2 lbm/s. Find the turbine heat transfer
and power output.
9.163E Helium gas enters a steady-flow expander at 120
lbf/in. 2 , 500 F, and exits at 18 lbf/in. 2 . The mass
flow rate is 0.4 lbm/s, and the expansion process
can be considered as a reversible polytropic
process with exponent, n ~ 1.3. Calculate the
power output of the expander.
9.164E A compressor is used to bring saturated water
vapor at 103 lbf/in. 2 up to 2000 lbf/in. 2 , where
the actual exit temperature is 1200 F. Find the
isentropic compressor efficiency and the en-
tropy generation.
9,1 65E A small air turbine with an isentropic efficiency
of 80% should produce 120 Btu/lbm of work.
The inlet temperature is 1800 R, and it exhausts
to the atmosphere. Find the required inlet pres-
sure and the exhaust temperature.
9.166E Air enters an insulated compressor at ambient
conditions, 14.7 lbf/in. 2 , 70 F, at the rate of 0.1
lbm/s and exits at 400 F. The isentropic effi-
ciency of the compressor is 70%. What is the
exit pressure? How much power is required to
drive the compressor?
9.167E A watercooled air compressor takes air in at 70
F, 14 lbf/in. 2 , and compresses it to 80 lbf/in. 2 .
The isothermal efficiency is 80%, and the actual
compressor has the same heat transfer as the
ideal one. Find the specific compressor work
and the exit temperature.
9.168E A nozzle is required to produce a steady stream
of R- 134a at 790 ft/s at ambient conditions, 15
lbf/in. 2 , 70 F. The isentropic efficiency may be
assumed to be 90%. What pressure and temper-
ature are required in the line upstream of the
nozzle?
9.169E Redo Problem 9.1 59E if the water pump has
an isentropic efficiency of 85% (hose, nozzle
included).
Computer Design, and Open-Ended Problems H 341
9.170E Repeat Problem 9.160E for a pump/compressor
isentropic efficiency of 70%.
9.171E A rigid 35 ft 3 tank contains water initially at 250
F, with 50% liquid and 50% vapor, by volume.
A pressure-relief valve on the top of the tank is
set to 140 lbf/in. 2 . (The tank pressure cannot ex-
ceed 140 lbf/in. 2 -— water will be discharged in-
stead.) Heat is now transferred to the tank from
a 400 F heat source until the tank contains satu-
rated vapor at 140 lbf/in. 2 . Calculate the heat
transfer to the tank and show that this process
does not violate the second law.
9.172E Air at 1 atm, 60 F, is compressed to 4 atm,
after which it is expanded through a nozzle
back to the atmosphere. The compressor and
the nozzle both have efficiency of 90%, and
kinetic energy in/out of the compressor can be
neglected. Find the actual compressor work
and its exit temperature, and find the actual
nozzle exit velocity.
Computer, design, and open-Ended Problems
9.173 Use the menu-driven software to get the proper-
ties for the calculation of the isentropic effi-
ciency of the pump in the steam power plant of
Problem 6.99.
9.174 Write a program to solve the general case of
Problem 9.27, in which the states, velocities, and
area are input variables. Use a constant specific
heat and find diffuser exit area, temperature, and
pressure.
9.175 Write a program to solve Problem 9.118 fn which
the inlet and exit flow states are input variables.
Use a constant specific heat and let the program
calculate the split of the mass flow and the overall
entropy generation.
9.176 Write a program to solve the general version of
Problem 9.84. Initial state, flow rate, and final
pressure are input variables. Compute the re-
quired pump power from the assumption of con-
stant specific volume equal to the inlet state
value.
9.177 Write a program to solve Problem 9.129 with the
final bottle pressure as an input variable. Print out
the temperature right after charging and the tem-
perature, pressure, and heat transfer after state 3 is
reached.
9.178 Consider a small air compressor taking atmos-
pheric air in and compressing it to 1 MPa in a
steady flow process. For a maximum flow rate of
0. 1 kg/s, discuss the necessary sizes for the piping
and the motor to drive the unit.
9.179 Small gasoline engine or electric motor-driven air
compressors are used to supply compressed air to
power tools, machine shops, and so on. The com-
pressor charges air into a tank that acts as a stor-
age buffer. Find examples of these and discuss
their sizes fn terms of tank volume, charging pres-
sure, engine, or motor power. Also find the time it
will take to charge the system from startup and its
continuous supply capacity.
9.180 A coflowing heat exchanger receives air at 800 K,
15 MPa, and water at 15°C, 100 kPa. The two
flows exchange energy as they flow alongside
each other to the exit, where the air should be
cooled to 350 K. Investigate the range of water
flows necessary per kilogram per second airflow
and the possible water exit temperatures, with
the restriction that the minimum temperature dif-
ference between the water and air should be
25°C. Include an estimation for the overall en-
tropy generation in the process per kilogram of
airflow.
9.181 Consider a geothermal supply of hot water avail-
able as saturated liquid at P x = 1.5 MPa. The liq-
uid is to be flashed (throttled) to some lower
pressure, P 2 . The saturated liquid and saturated
vapor at this pressure are separated, and the vapor
is expanded tlirough a reversible adiabatic turbine
to the exhaust pressure, P 3 = 10 kPa. Study the
turbine power output per unit initial mass, m { as a
function of the pressure, P 2 .
9.182 A reversible adiabatic compressor receives air at
the state of the surroundings, 20°C, 100 kPa. It
should compress the air to a pressure of 1 .2 MPa
in two stages with a constant pressure intercooler
between the two stages, investigate the work
input as a function of the pressure between the
two stages assuming the intercooler brings the air
down to 50°C.
9.183 (Adv.) Investigate the optimal pressure, P 2 , for a
constant pressure intercooler between two stages in
342 ■ CHAPTER NINE SECOND-LAW ANALYSIS FOR A CONTROL VOLUME
a compressor. Assume the compression process in
each stage follows a polytropic process and that the
intercooler brings the substance to the original inlet
temperature, T v Show that the minimal work for
the combined stages arises when
p 2 = (W a
where P 3 is the final exit pressure.
9.184 (Adv.) Reexamine the previous problem when the
intercooler cools the substance to a temperature,
T 2 > 7\, due to finite heat-transfer rates. What is
the effect of having isentropic efficiencies for the
compressor stages less than 100% on the total
work and selection of P{1
9.185 Investigate the sizes of turbochargers and super-
chargers available for automobiles. Look at their
boost pressures and check if they also have inter-
coolers mounted. Analyze an example with re-
spect to the power input and the air it can deliver
to the engine and estimate its isentropic efficiency
if enough data are found.
Irreversibility and
Availability
We now turn our attention to irreversibility and availability, two additional concepts that
have found increasing use in recent years. These concepts are particularly applicable in
the analysis of complex thermodynamic systems, for with the aid of a digital computer,
irreversibility and availability are very powerful tools in design and optimization studies
of such systems.
10,1 Available Energy, Reversible work,
and Irreversibility
In the previous chapter, we introduced the concept of the efficiency of a device, such as a
turbine, nozzle, or compressor (perhaps more correctly termed a first-law efficiency,
since it is given as the ratio of two energy terms). We proceed now to develop concepts
that include more meaningful second-law analysis. Our ultimate goal is to use this analy-
sis to manage our natural resources and environment better.
We first focus our attention on the potential for producing useful work from some
source or supply of energy. Consider the simple situation shown in Fig. 10.1a, in which
there is an energy source Q in the form of heat transfer from a very large and, therefore,
constant-temperature reservoir at temperature T. What is the ultimate potential for pro-
ducing work?
To answer this question, we imagine that a cyclic heat engine is available, as shown
in Fig. 10.16. To convert the maximum fraction of Q to work requires that the engine be
completely reversible, that is, a Carnot cycle, and that the lower-temperature reservoir be
at the lowest temperature possible, often, but not necessarily, at the ambient temperature.
From the first and second laws for the Carnot cycle and the usual consideration of all the
Q's as positive quantities, we find
i.e. = Q ~ Qo
T T
so that
(10.1)
343
344 m Chapter Ten irreversibility and Availability
FIGURE 10.1
Constant-temperature
energy source.
\ Reservoir at T /
\ Reservoir at T /
Cyclic
heat
engine
Go
<«0
/Environment at T ^
(b)
We might say that the fraction of Q given by the right side of Eq. 10.1 is the available
portion of the total energy quantity fi. To carry this thought one s*p further, consider (he
situation show on the TS diagram in Fig. 10.2. The total shaded area is fi. The Portion
of Q that is below T 0i the environment temperature, cannot be converted into work by the
heat engine and must instead be throw away. This portion is therefore the unavailable
portion of total energy fi, and the portion lying between the two temperatures T and T is
the available energy. . .. , ]o
Let us next consider the same situation, except that the heat transfer fi is available
from a constant-pressure source, for example, a simple heat exchanger as shown m Fig.
10 3a The Camot cycle must now be replaced by a sequence of such engines, with the re-
sult shown in Fig. 103b. The only difference between the first and second examples is
that the second includes an integral, which corresponds to AS.
(10.2)
(10.3)
Substituting into the first law, we have
Note that this AS quantity does not include the standard sign convention. It corresponds to
the amount of change of entropy shown in Fig. 10.36. Equation 10.2 specifies the avail-
able portion of the quantity fi. The portion unavailable for producing work m this circum-
stance lies below T Q in Fig. 1036.
FIGURE 10.2 TS
diagram for constant-
temperature energy
source.
Available
energy
Unavailable
' energy
Available Energy, Reversible Work, and Irreversibility M 345
FIGURE 10.3
Changing-temperature
energy source.
The Steady-State Process
We now proceed to extend our analysis to real, irreversible processes. In doing so, we will
consider first the case of the steady-state control volume process, since the vast majority
of applications of this type of analysis refer to components of industrial systems such as
power plants or refrigerators. It should be kept in mind that a parallel development for a
thermodynamic system analysis or for transient control volume processes will also be
made, giving analogous results.
Consider the real steady-state process shown in Fig. 10.4, in which a control vol-
ume has a single fluid stream entering at state /, has a single stream exiting at state e, re-
ceives an amount of heat q (per unit flow mass) from a reservoir at temperature T m and
does an amount of work per unit mass w. The first law is, assuming no changes in kinetic
or potential energies,
q = iK- + w (10.4)
This real process is irreversible, such that
J; = ~ *) - ~ = s m > o (10,5)
We wish to establish a quantitative measure in energy terms of the extent or degree
to which any particular real process is irreversible. This can be accomplished by compari-
son with a control volume undergoing a steady-state process with the same inlet state /,
the same exit state e s the same amount of heat transfer with the reservoir, with everything
being reversible. That is, the control volume undergoes a steady-state change with inlet h h
j go.
h £ , s t -
i
Control
volume
FIGURE 10.4 A real
irreversible process.
V
346 H CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
s, and exit h ei s e with heat transfer q leaving the reservoir at T H . If this entire process is re-
versible, then the net rate of entropy change must equal zero. Noting that, all the terms in
Eq. 10.5 are the same as for the real irreversible process, we conclude that there must be
an additional negative term in the second law for the reversible case. This can only be a
heat transfer from the surroundings, divided by the temperature at which it leaves the sur-
roundings. To make this heat transfer as small as possible, it should come from the lowest
possible temperature, usually the ambient temperature T . This is the situation shown in
Fig. 10.5, such that the second law for the reversible process is
4% = fc-„)-f "f =0 00.6)
m at *h J o
Since any reversible heat transfer must occur over only an infinitesimal tempera-
ture difference, we recognize that both heat transfers shown in Fig. 10.5 must be trans-
ferred through reversible heat engines or heat pumps. These are located inside the
control volume boundary of Fig. 10.5, which includes the original control volume of Fig.
10.4 plus any necessary heat engines and pumps. Only the two fluid flow streams, the net
work, and the two heat transfers cross this extended control surface. Equation 10.6 can
be rewritten as
The first law for the reversible process of Fig. 10.5 is
w ™ = q + q™ - (h e - A,) (10.8)
and substituting Eq. 1 0.7, we get
w rev - T Q (s e - Sl ) - (h e -ht) + q(l- Y HJ
(10.9)
This expression establishes the theoretical upper limit for the work per unit mass
flow that could be produced by a control volume undergoing a steady-state process from i
to e in which heat g is transferred from a reservoir at T H , with all processes occurring in
the environment T Q . The difference between this quantity and the work actually done in
the real process of Eqs. 10.4 and 10.5 is a measure of the extent of the irreversibility i (per
unit mass flow) of the real process. That is,
i = /«-w (10.10)
Control
volume
FIGURE 10,5 An
ideal reversible process.
Available Energy, Reversible Work, and Irreversibility H 347
The irreversibility z of the real process can also be expressed in another form, by
substituting Eqs. 10.4 and 10.9 into 10.10, which results in
J_ ^net/real
m
dt
(10.11)
And we note that the irreversibility of a real process is another way of expressing the sec-
ond law for that process, in energy units instead of entropy units and directly proportional
to the entropy generation.
EXAMPLE 10.1 A feedwater heater has 5 kg/s water at 5 MPa and 40°C flowing through it, being heated
from two sources as shown in Fig. 10.6. One source adds 900 k\V from a 100°C reser-
voir, and the other source transfers heat from a 200°C reservoir such that the water exit
condition is 5 MPa, 180 o C. Find the reversible work and the irreversibility.
Conti'ol volume: Feedwater heater extending out to the two reservoirs.
Inlet state: P h T- t known; state fixed.
Exit state: P e , T e known; state fixed.
Process: Constant-pressure heat addition with no change in kinetic or
potential energy.
Model: Steam tables.
Analysis
This control volume has a single inlet and exit flow with two heat-transfer rates coming
from reservoirs different from the ambient surroundings. There is no actual work or ac-
tual heat transfer with the surroundings at 25°C, For the actual feedwater heater, the en-
ergy equation becomes
h t + q\ + q 2 ^ K
The reversible work for the given change of state is, from Eq. 1 0.9, with heat transfer q x
from reservoir T { and heat transfer q 2 from reservoir T 2 ,
+ ?2 1-
FIGURE 10.6 The
feedwater heater for
Example 10.1.
©
1* M z
©
348 Wi CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
From Eq. 10.10, since the actual work is zero, we have
i = - w = rf<
Solution
From the steam tables the inlet and exit state properties are
h, - 171.95 kJ/kg, s, = 0.5705 kj/kg K
h e = 765.24 kJ/kg, s e = 2.1341 kJ/kgK
The second heat transfer is found from the energy equation as
q 2 = h e - ft, ~q,= 765.24 - 171.95 - 900/5 = 413.29 kJ/kg
The reversible work is
EXAMPLE 10.2 Consider an air compressor that receives ambient air at 100 kPa and 25°C. It compresses
the air to a pressure of 1 MPa, where it exits at a temperature of 540 K. Since the air and
compressor housing are hotter than the ambient surroundings, 50 kJ per kilogram air
flowing through the compressor are lost. Find the reversible work, and the irreversibility
in the process.
Confrol volume: The air compressor.
Sketch: Fig. 10.7.
Inlet state: P {i T f known; state fixed.
Exit state: P e) T e known; state fixed.
Process: Nonadiabatic compression with no change in kinetic or
potential energy.
Model: Ideal gas.
Analysis
This steady-state process has a single inlet and exit flow so all quantities are done on a
mass basis as specific quantities. From the ideal gas air tables, we obtain
= 298.2(2.1341 - 0.5705) - (765.24 - 171.95)
= 466.27 - 593.29 + 36.17 + 152.84 - 62.0 kJ/kg
The irreversibility is
/ = w ™ = 62-0 kJ/kg
ht = 298.6 kJ/kg, 4, = 6.8631 kJ/kg K
h e = 544.7 kJ/kg, s° Tt = 7.4664 kJ/kg K
Available Energy, Reversible Work, and Irreversibility
h 349
FIGURE 10.7
Illustration for Example
10.2.
so the energy equation for the actual compressor gives the work as
^ —50 kJ/kg
w = h t - K + q = 298.6 - 544.7 - 50 = -296.1 kJ/kg
The reversible work for the given change of state is, from Eq. 10.9, with T H = T
T^-sd-iK-hd+qll- —
= 298.2(7.4664 - 6.8631 - 0.287 In 10) - (544.7 - 298.6) 4-
- -17.2 - 246.1 - -263.3 kJ/kg
FromEq. 10.10, we get
i = w mv — w
= -263.3 - (-296.1) = 32.8 kJ/kg
EXAMPLE 10.2E Consider an air compressor that receives ambient air at 14.7 Ibf/in. 2 , 80 F. It compresses
the air to a pressure of 1 50 lbf/in. 2 , where it exits at a temperature of 960 R. Since the air
and compressor housing are hotter than the ambient, it loses 22 Btu/Ibm air flowing
through the compressor. Find the reversible work and the irreversibility in the process.
Control volume: The air compressor.
Inlet state: P u 7) known; state fixed.
Exit state: P e , T e known; state fixed.
Process: Nonadiabatic compression with no change kinetic or potential
energy.
Model: Ideal gas.
Analysis
The steady-state process has a single inlet and exit flow so all quantities are done on a
mass basis as specific quantities. From the ideal gas air tables, we obtain
h t = 129.18 Btu/lbm s° T; = 1.6405 Btu/Ibm R
h e = 231.20 Btu/lbm s\ = 1.7803 Btu/lbmR
350 B Chapter Ten Irreversibility and Availability
so the energy equations for the actual compressor gives the work as
q = -22 Btuflbm
w = h s - h e + q = 129.18 - 231.20 - 22 = - 124.02 Btu/lbm
The reversible work for the given change of state is, from Eq. 10.9, with T H = T Q
= 539.7(1.7803 - 1.6405 - 0.06855 In 10.2) - (231.20 - 129.18) +
- -10.47 - 192.02 - -112.49 Btu/lbm
FromEq. 10.10, we get
The expression for reversible work for the steady-state process, Eq. 10.9, was de-
rived without including kinetic and potential energy terms. Whenever necessary, espe-
cially in nozzles and diffusers where obtaining kinetic energy change is the reason for
building the device, these terms can be included along with the enthalpy terms of the fluid
stream in and out of the control volume. Another way of including these terms would be
to say that the enthalpies in Eq. 10.9 are the total enthalpies, as used in Eq. 6.8. There are
also steady-state processes involving more than one fluid stream entering or exiting the
control volume. In such cases, it is necessary to rewrite Eq. 10.9 on a rate basis, including
the mass flow rates of the different streams involved in the process.
The Control Mass Process
Consider the real process shown in Fig. 10.8, in which a control mass receives an amount of
heat x Q t from a reservoir at temperature T Hi undergoes a change of state from 1-2, and does
an amount of work X W 2 . The first law is, assuming no changes in kinetic or potential energies,
i — w
.rev
= -112.49 - (-124.02) = 11.53 Btu/lbm
(10.12)
This real process is irreversible, such that
= A£
+ ASL
rp 1^2gen u
1 H
(10.13)
FIGURE 10.8 A real
irreversible process.
AVAILABLE ENERGY, REVERSIBLE WORK, AND IRREVERSIBILITY H 351
As with the steady-state process, we wish to compare this irreversible process with
an ideal reversible process for the same change of state from U u S x to U 2i S 2 with heat
transfer X Q 2 from reservoir at T H . This is represented art Fig. 10.9, in which the extended
control surface includes the original control mass plus any necessary heat engines and
pumps required for reversible heat transfer, analogous to the case for the steady-state
process analyzed earlier. The second law for the reversible process is
A-W - (S 2 - S x ) - - ^- = (10.14)
which can be rewritten as
£T= US^Sd-^Y- (10.15)
The first law for the reversible process of Fig. 10.9 is
xWT = ,& + Qt - (U 2 - u,) (10.16)
and substituting Eq. 10.15,
JVT = US 2 - S { ) - (U 2 - U x ) + X Q 2 [\ - (10.17)
This expression establishes the theoretical upper limit for the work that could be produced
by a control mass undergoing the change of state 1-2 in which heat X Q 2 is transferred
from a reservoir at T H , all occurring in the environment T Q . The difference between this
quantity and the work actually done in the real process of Eqs. 10,12 and 10.13 is a mea-
sure of the extent of the irreversibility I of the real process, or
A = i0T-i»2 (10.18)
The irreversibility X I 2 of the real process can also be expressed in another form, by substi-
tuting Eqs. 10.12 and 10.17 into 1 0. 1 8 which results in
To
ih " T Q {S 2 — S x ) — — iQ 2
1 H
= T [AS^{\ = J 01 5 2gen (10.19)
A
Control
surface
FIGURE 10,9 An
ideal reversible process.
352 M Chapter Ten irreversibility and Availability
EXAMPLE 10.3 An insulated rigid tank is divided into two parts A and B by a diaphragm. Each part has
a volume of 1 m 5 . Initially, part A contains water at room temperature, 20°C, with a
quality of 50%, while part B is evacuated. The diaphragm then ruptures and the water
fills the total volume. Determine the reversible work for this change of state and the irre-
versibility of the process.
Control ?nass: Water
Initial state: T u x x known; state fixed.
Final state: V 2 known.
Process: Adiabatic, no change in kinetic or potential energy.
Model: Steam tables.
Analysis
There is a boundary movement for the water, but since it occurs against no resistance,
there is no work done. Therefore, the first law reduces to
m{u 2 — Ui) —
From Eq. 10.17 with no change in internal energy and no heat transfer,
' jr? = T a (S 2 - S{) = 7>(s 2 - *i)
From Eq. 10.18
Solution
From the steam tables at state 1,
«i = 1243.5 kJ/kg v x - 28.895 mVkg s x = 4.4819 kJ/kgK
Therefore,
v 2 = V 2 /m = 2 X vi = 57.79 u 2 = u x = 1243.5
These two independent properties, v 2 and u 2i fix state 2. The final temperature T 2 must
be found by trial and error in the steam tables.
For T 2 = 5°C and v 2 =>x = 0.3928, u - 948.5 kJ/kg
For r 2 =10°C and v 2 = 0.5433, «=1317kJ/kg
so the final interpolation in u gives a temperature of 9°C. If the software is used, the final
state is interpolated to be
r 2 = 9.1°C x 2 = 0.513 s 2 = 4.644 kJ/kgK
with the given u and v. Since the actual work is zero we have
= 293.2(l/28.895)(4.644 - 4.4819) = 1.645 kJ
Available Energy, Reversible Work, and Irreversibility H 353
For processes in which kinetic and potential energy changes are significant, the de-
velopment of the expressions for work, reversible work, and irreversibility are all the
same, substituting E = U + KE 4- PE for U in any equation involving energy.
The Transient Process
The transient process has a control volume change from state 1 to state 2 with possible
mass flow in at state i and/or flow out at state e. The procedure for developing an expres-
sion for reversible work for this process is analogous to the previous examples followed for
the steady state and the control mass. In this case, assuming no kinetic or potential energy
terms are included, the equation for reversible work will contain control volume entropy
and energy terms of the same form as in Eq. 10.17 (but recognizing that the masses at
states 1 and 2 are different), and will also contain entropy and enthalpy flow terms the
same as in Eq. 10.9 (each one including the appropriate mass flow). The result is
= T {m 2 s 2 - m^i) - {m 2 u 2 - ?n 1 u{)
+ T (m e s e - jtifr) ~ {m t h t - mfc)
+ 0,v.(l -jrj (10.20)
This expression can also be grouped as
^ = *n&, ~ V;) " mlK " T s e )
+ m x {u x - - m 2 (u 2 - T s 2 ) + £, v . ^1 - (10.21)
As in the previous developments, A, and h e can e replaced by h 10Ti and /i T0Te , and u x and u 2
can be replaced by e x and e 2> whenever kinetic and potential energies are significant. Also,
summations can be added to the flow terms in cases where there is more than one flow
stream in or out of the control volume.
The irreversibility for the transient process is found from the general definition.
4v. = W™. ~ ffW (10.22)
which, by substitution of Eq. 1 0.20 and the first law, can also be expressed as
(m 2 s 2 - m^) -1- m e s e - mp, - %
T [AS^ + AS SUJT ] = r [A^ et/lE3l ] = (10.23)
EXAMPLE 10.4 A 1 -m 3 rigid tank, Fig. 10. 10, contains ammonia at 200 kPa and the ambient temperature
20°C. The tank is comiected with a valve to a line flowing saturated liquid ammonia at
- 10°C. The valve is opened, and the tank is charged quickly until the flow stops and the
valve is closed. As the process happens very quickly, there is no heat transfer. Deter-
mine the final mass in the tank and the irreversibility in the process.
Control volume: The tank and the valve.
Initial state: r 5 , P l known; state fixed.
354 a
FIGURE 10.10
Ammonia tank and line
for Example 10.4.
y///////////////////////.
Inlet state:
Final state:
Process:
Model:
Ammonia
line
supply
T h Xj known; state fixed.
I J i known.
Adiabatic, no kinetic or potential energy change.
Ammonia tables.
Analysis
Since the line pressure is higher than the initial pressure inside the tank, flow is going
into the tank and the flow stops when the tank pressure has increased to the line pres-
sure. The continuity, energy, and entropy equations are
?n 2 — mi = m t
m 2 u 2 — — mA- ~ ( m 2 ~~ m \Vh
m 2 s 2 - m x s x ~ + { S 2gi!l
where kinetic and potential energies are zero for the initial and final states and neglected
for the inlet flow.
Solution
From the ammonia tables, the initial and line state properties are
u, = 0.6995 m 3 /kg a, = 1369.5 kJ/kg s, = 5.927 kJ/kg K
h ( = 134.41 kJ/kg St - 0.5408 kJ/kgK
The initial mass is therefore
m, = 1%! = 1/0.6995 = 1.4296 kg
It is observed that only the final pressure is known, so one property is needed. The un-
knowns are the final mass and final internal energy in the energy equation. Since only
one property is unknown, the tw.o quantities are not independent. From the energy equa-
tion we have
m 2 (u 2 - /*,) = wi(«i h,)
from which it is seen that u 2 > h t and the state therefore is two-phase or superheated
vapor. Assume that the state is two phase, then
m 2 - v/v 2 = 1/(0.001534 + x 2 X 0.41684)
u 2 = 133.964 I x 2 X 1175.257
Availability and Second-Law Efficiency 1 355
so the energy equation is
133.964 + x 2 X 1175.257 - 134.41
0.001534 + ^X 0.041684 ~= ^296(1369.5 - 134.41) = 1765.67 kJ
This equation is solved for the quality and the rest of the properties to give
x 2 = 0.007182 v 2 = 0.0045276 m 3 /kg s 2 = 0.5762 kj/kg
Now the final mass and the irreversibility are found
m 2 = F/y 2 = 1/0.0045276 = 220.87 kg
i$g« = m 2 s 2 - m x s x - mfr = 127.265 - 8.473 - 118.673 = 0.119 kJ/IC
4v. = 7* 1( 5 2gen - 293.15 X 0.119 = 34.885 kJ
10.2 Availability and
second-law efficiency
What is the maximum reversible work that can be done by a given mass in a given state?
In the previous section, we developed expressions for the reversible work for a given
change of state for a control mass and control volume undergoing specific types of
processes. For any given case, what final state will give the maximum reversible work?
The answer to this question is that, for any type of process, when the mass comes
into equilibrium with the environment, no spontaneous change of state will occur and the
mass will be incapable of doing any work. Therefore, if a mass in a given state undergoes
a completely reversible process until it reaches a state in which it is in equilibrium with
the environment, the maximum reversible work will have been done by the mass. In this
sense, we refer to the availability at the original state in terms of the potential for achiev-
ing the maximum possible work by the mass.
If a control mass is in equilibrium with the surroundings, it must certainly be in pres-
sure and temperature equilibrium with the surroundings, that is, at pressure P and tempera-
ture T . It must also be in chemical equilibrium with the surroundings, which implies that no
further chemical reaction will take place. Equilibrium with the surroundings also requires
that the system have zero velocity and minimum potential energy. Similar requirements can
be set forth regarding electrical and surface effects if these are relevant to a given problem.
The same general remarks can be made about a quantity of mass that undergoes a
steady-state process. With a given state for the mass entering the control volume, the re-
versible work will be a maximum when this mass leaves the control volume in equilibrium
with the surroundings. This means that as the mass leaves the control volume, it must be at
the pressure and temperature of the surroundings, be in chemical equilibrium with the sur-
roundings, and have minimum potential energy and zero velocity. (The mass leaving the
control volume must of necessity have some velocity, but it can be made to approach zero.)
Let us first consider the availability associated with a steady-state process. For a
control volume with a single-flow stream, the reversible work is given by Eq. 10.9. In-
cluding kinetic and potential energies, this expression is rewritten in the form,
^ rev = (Ajari - W - (h mje - T s e ) + q ^\-^j
356 H CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
From the discussion of the heat engine that led to Eq. 10.1, it is clear that the last term in
the expression for the reversible work is the contribution to the net reversible work from
the heat transfers. These can be viewed as transfer of availability associated with g, which
gives a potential to do work as in a heat engine. Such contributions are separate from the
availability in the flow itself. The steady-state flow reversible work will be maximum, rel-
ative to the surroundings, when the mass leaving the control volume is in equilibrium
with the surroundings. The state in which the fluid is in equilibrium with the surroundings
is designated with subscript 0, and the reversible work will be maximum when h e = h ,
s e = s a> V e = 0, and Z e = Z„. This maximum reversible work per unit mass flow without
the additional heat transfers is the flow availability or exergy and is assigned the symbol ip
ijf=(h-T^ + ^+gzj- (h Q - + gZ,) (1 0.24)
It is written without subscript for the inlet state to indicate that this is the flow availability
associated with a substance in any state as it enters the control volume in a steady-state
process. The reversible work is therefore seen to be equal to the decrease in flow avail-
ability plus the reversible work that can be extracted from heat engines operating with the
heat transfer at T H and the ambient.
Irreversibility is as usual defined as the difference between the reversible work and
the actual work. If we write this as a general expression on a rate basis, including summa-
tions to account for the possibility of more than one flow stream and also more than one
heat transfer, the result is
L. - (s wf> t ~ 2 ™a) + 2 (i - y) Qc,,j - (io-25)
In this form, the irreversibility is equal to the decrease in the availability of the mass flows
plus the decrease of availability of each heat transfer rate j at reservoir 3} minus the in-
crease in availability of the surroundings that receive the actual work. The rate of irre-
versibility is thus seen to be the rate of destruction of availability, which is also directly,
proportional to the net rate of entropy increase, as noted in Eq. 1 0. 1 1.
For a control mass, a similar consideration of the maximum reversible work will
lead to a nonftow availability concept. In this case the volume may change, and some
work is exchanged with the ambient, which is not available as useful work. The reversible
work for a control mass is given by Eq. 10.17. Including kinetic and potential energies
(e instead of u% this expressions is rewritten as
lW f = (e, - Ttfi) - (e 2 - V 2 ) + x q 2 (a -
which is the maximum between the two given states. This work is available if the final
state is in equilibrium with the surroundings, for which we must have e 2 = e = u$ + gZ ,
the kinetic energy being zero, and s 2 = s . The work done against the surroundings, w^, is
w^r = A(t> " v{) = -P (t>i - f o)
such that the maximum available work is
w avail — '"max "W
= {e- Tos) ~ {e - 7> ) + P (v - v ) + ift ( 1 " ^) ( 10 - 26 )
Availability and Second-Law efficiency S 357
The subscript is again dropped to indicate that this is the maximum available work at a
given state having also x q 2 available from a source at T H . The nonflow availability is
defined as the maximum available work from a state without the heat transfers in-
cluded as
Sometimes the definition excludes the kinetic and potential energies, in which case u
should be used instead of e.
The irreversibility may again be related to the changes in availability through the
difference between the reversible work and the actual work. The reversible work from
above is expressed with the availability from which the actual work is subtracted to
give
including the possibility of having heat transfer with more than one reservoir. The irre-
versibility is then equal to the decrease in availability of the control mass plus the de-
crease in availability of the heat transfers at reservoirs 7} minus the increase in availability
of the surroundings that received the actual work. It is noted again that the irreversibility
expresses the net destruction of availability of the control mass and surroundings, which
is proportional to the net entropy increase, as given in Eq, 10.19.
The less the irreversibility associated with a given change of state, the greater the
amount of work that will be done (or the smaller the amount of work that will be re-
quired). This relation is significant in at least two regards. The first is that availability
is one of our natural resources. This availability is found in such forms as oil reserves,
coal reserves, and uranium reserves. Suppose we wish to accomplish a given objective
that requires a certain amount of work. If this work is produced reversibly while draw-
ing on one of the availability reserves, the decrease in availability is exactly equal to
the reversible work. However, since there are irreversibilities in producing this re-
quired amount of work, the actual work will be less than the reversible work, and the
decrease in availability will be greater (by the amount of the irreversibility) than if this
w r ork had been produced reversibly. Thus, the more irreversibilities we have in all our
processes, the greater will be the decrease in our availability reserves. 1 The conserva-
tion and effective use of these availability reserves is an important responsibility for
all of us.
The second reason that it is desirable to accomplish a given objective with the
smallest irreversibility is an economic one. Work costs money, and in many cases a given
objective can be accomplished at less cost when the irreversibility is less. It should be
noted, however, that many factors enter into the total cost of accomplishing a given objec-
tive, and an optimization process that considers many factors is often necessary to arrive
at the most economical design. For example, in a heat-transfer process, the smaller the
'In many popular talks, reference is made to our energy reserves. From a thermodynamic point of
view, availability reserves would be a much more acceptable term. There is much energy in the at-
mosphere and the ocean, but relatively little availability.
<f> = (e-T4s)- {e - 7> ) + P (y - u )
- (e + P c v - 7» - (e Q + P v - T s )
(10.27)
x h = m{4>x -4>2) + 2h-yjQj- diV? ~ Po(V 2 ~ V t )) (10.28)
358 M CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
FIGURE 10.11
Irreversible turbine.
temperature difference across which the heat is transferred, the less the irreversibility.
However, for a given rate of heat transfer, a smaller temperature difference will require a
larger (and therefore more expensive) heat exchanger. These various factors must all be
considered in developing the optimum and most economical design.
In many engineering decisions, other factors, such as the impact on the environment
(for example, air pollution and water pollution) and the impact on society must be consid-
ered in developing the optimum design.
Along with the increased use of availability analysis in recent years, a term called the
second-law efficiency has come into more common use. This term refers to comparison of
the desired output of a process with the cost, or input, in terms of the thermodynamics avail-
ability. Thus, the isentropic turbine efficiency defined by Eq. 9.27 as the actual work output
divided by the work for a hypothetical isentropic expansion from the same inlet state to the
same exit pressure might well be called a first-law efficiency, in that it is a comparison of
two energy quantities. The second-law efficiency, as just described, would be the actual
work output of the turbine divided by the decrease in availability from the same inlet state to
the same exit state. For the turbine shown in Fig. 10.1 1, the second-law efficiency is
law
(10.29)
In this sense, this concept provides a rating or measure of the real process in terms of the
actual change of state and is simply another convenient way of utilizing the concept of
thermodynamic availability. In a similar manner, the second-law efficiency of a pump or
compressor is the ratio of the increase in availability to the work input to the device.
EXAMPLE 10.5 An insulated steam turbine, Fig. 10.12, receives 30 kg of steam per second at 3 MPa,
350°C. At the point in the turbine where the pressure is 0.5 MPa, steam is bled off for
processing equipment at the rate of 5 kg/s. The temperature of this steam is 200°C. The
balance of the steam leaves the turbine at 15 kPa, 90% quality. Determine the availabil-
ity per kilogram of the steam entering and at both points at which steam leaves the tur-
bine, the isentropic efficiency and the second-law efficiency for this process.
Control volume: Turbine. .
Inlet state: /',, 7', known; state fixed.
Exit state: P 2> T 2 known; P 3> x 3 known; both states fixed.
Availability and second-Law Efficiency M 359
FIGURE 10.12
Sketch for Example 10.5.
30 kg/s
Control
surface
5 kg/s
0.5 MPa, 200°C
25 kg/s
15kPa, 90% quality
Process: Steady state.
Model: Steam tables.
Analysis
The availability at any point for the steam entering or leaving the turbine is given by Eq.
10.24,
4> = (h - h ) - T (s ~s )+^ + g(Z ~ Z )
Since there are no changes in kinetic and potential energy in this problem, this equation
reduces to
<A = (h ~ h ) ~ T (s - s )
For the ideal isentropic turbine,
For the actual turbine,
W= - m 2 h 2 ~ m 3 k 3
Solution
At the pressure and temperature of the surroundings, 0.1 MPa, 25°C, the water is a
slightly compressed liquid, and the properties of the water are essentially equal to those
for saturated liquid at 25°C. ■
h = 104.9 kJ/kg s = 0.3674 kj/kg K
FromEq. 10.24
ip x = (3115.3 - 104.9) - 298.15(6.7428 - 0.3674) - 1109.6 kJ/kg
(A 2 = (2855.4 - 104.9) - 298.15(7.0592 - 0.3674) - 755.3 kJ/kg
tft 3 = (2361.8 - 104.9) - 298.15(7.2831 - 0.3674) = 195.0 kJ/kg
m^i - m 2 iff 2 ~ = 30(1109.6) - 5(755.3) - 25(195.0) = 24 637 kW
360 B CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
For the ideal isentropic turbine,
S2j = 6.7428 = 1.8606 + X 4.9606, ^ = 0.9842
^ = 640.2 + 0.9842 X 2108.5 = 2715.4
= 6.7428 = 0.7549 + X 7.2536, xj, '='0.8255
h y = 225.9 + 0.8255 X 2373.1 - 2184.9
W s = 30(3115.3) - 5(2715.4) - 25(2184.9) - 25 260 kW
For the actual turbine,
W= 30(3115.3) - 5(2855.4) - 25(2361.8) =' 20 137 kW
The isentropic efficiency is
= iii = 0797
and the second-law efficiency is
20 137 a on
^hw™ 24637 "
For a device that does not involve the production or the input of work, the defini-
tion of second-law efficiency refers to the accomplishment of the goal of the process rel-
ative to the process input, in terms of availability changes or transfers. For example, in a
heat exchanger, energy is transferred from a high-temperature fluid stream to a low-
temperature fluid stream, as shown in Fig. 10.13, in which case the second-law effi-
ciency is defined as
(I030)
The previous expressions for the second-law efficiency can be presented by a single ex-
pression. First, notice that the actual work from Eq. 10.25 is
where * MUtM is the total rate of availability supplied from all sources; flows, heat trans-
fers, and work inputs. In other words, the outgoing availability, IF CV ., equals the incoming
availability less the irreversibility. Then for all cases we may write
— ^ wafted
(10.32)
FIGURE 10.13 A
two-fluid heat exchanger.
Low-r fluid in
©
_yyyY\ ( A---4<5 — High-r fluid in
Availability and Second-Law efficiency M 361
and the wanted quantity is "then expressed as availability whether it actually is a work
term or a heat transfer. We can verify that this covers the turbine, Eq. 10.29, the pump
or compressor, where work input is the source, and the heat exchanger efficiency in Eq.
EXAMPLE 10.6 In a boiler, heat is transferred from the products of combustion to the steam. The tem-
perature of the products of combustion decreases from 1 100°C to 550°C, while the pres-
sure remains, constant at 0.1 MPa. The average constant-pressure specific heat of the
products of combustion is 1.09 kJ/kg K. The water enters at 0.8 MPa, 150°C, and leaves
at 0.8 MPa, 250°C. Determine the second-law efficiency for this process and the irre-
versibility per kilogram of water evaporated.
Control volume:
Sketch:
Inlet states:
Exit states:
Process:
Diagram:
Model:
Overall heat exchanger.
Fig. 10.14.
Both known, given in Fig. 10.14.
Both known, given in Fig. 10.14.
Overall, adiabatic.
Fig. 10.15.
Products — ideal gas, constant specific heat. Water-
steam tables.
Analysis
For the products, the entropy change for this constant-pressure process is
( S e s i)pia& ~ C pl> In -
For this control volume we can write the following governing equations:
Continuity equation:
First law (a steady-state process):
(a)
(b)
(c)
Control surface
F *GURE 10.14
S ketch for Example 10.6.
Products
1100°C
H 9
150°C
362 ■ Chapter ten irreversibility and Availability
FIGURE 10.15
Temperature-entropy
diagram for Example
10.6.
Second law (the process is adiabatic for the control volume shown):
calculate the ratio of the mass flow of products to the
Solution
From Eqs. a, b, and c, we can <
flow of water.
fflprod _ ih - h i)n l0 = 2950 - 632.2 = 3 m
ri^"(A,-AeW 1.09(1100 - 550)
The increase in availability of the water is, per kilogram of water,
= (2950 - 632.2) - 298.15(7.0384 - 1.8418)
= 768.4 kJ/kgH 2
The decrease in availability of the products, per kilogram of water, is
- ^ = [(/r 3 - h 4 ) ~ Us, - s 4 )]
- 3.866
1.09(1100 - 550) - 298.15 I 1.09 In
1373.15
= 1674.7 kJ/kgH 2
Therefore, the second-law efficiency is, from Eq. 10.30,
— 768.4 „ ft
Vlnihw ~ 1674.7
Energy Balance Equation m 363
FromEq. 10.25, the process irreversibility per kilogram of water is
~ * Z — fa ~ ^ ~ — fa
- (-768.4 + 1674.7) = 906.3 kJ/kgH 2
It is also of interest to determine the net change of entropy. The change in the en-
tropy of the water is
- sJ Hl0 = 7.0384 - 1.8418 - 5.1966 kJ/kg H 2 K
The change in the entropy of the products is
<* 4 " = "3.866 (l.09 In —^—j = -2.1564 kJ/kg H a O K
Thus, there is a net increase in entropy during the process. The irreversibility could also
have been calculated from Eq. 10.11:
/ = 2 - 2 mfc - 4.v. = T S gta
For the control volume selected, £> c v = 0, and therefore
- 298.15(5.1966) + 298.15(-2.1564)
= 906.3 kJ/kg H 2
These two processes are shown on the T-s diagram of Fig. 10.15. Line 3-4 represents
the process for the 3.866 kg of products. Area 3-4-e-</-3 represents the heat trans-
ferred from the 3.866 kg of products of combustion, and area 3^~e-f~3 represents
the decrease in availability of these products. Area l- a -6-2-/i-c-l represents
the heat transferred to the water, and this is equal to area 3-4-c-rf-3, which repre-
sents the heat transferred from the products of combustion. Area l-a-b-2~g-e-\
represents the increase in availability of the water. The difference between area
3-4-e-/-3 and area \-a-b~2-g-e-l represents the net decrease in availability. It is
readily shown that this net change is equal to f-g-h~d~f s or T Q (A$X H . Since the
actual work is zero, this area also represents the irreversibility, which agrees with our
calculation.
103 Exeegy Balance Equation
The previous treatment of availability or exergy in different situations was done sepa-
rately for the steady-flow, the control mass, and the transient processes. For each case, an
actual process was compared to an ideal counterpart, which led to the reversible work and
the irreversibility. When the reference was made with respect to the ambient state, we
364 S Chapter. Ten Irreversibility and availability
found the flow availability, 4> in Eq. 10.24, and the no-flow availability, <f> m Eq. 10.27.
We want to show that these forms of availability are consistent with one another The
whole concept is unified by a formulation of the exergy for a general control volume from
which we wilt recognize all the previous forms of availability as special cases of the more
general form. , .
In this analysis, we start out with the definition of exergy, <X> = m<f>, as the maxi-
mum available work at a given state of a mass from Eq. 10.27, as
$ - m<f> = m(e - e ) + PM" " u o) - " %> (10- 33 )
Here subscript "0" refers to the ambient state with zero kinetic energy, the dead state,
from which we take our reference. Because the properties at the reference state are con-
stants, the rate of change for $ becomes
= ±e + p f - r„ d -f - (/,„ - W f 0*w>
and we used, /i„ = e„ + J>„«o. to shorten the expression. Now we substitute the rate of
change of mass from the continuity equation, Eq. 6.1,
the rate of change of total energy from the energy equation, Eq. 6.8,
dt dt
and the rate of change of entropy from the entropy equation, Eq. 9.2
into the rate of exergy equation Eq. 10.34. When that is done, we get
^ = 2 ficv. - ^c.v. +" 2 mK* i - 1 "A*
Now collect the terms relating to the heat transfer together and those relating to the flow
together and group them as
Y~ = 2 ( 1 " y) ^ e v - Transfer by heat at T
_iv _i_ p 4L Transfer by shaft/boundary work
+ 2 m&t - 2 A Transfer by flow
~" ^tAgen Exergy destruction (10.36)
Energy Balance Equation
m 365
Here we recognize the appearance of the specific flow exergies, ip from Eq. 10.24,
*Pi = t - h - T Q (Si - s )
$e = Kt s -h~ T (s e - s Q ) (10.37)
as the terms associated with the flow. The rate equation for exergy can verbally be stated
as all the other balance equations
Rate of exergy storage - Transfer by heat + Transfer by shaft/boundary work
+ Transfer by flow - Exergy destruction
and we notice that all the transfers take place with some surroundings and thus do not add
up to any net change when the total world is considered. Only the exergy destruction due
to entropy generation lowers the overall exergy level, and we can thus identify the regions
in space where this occurs as the locations that have entropy generation. The exergy de-
struction is identical to the previously defined term, irreversibility.
Let us briefly consider some of the special cases covered in Section 10.1. First look
at the steady-flow control volume with a single flow in and out. For steady state we have
no storage of mass, energy, entropy, or exergy, and the volume is constant, so solve for
the work term in Eq. 10.36 as
Kv. = 2 - y) + ~ ^ ~ T ^ ( 103S )
Now divide by the mass flow rate to give the specific work
W = 2 ~ y) ?c.v. + Vgen (10.39)
The reference state offset in the two specific flow exergies drop out as
& " & = h m , - // - Tfa ~ So) - [h tot e -h Q - T (s e - so)]
and the specific work becomes
w = 2 - y) + h ™ i ~ A «* * ~ T ^ s i ~ s e) - T o* g ™
= w rev - Ttf^ (10.40)
It reduces to the expression for reversible work in Eq. 10.9 when there is a single heat-
transfer term and no entropy generation. We can also see how much lower the actual spe-
cific work is due to the entropy generation; namely, it equals the reversible specific work
minus the specific exergy destruction.
EXAMPLE 10,7 Let us look at the flows and fluxes of exergy for the feedwater heater in Example 10.1.
The feedwater heater has a single flow, two heat transfers, and no work involved. When
we do the balance of terms in Eq. 10.39 and evaluate the flow exergies from Eq. 10.37,
we need the reference properties (take saturated liquid instead of 100 kPa at 25°C):
Table B.I.I: h a = 104.87 kj/kg, s = 0.3673 kJ/kgK
CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
FIGURE 10.16
Fluxes, flows, and
destruction of exergy in
the feedwater heater.
The flow exergies become
= 171.97 - 104.87 - 298.2 X (0.5705 - 0.3687) = 6.92 kJ/kg
& = Awe - T<>{s e - s )
= 765.25 - 104.87 - 298.2 X (2.1341 - 0.3687) = 133.94 kJ/kg
and the exergy fluxes from each of the heat transfers are
(l-g^-(^ll) 180 = 36 - 17kJ/kg
1 " t) qi = { l ~ ^) 413 * 28 = 152,84 ^
The destruction of exergy is then the balance (w = 0) of Eq. 10.39 as
= 36.17 + 152.84 + 6.92 - 133.94 = 62.0kJ/kg
We can now express the heater's second-law efficiency as
h.v. 36.17 + 152.84 - 62.0
4>
VlnA law ~
~ source
The exergy fluxes are shown in Fig. 10.16, and the second-law efficiency shows that
there is a potential for improvement. We should lower the temperature difference be-
tween the source and the water flow by adding more energy from the low-temperature
source, thus decreasing the irreversibility.
36.17 + 152.84
= 0.67
The no flow steady-state control volume of a heat engine, heat pump, electric heat-
ing element, or refrigerator has no storage, and the work is given by Eq. 10.38 with m -
'gen
(10.41)
Energy Balance Equation El 367
The work equals the Carnot heat engine work reduced by the exergy destruction of any ir-
reversible process occurring inside the control volume. For the case of a single heat trans-
fer and a reversible process, the work reduces to that given in Eq. 10.1 except it is
expressed as a rate in Eq. 10.41.
EXAMPLE 10.8
Assume a 500 W heating element in a stove with an element surface temperature of
1000 K. On top of the element is a ceramic top with a top surface temperature of 500
K, where both of these are shown in Fig. 10,17. Let us disregard any heat transfer
downwards and follow the flux of exergy and find the exergy destruction in the
process.
Solution
Take just the heating element as a control volume in steady state with electrical work
going in and heat transfer going out.
Energy Eq.: 0=^,^-0^
Entropy Eq.: = -f* +
air &
(1 j ) Qout ( ^electrical) — ^gen
Exergy Eq.:
From the balance equations we get
0out = Electrical = 500 W
4« = QoJTsrt = 500 W/1000 K = 0.5 W/K
^de^ction = Tjs^ = 298.15 K X 0.5 W/K = 149 W
so the heating element receives 500 W of exergy ftux, destroys 149 W, and gives out the
balance of 35 1 W with the heat transfer at 1000 K.
FIGURE 10.17 The
electric heating element
and ceramic top of a
stove.
C.V..
c.v.
368 M CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
FIGURE 10.18 The
fluxes and destruction
terms of exergy.
If-el - ^source
>
top flux
^destr. ceramic
'^destr. element
Take a second control volume from the heating element surface to the ceramic stove
top. Here heat transfer comes in at 1000 K and leaves at 500 K withno work vnvolved.
Energy Eq.:
=
2out
Entropy Eq.:
=
£?out
^top
+ S&n
Exergy Eq,:
=
(-
fin"
Mop
From the energy equation we see that the two heat transfers are equal, and the entropy
generation then becomes
1
top
5UU V500
1000
W/K - 0.5 W/K
The terms in the exergy equation become
0-fl
298.15 ^ 500 w - ( 1 - ) 500 W - 298.15 K X 0.5 W/K
1000 / \
Ul
= 351 W- 202 W- 149 W
This means that the top layer receives 35 1 W of exergy from the electric heating element
and gives out 202 W from the top surface, having destroyed 149 W of exergy m the
process. The flow of exergy and its destruction are illustrated mFig. 10.18.
A control mass going through a process starting at state 1 and ending at state 2 is
described in terms of finite changes by integration of the rate equation over time. With no
mass flow rates, time integration of Eq. 10.36 gives
* 2 - *. - } 2 (i - y) 2- * - M + Pf$7t ~ vt - r 0l s 2gw (io.42)
The reference state properties (e, v, and s ) drop out in the storage term, which becomes
With a single heat transfer taking place at a constant T Hi the available work becomes
l W 2 - Po(V 2 - V x ) =(i~t) & - (** " *») - (10.44)
KEY CONCEPTS AND FORMULAS H 369
so it is equal to the exergy from the heat transfer lowered by the increase in the stored ex-
ergy and the amount of exergy destroyed in the process. This result for a reversible
process is identical to the reversible work inEq. 10.17 minus the work to the ambient.
NUMMARY
Work out of a Carnot-cycle heat engine is the available energy in the heat transfer from
the hot source; the heat transfer to the ambient is unavailable. When an actual device is
compared to an ideal device with the same flows and states in and out, we get to the con-
cept of reversible work and exergy (availability). The reversible work is the maximum
work we can get out of a given set of flows and heat transfers or, alternatively, the mini-
mum work we have to put into the device. The comparison between the actual work and
the theoretical maximum work gives a second-law efficiency. When exergy (availability)
is used, the second-law efficiency can also be used for devices that do not involve shaft-
work such as heat exchangers. In that case, we compare the exergy given out by one flow
to the exergy gained by the other flow, giving a ratio of exergies instead of energies used
for the first-law efficiency. Any irreversibility (entropy generation) in a process does de-
stroy exergy (availability) and is undesirable. The concept of available work can be used
to make a general definition of exergy as being the reversible work minus the work that
must go to the ambient. From this definition we can construct the exergy balance equation
and apply it to different control volumes. From a design perspective we can then focus on
the flows and fluxes of exergy and improve the processes that destroy exergy.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Understand the concept of available energy.
• Understand that energy and availability are different concepts.
• Be able to conceptualize the ideal counterpart to an actual system and find the re-
versible work and heat transfer in the ideal system.
• Understand the difference between a first-law and a second-law efficiency.
• Relate the second-law efficiency to the transfer and destruction of availability.
• Be able to look at flows (fluxes) of exergy.
• Determine irreversibilities as the destruction of exergy.
• Know that destruction of exergy is due to entropy generation.
• Know that transfers of exergy do not change total or net exergy in the world.
• Know that the exergy equation is based on the energy and entropy equations and
thus does not add another law.
^£°™ r EPTS A ™ laWe ™* fr om heat W -Q { 1 - £
and Formulas ^ V t s
Reversible flow work with extra q$ v
from ambient at T Q and q in at T H q 1 ^ = T Q (s e - j,) - q
370 B CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
Flow irreversibility
Reversible work CM.
Irreversibility CM.
Second-law efficiency
Exergy, flow availability
Exergy, stored
Exergy transfer by heat
Exergy transfer by flow
Exergy rate Eq.
Exergy Eq. CM.
i = >v iev - w = qT = T S s Jm - T s gsa
7 )2ndlaw
Applied ^supplied
r
^transfer " 9 \ 1 7^
■> transfer tot
10.1 Can I have any energy transfer as heat transfer
that is 100% available?
10.2 Is energy transfer as work 100% available?
10.3 We cannot create or destroy energy, but how
about available energy?
10 4 Energy can be stored as internal energy, potential
energy, or kinetic energy. Are those energy forms
all 100% available?
10.5 Is all the energy in the ocean available?
10.6 Does a reversible process change the availability
if there is no work involved?
10.7 Is the reversible work between two states the
same as ideal work for the device?
10.8 When is the reversible work the same as the isen-
tropic work?
10.9 If 1 heat some cold liquid water to T , do I in-
crease its availability?
10.10 Are reversible work and availability (exergy)
connected?
10 11 Consider availability (exergy) associated with a
flow The total exergy is based on the thermody-
namic state and the kinetic and potential energies.
Can they all be negative?
10 12 A flow of air at 1000 kPa, 300 K, is throttled to
500 kPa. What is the irreversibility? What is the
drop in flow availability?
10 13 A steam turbine inlet is at 1200 kPa, 500»C. The
actual exit is at 300 kPa, with an actual work of
407 kJ/kg. What is its second-law efficiency?
10 14 A heat exchanger increases the availability of 3
kg/s water by 1650 kJ/kg using 10 kg/s air com-
ing in at 1400 K and leaving with 600 kJ/kg less
availability. What are the irreversibility and the
second-law efficiency?
10 15 A heat engine receives 1 kW heat transfer at 1000
K and gives out 600 W as work with the rest as
heat transfer to the ambient. What are the fluxes
of exergy in and out?
Homework Problems II 371
10.16 A heat engine receives 1 k\V heat transfer at 1000
K and gives out 600 W as work with the rest as
heat transfer to the ambient. Find its first- and
second-law efficiencies.
10.17 Is the exergy equation independent of the energy
and entropy equations?
10.18 A heat pump has a coefficient of performance of 2
using a power input of 2 kW. lis low temperature
is 20°C and the high temperature is 80°C, with an
* ambient at T . Find the fluxes of exergy associ-
ated with the energy fluxes in and out.
10.19 Use the exergy balance equation to find the effi-
ciency of a steady-state Camot heat engine oper-
ating between two fixed temperature reservoirs.
10.20 Find the second-law efficiency of the heat pump
in Problem 10.18.
Homework Problems
Available Energy, Reversible Work
10.21 Find the availability of 100 kW delivered at 500
K when the ambient temperature is 300 K.
10.22 A control mass gives out 10 kJ of energy in the
form of
a. Electrical work from a battery.
b. Mechanical work from a spring.
c. Heat transfer at 500°C.
Find the change in availability of the control mass
for each of the three cases.
10.23 A heat engine receives 5 kW at 800 K and 10 kW
at 1000 K, rejecting energy by heat transfer at 600
K. Assume it is reversible and find the power out-
put. How much power could be produced if it
could reject energy at T = 298 K?
10.24 The compressor in a refrigerator takes refrigerant
R-134a in at 100 kPa, -20°C, and compresses it
to 1 MPa, 40°C. With the room at 20°C find the
minimum compressor work.
10.25 Find the specific reversible work for a steam tur-
bine with inlet at 4 MPa and 500°C and an actual
exit state of 100 kPa, x= 1.0 with 25°C ambient
surroundings.
10.26 Calculate the reversible work out of the two-stage,
turbine shown in Problem 6.82, assuming the am-
bient is at 25°C. Compare this to the actual work
that was found to be 18.08 MW.
10.27 A household refrigerator has a freezer at T F and a
cold space at T c from which energy is removed and
rejected to the ambient at T A as shown in Fig.
P10.27. Assuming that the rate of heat transfer from
the cold space, Q c , is the same as from the freezer,
Q Fi find an expression for the minimum power into
the heat pump. Evaluate this power when T A =
20°C, T c = 5°C, T F = - 10°C, and Q F = 3 kW.
w
5>
I
\
Qa
o
/
\
Qf
<
Qc
\
FIGURE P10.27
10.28 Find the specific reversible work for an R-134a
compressor with inlet state of ~20°C, 100 kPa
and an exit state of 600 kPa, 50°C. Use a 25°C
ambient temperature.
10.29 An air compressor takes air in at the state of the
surroundings, 100 kPa, 300 K. The air exits at 400
kPa, 200°C, at the rate of 2 kg/s. Determine the
minimum compressor work input.
10.30 A steam turbine receives steam at 6 MPa, 800°C.
It has a heat loss of 49.7 kJ/kg and an isentropic
efficiency of 90%. For an exit pressure of 15 kPa
and surroundings at 20°C, find the actual work
and the reversible work between the inlet and the
exit.
10.31 An air compressor receives atmospheric air at
T = 17°C, 100 kPa, and compresses it up to 1400
kPa. The compressor has an isentropic efficiency
of 88%, and it loses energy by heat transfer to the
atmosphere as 10% of the isentropic work. Find
the actual exit temperature and the reversible
work.
372 M CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
10.32 Air flows through a constant pressure heating de-
vice, shown in Fig. P10.32. It is heated up in a re-
versible process with a work input of 200 kJ/kg
air flowing. The device exchanges heat with the
ambient at 300 K. The air enters at 300 K, 400
kPa. Assuming constant specific heat, develop an
expression for the exit temperature and solve for
it by iterations.
Tn, Pa
®
A
96
FIGURE P10.32
"\
10.33 A piston/cylinder has forces on the piston so it
keeps constant pressure. It contains 2 kg of ammo-
nia at 1 MPa, 40°C, and is now heated to 100°C by
a reversible heat engine that receives heat from a
200°C source. Find the work out of the heat engine.
10.34 A rock bed consists of 6000 kg granite and is at
70°C. A small house with lumped mass of 12 000
kg wood and 1000 kg iron is at 15°C. They are
now brought to a uniform final temperature with
no external heat transfer by connecting the house
and rock bed through some heat engines. If the
process is reversible, find the final temperature
and the work done during the process.
10.35 An airflow of 5 kg/min at 1500 K, 125 kPa, goes
through a constant pressure heat exchanger, giving
energy to a heat engine shown in Fig. P10.35. The
air exits at 500 K, and the ambient is at 298 K, 100
kPa. Find the rate of heat transfer delivered to the
engine and the power the engine can produce.
Irreversibility
10.36 Calculate the irreversibility for the condenser in
Problem 9.53 assuming an ambient temperature at
17°C.
10.37 A constant-pressure piston/cylinder contains 2 kg
of water at 5 MPa and 100°C. Heat is added from
a reservoir at 700°C to the water until it reaches
700°C. We want to find the total irreversibility in
the process.
10.38 Calculate the reversible work and irreversibility
for the process described in Problem 5.97, assum-
ing that the heat transfer is with the surroundings
at20°C.
10.39 A supply of steam at 100 kPa, 150°C, is needed in
a hospital for cleaning purposes at a rate of 15
kg/s. A supply of steam at 150 kPa, 250°C, is
available from a boiler, and tap water at 100 kPa,
15°C, is also available. The two sources are then
mixed in a mixing chamber to generate the de-
sired state as output. Determine the rate of irre-
versibility of the mixing process.
10.40 The throttle process in Example 6,5 is an irre-
versible process. Find the reversible work and ir-
reversibility assuming an ambient temperature at
25°C.
10.41 Two flows of air both at 200 kPa of equal flow
rates mix in an insulated mixing chamber. One
flow is at 1500 K, and the other is at 300 K. Find
the irreversibility in the process per kilogram of
air flowing out.
10.42 Fresh water can be produced from saltwater by
evaporation and subsequent condensation. An ex-
ample is shown in Fig. P10.42 where 150-kg/s
saltwater, state 1, comes from the condenser in a
large power plant. The water is throttled to the
saturated pressure in the flash evaporator, and the
vapor, state 2, is then condensed by cooling with
sea water. As the evaporation takes place below
atmospheric pressure, pumps must bring the liq-
uid water flows back up to P . Assume that the
saltwater has the same properties as pure water,
the ambient is at 20°C, and there are no external
heat transfers. With the states as shown in the fol-
lowing table find the irreversibility 'in the throt-
tling valve and in the condenser.
FIGURE P10.35
State
7/fC]
1 2
30 25
3
25
4 5 6
— 23 —
7
17
20
Homework Problems B 373
brine
FIGURE P10.42
Fresh
water
10.43 Calculate the irreversibility for the process de-
scribed in Problem 6.133, assuming that heat
transfer is with the surroundings at 17°C.
10.44 A 2-kg piece of iron is heated from room temper-
ature 25°C to 400°C by a heat source at 600°C.
What is the irreversibility in the process?
10.45 Air enters the turbocharger compressor (see Fig.
PI 0.45) of an automotive engine at 100 kPa,
30°C, and exits at 170 kPa. The air is cooled by
50°C in an intercooler before entering the engine.
The isentropic efficiency of the compressor is
75%. Determine the temperature of the air enter-
ing the engine and the irreversibility of the
compression-cooling process.
10.46 A 2 kg/s flow of steam at 1 MPa and 700°C
■ should be brought to 500°C by spraying in liquid
water at 1 MPa and 20°C in a steady flow. Find
the rate of irreversibility, assuming that the sur-
roundings are at 20°C.
10.47 A car air-conditioning unit has a 0.5-kg aluminum
storage cylinder that is sealed with a valve, and it
contains 2 L of refrigerant R-134a at 500 kPa and
both are at room temperature 20°C. It is now in-
stalled in a car sitting outside where the whole
system cools down to ambient temperature at
— 10°C. What is the irreversibility of this process?
10.48 The high-temperature heat source for a cyclic heat
engine is a steady-flow heat exchanger where
R-134a enters at 80°C, saturated vapor, and exits
at 80°C, saturated liquid at a flow rate of 5 kg/s.
Heat is rejected from the heat engine to a steady-
flow heat exchanger where air enters at 150 kPa
and ambient temperature 20°C, and exits at 125
kPa, 70°C. The rate of irreversibility for the over-
all process is 175 kW. Calculate the mass flow
rate of the air and the thermal efficiency of the
heat engine.
80°C
sat. vapor
80°C
sat. liquid
20°C
125 kPa
yf\f\f\f\f**- — - — | — Air
70°C
FIGURE P10.4S
FIGURE P10.48
10.49 A rigid container with volume 200 L is divided
into two equal volumes by a partition. Both sides
contain nitrogen, one side is at 2 MPa, 300°C, and
the other at 1 MPa, 50°C. The partition ruptures,
and the nitrogen comes to a uniform state at
100°C. Assuming the surroundings are at 25°C,
find the actual heat transfer and the irreversibility
in the process.
374 H CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
10.50 A rock bed consists of 6000 kg granite and is at
70°C. A small house with lumped mass of 12 000
kg wood and 1000 kg iron is at 15°C. They are now
brought to a uniform final temperature by circulat-
ing water between the rock bed and the house. Find
the final temperature and the irreversibility in the
process, assuming an ambient at 15°C.
Availability (exergy)
10.51 A steady stream of R-22 at ambient temperature,
10°C, and at 750 kPa enters a solar collector. The
stream exits at 80°C and 700 kPa. Calculate the
change in availability of the R-22 between these
two states.
10.52 Consider the springtime melting of ice in the
mountains, which gives cold water running in a
river at 2°C while the air temperature is 20°C.
What is the availability of the water relative to the
ambient temperature?
A geothermal source provides 10 kg/s of hot
water at 500 kPa and 150°C flowing into a flash
evaporator that separates vapor and liquid at 200
kPa. Find the three fluxes of availability (inlet and
two outlets) and the irreversibility rate.
10.53
-4— *»- Vapor
■ Liquid
10.58 A steady combustion of natural gas yields 0.15
kg/s of products (having approximately the same
properties as air) at 1100°C and 100 kPa. The
products are passed through a heat exchanger and
exit at 550°C. What is the maximum theoretical
power output from a cyclic heat engine operating
on the heat rejected from the combustion prod-
ucts, assuming that the ambient temperature is
20°C?
10.59 Find the change in availability from inlet to exit
of the condenser in Problem 9.42.
10.60 Refrigerant R-12 at 30°C, 0.75 MPa, enters a
steady-flow device and exits at 30°C, 100 kPa.
Assume the process is isothermal and re-
versible. Find the change in availability of the
refrigerant.
10.61 An air compressor is used to charge an initially
empty 200-L tank with air up to 5 MPa. The air
inlet to the compressor is at 100 kPa, 17°C, and
the compressor isentropic efficiency is 80%. Find
the total compressor work and the change in
availability of the air.
10.62 Water as saturated liquid at 200 kPa goes through
a constant pressure heat exchanger as shown in
Fig. PI 0.62. The heat input is supplied from a
reversible heat pump extracting heat from the
surroundings at 17°C. The water flow rate is
2 kg/min and the whole process is reversible, that
is, there is no overall net entropy change. If the
heat pump receives 40 kW of work, find the water
exit state by iteration and the increase in avail-
ability of the water.
FIGURE P10.53
10.54 Find the availability at all four states in the
power plant of Problem 9.42 with an ambient
temperature at 298 K.
10.55 Air flows at 1500 K and 100 kPa through a con-
stant-pressure heat exchanger giving energy to a
heat engine and comes out at 500 K. At what
constant temperature should the same heat trans-
fer be delivered to provide the same availability?
10.56 Calculate the change in availability (kW) of the
two flows in Problem 9.61.
10.57 Nitrogen flows in a pipe with a velocity of 300
m/s at 500 kPa and 300°C. What is its availabil-
ity with respect to ambient surroundings at 100
kPa and20°C?
L surr
FIGURE P10.62
HOMEWORK PROBLEMS H 375
10.63 An electric stove has one heating element at
300°C getting 500 W of electric power. It trans-
fers 90% of the power to 1 kg water in a kettle
initially at ambient 20°C, 100 kPa; the remaining
10% leaks to the room air. The water at a uniform
T is brought to the boiling point. At the start of
the process, what is the rate of availability trans-
fer by (a) electrical input, (b) from heating ele-
ment, and (c) into the water at r^, er ?
10.64 Calculate the availability of the water at the initial
and final states of Problem 8.70, and the irre-
versibility of the process.
10.65 A 10-kg iron disk brake on a car is initially at
10°C. Suddenly the brake pad hangs up, increas-
ing the brake temperature by friction to 110°C
while the car maintains constant speed. Find the
change in availability of the disk and the energy
depletion of the car's gas tank due to this process
alone. Assume that the engine has a thermal effi-
ciency of 35%.
10.66 A 1-kg block of copper at 350°C is quenched in a
10-kg oil bath initially at ambient temperature of
20°C. Calculate the final uniform temperature (no
heat transfer to/from ambient) and the change of
availability of the system (copper and oil).
10.67 Calculate the availability of the system (aluminum
plus gas) at the initial and final states of Problem
8.137, and also the process irreversibility.
10.68 A wooden bucket (2 kg) with 10 kg hot liquid
water, both at 85°C, is lowered 400 m down into a
mineshaft. What is the availability of the bucket
and water with respect to the surface ambient at
20°C?
Device Second-Law Efficiency
10.69 Air enters a compressor at ambient conditions, 100
kPa and 300 K, and exits at 800 kPa. If the isen-
tropic compressor efficiency is 85%, what is the
second-law efficiency of the compressor process?
10.70 A compressor takes in saturated vapor R-134a
at -20°C and delivers it at 30°C and 0.4 MPa.
Assuming that the compression is adiabatic,
find the isentropic efficiency and the second-
law efficiency.
10.71 A steam turbine has inlet at 4 MPa and 500°C and
actual exit of 100 kPa with* = 1.0. Find its first-
law (isentropic) and its second-law efficiencies.
10.72 The condenser in a refrigerator receives R-134a at
. 700 kPa and 50°C, and it exits as saturated liquid
at 25°C. The flow rate is 0.1 kg/s, and the con-
denser has air flowing in at an ambient tempera-
ture of 15°C and leaving at 35°C. Find the
minimum flow rate of air and the heat exchanger
second-law efficiency.
10.73 Steam enters a turbine at 25 MPa, 550°C and
exits at 5 MPa, 325°C at a flow rate of 70 kg/s.
Determine the total power output of the turbine,
its isentropic efficiency, and the second-law
efficiency.
10.74 A compressor is used to bring saturated water
vapor at 1 MPa up to 17.5 MPa, where the actual
exit temperature is 650°C. Find the irreversibility
and the second-law efficiency.
10.75 A flow of steam at 10 MPa, 550°C, goes through a
two-stage turbine. The pressure between the
stages is 2 MPa, and the second stage has an exit
at 50 kPa. Assume both stages have an isentropic
efficiency of 85%. Find the second-law efficien-
cies for both stages of the turbine.
10.76 The simple steam power plant shown in Problem
6.99 has a turbine with given inlet and exit states.
Find the availability at the turbine exit, state 6.
Find the second-law efficiency for the turbine, ne-
glecting kinetic energy at state 5.
10.77 A steam turbine inlet is at 1200 kPa, 500°C.
The actual exit is at 200 kPa, 300°C. What are
the isentropic efficiency and its second-law
efficiency?
10.78 Steam is supplied in a line at 3 MPa, 700°C. A
turbine with an isentropic efficiency of 85% is
connected to the line by a valve, and it ex-
hausts to the atmosphere at 100 kPa. If the
steam is throttled down to 2 MPa before enter-
ing the turbine, find the actual turbine specific
work. Find the change in availability through
the valve and the second-law efficiency of the
turbine.
10.79 Air flows into a heat engine at ambient conditions
100 kPa, 300 K, as shown in Fig. PI 0.79. Energy
is supplied as 1200 kJ/kg air from a 1500-K.
source, and in some part of the process a heat-
transfer loss of 300 kJ/kg air occurs at 750 K. The
air leaves the engine at 100 kPa, 800 K. Find the
first- and second-law efficiencies.
376 ■ CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
FIGURE P10.79
10.80 Air enters a steady-flow turbine at 1600 K and ex-
hausts to the atmosphere at 1000 K. The second-
law efficiency is 85%. What is the turbine inlet
pressure?
10.81 Calculate the second-law efficiency of the coun-
terflowing heat exchanger in Problem 9.61 with
an ambient temperature at 20°C.
10.82 Calculate the second-law efficiency of the coflow-
ing heat exchanger in Problem 9.62 with an ambi-
ent temperature at 17°C.
10.83 A heat exchanger brings 10 kg/s water from
100°C to 500°C at 2000 kPa using air coming
in at 1400 K and leaving at 460 K. What is the
second-law efficiency?
600 K
Wail
Q
400 K
Exergy Balance Equation
10.84 Find the specific flow exergy in and out of the
steam turbine in Example 9.1 assuming an ambi-
ent at 293 K. Use the exergy balance equation to
find the reversible specific work. Does this calcu-
lation of specific work depend on T 7
10.85 A counterflowing heat exchanger cools air at 600
K, 400 kPa, to 320 K using a supply of water at
20°C, 200 kPa. The water flow rate is 0.1 kg/s,
and the airflow rate is 1 kg/s. Assume this can be
done in a reversible process by the use of heat en- -
gines and neglect kinetic energy changes. Find
the water exit temperature and the power out of
the heat engine(s).
10.86 Evaluate the steady-state exergy fluxes due to a
heat transfer of 250 W through a wall with 600 K
on one side and 400 K on the other side, shown in
Fig. P10.86. What is the exergy destruction in the
wall?
FIGURE P10.86
10.87 A heat engine operating with an environment at
298 K produces 5 kW of power output with a
first-law efficiency of 50%. It has a second-law
efficiency of 80% and T L = 310 K. Find all the
energy and exergy transfers in and out.
10.88 Consider the condenser in Problem 9.42. Find the
specific energy and exergy that are given out, as-
suming an ambient at 20°C. Also find the specific
exergy destruction in the process.
10.89 The condenser in a power plant cools 10 kg/s
water at 10 kPa, quality 90%, so that it comes
out as saturated liquid at 100 kPa. The cooling
is done by ocean water coming in at ambient
15°C and returned to the ocean at 20°C. Find the
transfer out of the water and the transfer into the
ocean water of both energy and exergy (4
terms).
10.90 Use the exergy equation to analyze the compres-
sor in Example 6.10 to find its second-law effi-
ciency assuming an ambient at 20°C.
10.91 Consider the car engine in Example 7.1 and as-
sume the fuel energy is delivered at a constant
1500 K. The 70% of the energy that is lost is
40% exhaust flow at 900 K, and the remainder
30% heat transfer to the walls at 450 K goes on
to the coolant fluid at 370 K, finally ending up
in atmospheric air at ambient 20°C. Find all the
energy and exergy flows for this heat engine.
Also find the exergy destmction and where that
is done.
10.92 Estimate some reasonable temperatures to use and
find all the fluxes of exergy in the refrigerator
given in Example 7.2.
10.93 Use the exergy equation to evaluate the exergy
destruction for Problem 10.44.
10.94 Use the exergy balance equation to solve for the
work in Problem 10.33.
Homework problems M 311
Review Problems
10.95 A small air gun has 1 cm 3 air at 250 kPa, 27°C.
The piston is a bullet of mass 20 g. What is the
potential highest velocity with which the bullet
can leave?
10.96 Calculate the reversible work and irreversibility
for the process described hi Problem 5.134 as-
suming that the heat transfer is with the sur-
roundings at 20°C.
10.97 A piston/cylinder arrangement has a load on the
piston so it maintains constant pressure. It con-
tains 1 kg of steam at 500 kPa, 50% quality.
Heat from a reservoir at 700°C brings the steam
to 600°C. Find the second-law efficiency for this
process. Note that no formula is given for this
particular case, so determine a reasonable ex-
pression for it,
10.98 Consider the high-pressure closed feedwater
heater in the nuclear power plant described
in Problem 6,102. Determine its second-law
efficiency.
10.99 Consider a gasoline engine for a car as a steady
device where air and fuel enter at the surround-
ing conditions 25°C, 100 kPa, and leave the en-
gine exhaust manifold at 1000 K, 100 kPa, as
products assumed to be air. The engine cooling
system removes 750 kJ/kg air through the en-
gine to the ambient. For the analysis, take the
fuel as air where the extra energy of 2200 kJ/kg
of air released in the combustion process is
added as heat transfer from a 1 800-K reservoir.
Find the work out of the engine, the irreversibil-
ity per kilogram of air, and the first- and second-
law efficiencies.
10.100 Consider the nozzle in Problem 9.112. What is
the second-law efficiency for the nozzle?
10.101 Air in a piston/cy finder arrangement is at lit)
kPa, 25°C, with a volume of 50 L. It goes
through a reversible polytropic process to a final
state of 700 kPa, 500 K, and exchanges heat with
the ambient at 25°C through a reversible device.
Find the total work (including the external de-
vice) and the heat transfer from the ambient.
10.102 Consider the irreversible process in Problem
8.128, Assume that the process could be done
reversibly by adding heat engines/pumps be-
tween tanks A and B and the cylinder. The total
system is insulated, so there is no heat transfer to
or from the ambient. Find the final state, the
work given out to the piston, and the total work
to or from the heat engines/pumps.
10.103 Consider the heat engine in Problem 10.79. The
exit temperature was given as 800 K, but what
are the theoretical limits for this temperature?
Find the lowest and the highest, assuming the
heat transfers are as given. For each case give
the first- and second-law efficiency.
10.104 Air in a piston/cylinder arrangement, shown in
Fig. PI 0.1 04, is at 200 kPa, 300 K, with a vol-
ume of 0.5 m 3 . If the piston is at the stops, the
volume is 1 m 3 and a pressure of 400 kPa is re-
quired. The air is then heated from the initial
state to 1500 K by a 1900 K reservoir. Find the
total irreversibility in the process, assuming sur-
roundings are at 20°C.
1
Air:
A
102
/ T <*s FIGURE P10.104
10.105 A jet of air at 200 m/s flows at 25°C, 100 kPa,
towards a wall where the jet flow stagnates and
leaves at very low velocity. Consider the process
to be adiabatic and reversible. Use the exergy
equation and the second law to find the stagna-
tion temperature and pressure.
• 2
FIGURE P10.105
378 S CHAPTER TEN IRREVERSIBILITY AND AVAILABILITY
10 106 Consider the light bulb in Problem 8.123. What and the entire room including the bulb? The
10.106 Conner me 1 gn ^ ^ ^ gas lhe g| the
"enled? What is to exergy destruction in bulb, but it gets absorbed on to room waU,
the filament, the entire bulb including the glass,
English Unit problems
Concept Problems
10.107E A flow of air at 150 psia, 540 R 5 is throttled to
75 psia. What is the irreversibility? What is the
drop in flow availability?
10.108E A heat exchanger increases the availability of
6 Ibm/s water by 800 Btu/lbm using 20 lbm/s
air coming in at 2500 R and leaving with 250
Btu/lbm less availability. What is the irre-
versibility and the second-law efficiency?
10.109E A heat engine receives 3500 Btu/h heat trans-
fer at 1 800 R and gives out 2000 Btu/h as work
with the rest as heat transfer to the ambient.
What are the fluxes of exergy in and out?
10.110E A heat engine receives 3500 Btu/h heat trans-
fer at 1 800 R and gives out 2000 Btu/h as work
with the rest as heat transfer to the ambient.
Find its first- and second-law efficiencies.
10.111E A heat pump has a coefficient of performance
of 2 using a power input of 15000 Btu/h. Its
low temperature is T , and the high tempera-
ture is 180 F, with ambient at T . Find the
fluxes of exergy associated with the energy
fluxes in and out?
10.112E Find the second-law efficiency of the heat
pump in Problem 10.111.
English Unit Problems
10.113E A control mass gives out 1000 Btu of energy in
the form of
a. Electrical work from a battery
b. Mechanical work from a spring
c. Heat transfer at 700 F
Find the change in availability of the control
mass for each of the three cases.
10.114E The compressor in a refrigerator takes refriger-
ant R-134a in at 15 lbf/in. 2 , F, and com-
presses it to 125 lbf/in. 2 , 100 F. With the room
at 70 F, find the reversible heat transfer and the
minimum compressor work.
10.1 15E A heat engine receives 15 000 Btu/h at 1400 R
and 30 000 Btu/h at 1800 R, rejecting energy
by heat transfer at 900 R. Assume it is re-
versible and find the power output. How much
power could be produced if it could reject en-
ergy at T = 540 R?
10.116E Air flows through a constant-pressure heating
device as shown in Fig. P10.32. It is heated up
in a reversible process with a work input of 85
Btu/lbm air flowing. The device exchanges
heat with the ambient at 540 R. The air enters
at 540 R, 60 lbf/in. 2 . Assuming constant spe-
cific heat, develop an expression for the exit
temperature and solve for it.
10.117E A rock bed consists of 12 000 lbm granite and
is at 160 F. A small house with lumped mass of
24 000 lbm wood and 2000 lbm iron is at 60 F.
They are now brought to a uniform final tem-
perature with no external heat transfer by con-
necting the house and rock bed through some
heat engines. If the process is reversible, find
the final temperature and the work done during
the process.
10.118E A constant-pressure piston/cylinder contains 4
lbm of water at 1000 psia and 200 F. Heat is
added from a reservoir at 1300 F to the water
until it reaches 1300 F. We want to find the
total irreversibility in the process.
10.1 19E A supply of steam at 14.7 lbf/in. 2 , 320 F, is
needed in a hospital for cleaning purposes at a
rate of 30 lbm/s. A supply of steam at 20
lbf/in. 2 , 500 F, is available from a boiler, and
tap water at 14.7 lbf/in. 2 , 60 F, is also avail-
able. The two sources are then mixed in a mix-
ing chamber to generate the desired state as
output. Determine the rate of irreversibility of
the mixing process.
10.120E Fresh water can be produced from saltwater by
evaporation and subsequent condensation. An
example is shown in Fig. P10.42 where 300-
lbm/s saltwater, state 1, comes from the con-
English Unit Problems S 379
denser in a large power plant. The water, is
throttled to the saturated pressure in the flash
evaporator, and the vapor, state 2, is then con-
densed by cooling with sea water. As the evap-
oration takes place below atmospheric pressure,
pumps must bring the liquid water flows back
up to P . Assume that the saltwater has the
same properties as pure water, that the ambient
is at 68 F, and that there are no external heat
transfers. With the states as shown in the fol-
lowing list find the irreversibility in the throt-
tling valve and in the condenser.
State 1 2 3 4 5 6 7 8
T\F] 86 77 77 — 74 — 63 68
10.121E Calculate the irreversibility for the process de-
scribed in Problem 6.174, assuming that the
heat transfer is with the surroundings at 61 F.
10.122E A 4-lbm piece of iron is heated from room
temperature 77 F to 750 F by a heat source at
1100 F. What is the irreversibility in the
process?
10.123E Air enters the turbocharger compressor of an
automotive engine at 14.7 lbf/in. 2 , 90 F, and
exits at 25 lbf/in. 2 , as shown in Fig. P10.45. The
air is cooled by 90 F in an intercooler before en-
tering the engine. The isentropic efficiency of
the compressor is 75%. Determine the tempera-
ture of the air entering the engine and the irre-
versibility of the compression-cooling process.
10.124E A rock bed consists of 12 000 Ibm granite and
is at 160 F. A small house with lumped mass of
24 000 lbm wood and 2000 lbm iron is at 60 F.
They are now brought to a uniform final tem-
perature by circulating water between the rock
bed and the house. Find the final temperature
and the irreversibility in the process assuming
an ambient at 60 F.
10.125E A steady stream of R-22 at ambient tempera-
ture, 50 F, and at 1 10 lbf/in. 2 enters a solar col-
lector. The stream exits at 180 F, 100 lbf/in. 2 .
Calculate the change in availability of the R-22
between these two states.
10.126E Consider the springtime melting of ice in the
mountains, which gives cold water running in
a river at 34 F while the air temperature is 68
F. What is the flow availability of the water
relative to the temperature of the ambient?
10.127E A geothermal source provides 20 Ibm/s of hot
water at 80 lbf/in. 2 , 300 F flowing into a flash
evaporator that separates vapor and liquid at 30
lbf/in. 2 . Find the three fluxes of availability
(inlet and two outlets) and the irreversibility
rate.
10.128E An air compressor is used to charge an initially
empty 7-ft 3 tank with air up to 750 lbf/in. 2 . The
air inlet to the compressor is at 14.7 lbf/in 2 ,
60 F, and the compressor isentropic efficiency
is 80%. Find the total compressor work and the
change in energy of the air.
10.129E An electric stove has one heating element at
600 F getting 500 W of electric power. It trans-
fers 90% of the power to 2 lbm water in a ket-
tle initially at ambient 70 F, 1 atm; the rest,
10%, leaks to the room air. The water at a uni-
form T is brought to the boiling point. At the
start of the process, what is the rate of avail-
ability transfer by (a) electrical input, (b) from
heating element, and (c) into the water at
T ?
1 water 1
10.130E A 20-lbm iron disk brake on a car is at 50 F.
Suddenly the brake pad hangs up, increasing
the brake temperature by friction to 230 F
while the car maintains constant speed. Find
the change in availability of the disk and the
energy depletion of the car's gas tank due to
this process alone. Assume that the engine has
a thermal efficiency of 35%.
10.13 IE Calculate the availability of the system (alu-
minum plus gas) at the initial and final states of
Problem 8.183, and also the irreversibility.
10. 132 E A wood bucket (4 lbm) with 20 lbm hot liquid
water, both at 180 F, is lowered 1300 ft down
into a mineshaft. What is the availability of the
bucket and water with respect to the surface
ambient at 70 F?
10.133E A cofl owing (same direction) heat exchanger
has one line with 0.5 lbm/s oxygen at 68 F
and 30 psia entering, and the other line has
1.2 lbm/s nitrogen at 20 psia and 900 R enter-
ing. The heat exchanger is long enough so
that the two flows exit at the same tempera-
ture. Use constant heat capacities and find the
exit temperature and the second-law effi-
ciency for the heat exchanger assuming ambi-
ent at 68 F.
380 11 Chapter Ten Irreversibility and availability
10.134E A steam turbine has an inlet at 600 psia and
900 F and actual exit of 1 atm with x = 1.0.
Find its first-law (isentropic) and second-law
efficiencies.
10.135E A compressor is used to bring saturated water
vapor at 103 lbf/in. 2 up to 2000 lbf/in. 2 , where
the actual exit temperature is 1200 F. Find the
irreversibility and the second-law efficiency.
10.136E The simple steam power plant in Problem 6 J 67,
shown in Fig. P6.99 has a turbine with given inlet
and exit states. Find the availability at the turbine
exit, state 6. Find the second-law efficiency for
the turbine, neglecting kinetic energy at state 5.
10.137E Steam is supplied in a line at 400 lbf/in. 2 , 1200
F. A turbine with an isentropic efficiency of
85% is connected to the line by a valve, and it
exhausts to the atmosphere at 14.7 lbf/in. 2 . If
the steam is throttled down to 300 lbf/in. 2 be-
fore entering the turbine, find the actual turbine
specific work. Find the change in availability
through the valve and the second law effi-
ciency of the turbine.
10.138E Air flows into a heat engine at ambient condi-
tions 14.7 lbf/fn. 2 , 540 R, as shown in Fig.
P. 10.79. Energy is supplied as 540 Btu per lbm
air from a 2700 R source, and in some part of
the process a heat transfer loss of 135 Btu per
lbm air happens at 1350 R. The air leaves the
engine at 14.7 lbf/in. 2 , 1440 R. Find the first-
and second-law efficiencies.
10.139E A heat engine operating with an environment at
540 R produces 17 000 Btu/h of power output
with a first-law efficiency of 50%. It has a second-
law efficiency of 80% and T L ~ 560 R. Find all
the energy and exergy transfers in and out.
10.140E The condenser in a power plant cools 20 lbm/s
water at 120 F, quality 90%, so that it comes
out as saturated liquid at 120 F. The cooling is
done by ocean water coming in at 60 F and re-
turned to the ocean at 68 F. Find the transfer
out of the water and the transfer into the ocean
water of both energy and exergy (4 terms).
10.141E Calculate the reversible work and irreversibil-
ity for the process described in Problem 5.168
assuming that the heat transfer is with the sur-
roundings at 68 F.
10.142E A piston/cylinder arrangement has a load on
the piston so it maintains constant pressure. It
contains 1 lbm of steam at 80 lbf/in. 2 , 50%
quality. Heat from a reservoir at 1300 F brings
the steam to 1000 F. Find the second-law effi-
ciency for this process. Note that no formula is
given for this particular case, so determine a
reasonable expression for it.
10.143E Consider a gasoline engine for a car as a steady-
flow device where air and fuel enters at the sur-
rounding conditions 77 F, 14.7 lbf/in. 2 , and
leaves the engine exhaust manifold at 1 800 R,
14.7 lbf/in. 2 as products assumed to be air. The
engine cooling system removes 320 Btu/Ibm air
through the engine to the ambient. For the analy-
sis, take the fuel as air where the extra energy of
950 Btu/lbm of air released in the combustion
process is added as heat transfer from a 3240 R
reservoir. Find the work out of the engine, the ir-
reversibility per pound-mass of air, and the first-
and second-law efficiencies.
10.144E The exit nozzle in a jet engine receives air at.
2100 R, 20 psia, with negligible kinetic energy.
The exit pressure is 10 psia, and the actual exit
temperature is 1780 R. What is the actual exit
velocity and the second-law efficiency?
10.145E Air in a piston/cylinder arrangement, shown in
Fig. P. 10. 104, is at 30 lbf/in. 2 , 540 R, with a
volume of 20 ft 3 . If the piston is at the stops,
the volume is 40 ft 3 and a pressure of 60
lbf/in. 2 is required. The air is then heated from
the initial state to 2700 R by a 3400 R reser-
voir. Find the total irreversibility in the
process, assuming surroundings are at 70 F.
Computer, Design, and Open-Ended Problems
10.146 Use the software to get the properties of water as 4 going to the low-pressure turbine. Assume no
needed for consideration of the moisture separa- heat transfer to the surroundings at 20°C and
tor in Problem 6.102. Steam comes in at state 3 find the total entropy generation and irreversibil-
and leaves as liquid, state 9, with the rest at state ity in the process.
computer, Design and open -ended problems El 381
10.147 Use the software to get the properties of water as
needed and calculate the second-law efficiency
of the low-pressure turbine in Problem 6.102.
10.148 Write a program to solve the general case of
Problem 10.44, The initial state is to be the pro-
gram input variable, and the output should also
include the change in availability for both the
iron and the source.
10.149 Write a program to solve Problem 10.32. Use
constant specific heat and let the work input be a
program input variable.
10.150 The maximum power a windmill can possibly
extract from the wind is
Water flowing through Hoover Dam (see Prob-
lem 6.51) produces W = 0.8;«^ ter gh. Burning
1 kg of coal gives 24 000 kJ delivered at 900 K
to a heat engine. Find other examples in the lit-
erature and from problems in the previous chap-
ters with steam and gases into turbines. Make a
list of the availability (exergy) for a flow of 1
kg/s of substance with the above examples. Use
a reasonable choice for the values of the parame-
ters and do the necessary analysis.
10.151 Consider the condenser in the simple steam
power plant described in Problem 6.99. The
cooling water is lake water at 20°C, and it should
not be heated more than 5°C as it goes back to
the lake. Assume the heat-transfer rate inside the
condenser is 350 W/m 2 K so Q ~ 350 X A Afin
watts. Estimate the flow rate of the cooling water
and the needed interface area inside the con-
denser, A. Find the change in the availability of
the cooling water and the steam inside the con-
denser and compare. Discuss your estimates and
the size of the pump for the cooling water.
10.152 Consider the nuclear power plant shown in
Problem 6.102. Select one feedwater heater and
one pump and make an analysis of their perfor-
mance. Check the energy balances and do the
second-law analysis. Determine the change of
availability in all the flows and discuss measures
of performance for both the pump and the feed-
water heater.
10.153 Reconsider the use of the geothermal energy as
discussed in Problem 6.105. The analysis that was
done and the original problem statement specified
the turbine exit state as 10 kPa, 90% quality. Re-
consider this problem with an adiabatic turbine
having an isentropic efficiency of 85% and an exit
pressure of 10 kPa. Include a second-law analysis
and discuss the changes in availability. Describe
another way of using the geothermal energy and
make appropriate calculations.
10.154 An air gun should shoot a harpoon of mass 5 kg
out so that it has a velocity of 75 m/s as it leaves
the gun. The harpoon acts as the piston in a
cylinder, and air is trapped below the piston (end
of harpoon) that can be initially locked. The air is
charged so the initial state is at high pressure and
temperature. Determine sizes for the cylinder di-
ameter, cylinder length, air mass, and initial
(P,T) of the air. Make reasonable assumptions
about the process and include a determination of
the state of the air during the process.
10.155 Energy can be stored in many different forms.
Thermal energy can be stored as internal energy
in a mass like a rock bed, water, or metals. Me-
chanical energy (potential or kinetic) can be
stored in springs, rotating flywheels, elevated
masses, and the like. A tank with a compressed
gas that can drive a turbine is used. Batteries are
used in cars. Make a list with at least five differ-
ent ways of storing 1000 MJ of energy and size
the systems. Note how the energy is taken out
and find the availability for each case. Discuss
the various alternatives.
10.156 Find from the literature the amount of energy
that must be stored in a car to start the engine.
Size three different systems to provide that en-
ergy and compare those to an ordinary car bat-
tery. Discuss the feasibility and cost.
P OWER AND
REFRIGERATION
SYSTEMS
Some power plants, such as the simple steam power plant, which we have considered
several times, operate in a cycle. That is, the working fluid undergoes a series of
processes and finally returns to the initial state. In other power plants sucl : as he
Internal-combustion engine and the gas turbine, the working fluid does not go through
a thermodynamic cycle, even though the engine itself may operate m a mechanical
cycle. In this instance, the working fluid has a different composition or is in a differ-
ent state at the conclusion of the process than it had or was at the beginning. Such
equipment is sometimes said to operate on the open cycle the word cycle is really a
misnomer), whereas the steam power plant operates on a closed cycle. The same da-
tinction between open and closed cycles can be made regarding refrigeration devices.
For both the open- and closed-cycle apparatus, however, it is advantageous to analyze
the performance of an idealized closed cycle similar to the actual cycle. Such » proce-
dure is particularly advantageous for determining the influence of certain variables on
performance. For example, the spark-ignition internal-combustion engine is usually
approximated by the Otto cycle. From an analysis of the Otto cycle we conclude that
increasing the compression ratio increases the efficiency. This is also true for the ac-
tual engine, even though the Otto-cycle efficiencies may deviate significantly from
^ "TWsth^toTs'concenied with these idealized cycles for both power and refrigera-
tion apparatus. Both vapors and ideal gases are considered as working fluids. Anatten^pt
will be made to point out how the processes in actual apparatus deviate from the ideal
Consideration is also given to certain modifications of the basic cycles that are intended
to improve performance. These modifications include the use of devices such as regener-
ators multistage compressors and expanders, and mtercoolers. Vanous ; omb ^ c °f
these types of systems and also special applications, such as cogeneration of electrical
power and energy, combined cycles, topping and bottoming cycles, and binary cycle sys-
tems, are also discussed in this chapter and in the problems.
ll.i introduction to power Systems
In introducing the second law of thermodynamics in Chapter 7, we considered cyclic heat
engines consisting of four separate processes. We noted that these engines can be ^oper-
ated as steady-state devices involving shaft work, as shown m Fig. 7.16, or mstead as
382
Introduction to Power Systems 383
cylinder/piston devices involving boundary-movement work, as shown in Fig. 7.17. The
former may have a working fluid that changes phase during the processes in the cycle, or
may have a single-phase working fluid throughout. The latter type would normally have a
gaseous working fluid throughout the cycle.
For a reversible steady-state process involving negligible kinetic and potential en-
ergy changes, the shaft work per unit mass is given by Eq. 9.19,
w = - j v dP
For a reversible process involving a simple compressible substance, the boundary
movement work per unit mass is given by Eq. 4.3,
w = j P dv
The areas represented by these two integrals are shown in Fig. 11.1. It is of interest
to note that, in the former case, there is no work involved in a constant-pressure process,
while in the latter case, there is no work involved in a constant-volume process.
Let us now consider a power system consisting of four steady-state processes, as in
Fig. 7.16. We assume that each process is internally reversible and has negligible
changes in kinetic and potential energies, which results in the work for each process
being given by Eq. 9.19. For convenience of operation, we will make the two heat-trans-
fer processes (boiler and condenser) constant-pressure processes, such that those are
simple heat exchangers involving no work. Let us also assume that the turbine and pump
processes are both adiabatic, such that they are therefore isentropic processes. Thus, the
four processes comprising the cycle are as shown in Fig. 11.2. Note that if the entire
cycle takes place inside the two-phase liquid-vapor dome, the resulting cycle is the
Camot cycle, since the two constant-pressure processes are also isothermal. Otherwise,
this cycle is not a Carnot cycle. In either case, we find that the net work output for this
power system is given by
r2 r4 ri r3
ll Vt=-j vdP+O-j vdP + 0=-j vdP+ j vdP
and, since P 2 = P 3 and also Pj = P 4 , we find that the system produces a net work out-
put because the specific volume is larger during the expansion from 3 to 4 than it
is during the compression from 1 to 2. This result is also evident from the areas
p
FIGURE 11.1
Comparison of shaft work
and boundary-movement I I
work. v
384 fl Chapter, eleven power and refrigeration systems
FIGURE 11.2 Four-
process power cycle.
-/ vd P in Fig 11 2. We conclude that it would be advantageous to have this differ-
ence in specific volume be as large as possible, as, for example, the difference between
a vapor and a liquid. ,
If the four-process cycle shown in Fig. 11.2 were accomplished in a cylinder/piston
system involving boundary-movement work, then the net work output for this power sys-
tem would be given by
w„
= J 3 ' Pdv + J" ^ Pdu + j Pdu + Pdu
and from these four areas on Fig. 11.2, we note that the pressure is higher dunng any
given change in volume in the two expansion processes than in the two compression
processes, resulting in a net positive area and a net work output.
For either of the two cases just analyzed, it is noted from Fig. 1 1.2 that the net work
output of the cycle is equal to the area enclosed by the process lines 1-2-3-4-1, and this
area is the same for both cases, even though the work terms for the four individual
processes are different for the two cases.
In the next several sections, we consider the Rankine cycle, which is the ideal, tour-
steady-state process cycle shown in Fig. 1 1.2, utilizing a phase change between vapor and
liquid to maximize the difference in specific volume during expansion and compression.
This is the idealized model for a steam power plant system.
11.2 THE RANKINE CYCLE
We now consider the idealized four-steady-state-process cycle shown in Fig. 11.2, in
which state 1 is saturated liquid and state 3 is either saturated vapor or superheated vapor.
This system is termed the Rankine cycle and is the model for the simple steam power
plant. It is convenient to show the states and processes on a T-s diagram, as given in Fig.
1 1.3. The four processes are:
1- 2 : Reversible adiabatic pumping process in the pump
2- 3: Constant-pressure transfer of heat in the boiler
Reversible adiabatic expansion in the turbine (or other prime mover such as a steam
engine)
4-1: Constant-pressure transfer of heat in the condenser
The Ranking Cycle H 385
FIGURE 11.3 Simple
steam power plant that
operates on the Rankine
cycle.
TfiMfflNET
As mentioned earlier, the Rankine cycle also includes the possibility of superheat-
ing the vapor, as cycle 1-2-3 '-4 -1.
If changes of kinetic and potential energy are neglected, heat transfer and work may
be represented by various areas on the T-s diagram. The heat transferred to the working
fluid is represented by area a-2-2'-Z-b-a t and the heat transferred from the working fluid
by area a-l-4-b-a. From the first law we conclude that the area representing the work is
the difference between these two areas — area 1—2— 2'— 3^4— 1 . The thermal efficiency is
defined by the relation
-r, - w «'- area l-2~2'-3-4-l , 1lt .
Vth 9h areaa-2-2'-3-^ (1U)
For analyzing the Rankine cycle, it is helpful to think of efficiency as depending on
the average temperature at which heat is supplied and the average temperature at which
heat is rejected. Any changes that increase the average temperature at which heat is sup-
plied or decrease the average temperature heat is rejected will increase the Rankine-cycle
efficiency.
In analyzing the ideal cycles in this chapter, the changes in kinetic and potential en-
ergies from one point in the cycle to another are neglected. In general, this is a reasonable
assumption for the actual cycles.
It is readily evident that the Rankine cycle has a lower efficiency than a Carnot
cycle with the same maximum and minimum temperatures as a Rankine cycle because the
average temperature between 2 and 2' is less than the temperature during evaporation. We
might well ask, why choose the Rankine cycle as the ideal cycle? Why not select the
Carnot cycle l'-2'-3-4-I'? At least two reasons can be given. The first reason concerns
the pumping process. State 1' is a mixture of liquid and vapor. Great difficulties are en-
countered in building a pump that will handle the mixture of liquid and vapor at 1 ' and
deliver saturated liquid at 2', It is much easier to condense the vapor completely and han-
dle only liquid in the pump: The Rankine cycle is based on this fact. The second reason
concerns superheating the vapor. In the Rankine cycle the vapor is superheated at constant
pressure, process 3-3'. In the Carnot cycle all the heat transfer is at constant temperature,
and therefore the vapor is superheated in process 3-3". Note, however, that during this
process the pressure is dropping, which means that the heat must be transferred to the
vapor as it undergoes an expansion process in which work is done. This heat transfer is
also very difficult to achieve in practice. Thus, the Rankine cycle is the ideal cycle that
386 ■ Chapter eleven power and refrigeration systems
can be approximated in practice. In the following sections we will consider some varia-
Tns onTLrJcine cycle that enable it to approach more closely the effioency of the
before we discus, the influence of certain variables on the performance of the
Rankine cycle, we will study an example.
b0il T^Shtcye.e problems, we !e, ^denote the wojfcmto^punrp per
kilogram of fluid flowing and tl denote the heat rejected from the working flu.d per
kUOff T^olve"biem we consider, in succession, a control surface .round the
pump the bote, the turbine, and the condenser. For each the thermodynamic model »
ZS£tti£ and the process is steady state with negligible changes m kmet,c and
potential energies. First consider the pump:
Cofitrol volume: Pump.
Inlet state: P t known, saturated liquid; state fixed.
Exit state: P 2 known.
Analysis
From the first law, we have
The second law gives
and so
s 2 = s i
2
dP
Solution
Assuming the liquid to be incompressible, we have
Wp = v(P 2 - Pi) = (0.001 01)(2000 - 10) = 2.0kJ/kg
Aa =/,,+,„, = 191.8 + 2.0= 193.8 kJ/kg
Now consider the boiler:
Control volume: Boiler.
Inlet state: P 2 , h known; state fixed.
Exit state: P 3 known, saturated vapor; state fixed.
THE RANKING CYCLE H 387
Analysis
From the first law, we write
<hi ~ h ~h 2
Solution
Substituting, we obtain
q H = /i 3 - h 2 = 2799.5 - 193.8 = 2605.7 kJ/kg
Turning to the turbine next, we have:
Control volume: Turbine.
Inlet state: State 3 known (above).
Exit state: P 4 known.
Analysis
The first law gives
w, = h 3 — #4
The second law gives
Si = S 4
Solution
We can determine the quality at state 4 as follows:
s 3 = s 4 = 6.3409 = 0.6493 + * 4 7.5009, * 4 = 0.7588
h 4 = 191.8 + 0.7588(2392.8) = 2007.5 kJ/kg
w, = 2799.5 - 2007.5 = 792.0 kJ/kg
Finally, we consider the condenser.
Control volume: Condenser.
Inlet state: State 4 known (as given).
Exit state: State 1 known (as given).
Analysis
The first law gives
«t = h ~ h\
Solution
Substituting, we obtain
Hl = h -h x = 2007.5 - 191.8 = 1815.7 kJ/kg
388 m CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
We can now calculate the thermal efficiency:
_ w -< - frf-fr, „ w ' " w p _ 792,0 ; • 2.0 _ 30 3%
" </// " q H 1h 2605.7
We could also write an expression for thermal efficiency in terms of properties at vari-
ous points in the cycle:
_(/z 3 • h 2 ) (h, ~ hd _ (h 3 - h, { ) - (h 2 - h,)
1,th lt 3 -li 2 "~ ■ h 3 -h 2
- 2605.7 - 1815.7 = 792.0 - 2.0 30 3%
2605.7 2605.7
11.3 effect of pressure and temperature
on the Rankin e cycle
Let us first consider the effect of exhaust pressure and temperature on the Rankine
cycle. This effect is shown on the Ts diagram of Fig. 11.4. Let the exhaust pressure
drop from P 4 to P' 4> with the corresponding decrease in temperature at which heat is
rejected. The net work is increased by area l-4-4'-l'-2'-2-l (shown by the cross-
hatching). The heat transferred to the steam is increased by area a'-2'-2-a-a'. Since
these two areas are approximately equal, the net result is an increase in cycle efficiency.
This is also evident from the fact that the average temperature at which heat is rejected
is decreased. Note, however, that lowering the back pressure causes the moisture con-
tent of the steam leaving the turbine to increase. This is a significant factor because if
the moisture in the low-pressure stages of the turbine exceeds about 10%, not only is
there a decrease in turbine efficiency, but erosion of the turbine blades may also be a
very serious problem.
Next, consider the effect of superheating the steam in the boiler, as shown in Fig.
1 1.5. We see that the work is increased by area 3-3'-4'-4-3, and the heat transferred in
the boiler is increased by area 3-3'-6'-^-3. Since the ratio of these two areas is greater
than the ratio of net work to heat supplied for the rest of the cycle, it is evident that for
EFFECT OF PRESSURE AND TEMPERATURE ON THE RANKIN E CYCLE
389
FIGURE 11.5 Effect
of superheating on
Rankine-cycle efficiency.
given pressures, superheating the steam increases the Rankine-cycle efficiency. This in-
crease in efficiency would aiso follow from the fact that the average temperature at which
heat is transferred to the steam is increased. Note also that when the steam is superheated,
the quality of the steam leaving the turbine increases.
Finally the influence of the maximum pressure of the steam must be considered,
and this is shown in Fig. II .6. In this analysis the maximum temperature of the steam,
as well as the exhaust pressure, is held constant. The heat rejected decreases by area
U-4>-A-b-b' The net work increases by the amount of the single cross-hatching and
decreases by the amount of the double cross-hatching. Therefore, the net work tends to
remain the same, but the heat rejected decreases, and hence the Rankine-cycle effi-
ciency increases with an increase in maximum pressure. Note that in this instance too
the average temperature at which heat is supplied increases with an increase in pres-
sure. The quality of the steam leaving the turbine decreases as the maximum pressure
increases. . T
To summarize this section, we can say that the efficiency of the Rankine cycle can
be increased by lowering the exhaust pressure, by increasing the pressure during heat ad-
dition and by superheating the steam. The quality of the steam leaving the turbine is in-
creased by superheating the steam and decreased by lowering the exhaust pressure and by
increasing the pressure during heat addition.
390 H Chapter Eleven Power and refrigeration Systems
EXAMPLE 11.2 In a Rankine cycle, steam leaves the boiler and enters the turbine at 4 MPa and 400°C.
The condenser pressure is 10 kPa. Determine the cycle efficiency.
To determine the cycle efficiency, we must calculate the turbine work, the pump
work, and the heat transfer to the steam in the boiler. We do this by considering a con-
trol surface around each of these components in turn. In each case the thermodynamic
model is the steam tables, and the process is steady state with negligible changes in ki-
netic and potential energies.
Control volume: Pump.
Met state: P 1 known, saturated liquid; state fixed.
Exit stale: P 2 known.
Analysis
From the first law, we have
w p - h 2 - h t
The second law gives
Since s 2 = s lt
h
Solution
Substituting, we obtain
w p = v(P 2 ~ P{) = (0.001 01)(4000 - 10) - 4.0kJ/kg
h x = 191.8 kJ/kg
h 2 =191.8 + 4.0= 195.8 kJ/kg
For the turbine we have:
h t = vdP = v(P 2 - Pj)
Control volume: Turbine.
Inlet state: P 3 , T 3 known; state fixed.
Exit state: _P 4 known.
Analysis
The first law is
and the second law is
w ( = h 3 — h A
s 4 — J 3
Effect of
PRESSURE AMD TEMPERATURE ON THE RANKINE CYCLE M 391
Solution
Upon substitution we get
/i 3 = 3213.6 kJ/kg, s 3 = 6.7690 kJ/kg K
S3 = S4 = 6.7690 = 0.6493 + x 4 7.5009, x 4 = 0.8159
A 4 = 191.8 + 0.8159(2392.8) = 2144.1 kJ/kg
w t = A 3 - /i 4 = 3213.6 - 2144.1 - 1069.5 kJ/kg
Wna = W( _ Wp = 1069.5 - 4.0 = 1065.5 kJ/kg
Finally, for the boiler we have:
Control volume: Boiler.
Inlet state: P 2 , h 2 known; state fixed.
Exit state: State 3 fixed (as given).
Analysis
The first law is
Solution
Substituting gives
q H = h 3 - h 2
q u - h 3 - h 2 = 3213.6 - 195.8 = 3017.8 kJ/kg :
w net 1065.5 , KW
^ = "te" 3017^" 353/0
The net work could also be determined by calculating the heat rejected in the condenser,
q L , and noting, from the first law, that the net work for the cycle is equal to the net heat
transfer. Considering a control surface around the condenser, we have
q L = h~h = 2144.1 - 191.8 = 1952.3 kJ/kg
Therefore,
w net = q H -qi. = 3017.8 - 1952.3 = 1065.5 kJ/kg
EXAMPLE 11.2E In a Rankine cycle, steam leaves the boiler and enters the turbine at 600 tbtfin. 2 and
800 F. The condenser pressure is 1 lbf/in. 2 Determine the cycle efficiency.
To determine the cycle efficiency, we must calculate the turbine work, the pump
work, and the heat transfer to the steam in the boiler. We do this by considering a con-
trol surface around each of these components in turn. In each case the thermodynamic
model is the steam tables, and the process is steady state with negligible changes in ki.
netic and potential energies.
Control volume: Pump.
Inlet state: P x known, saturated liquid; state fixed. -
Exit state: P 2 known.
392 M Chapter eleven Power and Refrigeration systems
Analysis
From the first law, we have
w p = h 2 h x
The second law gives
s?. =-' *1 ,
Since s 2 ~ s u
h 2 -ht = j 2 vdP = v(P 2 ~Pi)
Solution
Substituting, we obtain
Wp = V (P 2 - P^ = 0.01614(600 - 1) X = 1.8 Btu/lbm
h { =69.70
/i 2 ~ 69.7 + 1.8 - 71.5 Btu/lbm
For the turbine we have
Control volume: Turbine.
Inlet state: P 3 , T 3 known; state fixed.
Exit state: P 4 known.
Analysis
The first law is
w s - h 3 - h 4
and the second law is
Solution
Upon substitution we get
/i 3 = 1407.6 s 3 = 1.6343
Sj = s 4 = 1.6343 = 1.9779 - (1 - .v) 4 1.8453
(1 -x) 4 = 0.1861
fi 4 = 1105.8 - 0.1861(1036.0) = 913.0
w, = h 3 - A 4 = 1407.6 - 913.0 = 494.6 Btu/lbm
wt = w, - w p = 494.6 - 1.8 = 492.8 Btu/lbm
the Reheat Cycle M 393
Finally, for the boiler we have
Control volume;
Inlet state;
Exit state:
Analysis
The first law is
Solution
Substituting gives
The net work could also be determined by calculating the heat rejected in the condenser,
q L , and noting, from the first law, that the net work for the cycle is equal to the net heat
transfer. Considering a control surface around the condenser, we have
q L = h 4 - h x = 913.0 - 69.7 = 843.3 Btu/lbm
Therefore,
w Mt = <lH~qL = 1336.1 ~ 843.3 = 492.8 Btu/lbm
11.4 The Reheat Cycle
In the last section we noted that the efficiency of the Rankine cycle could be increased
by increasing the pressure during the addition of heat. However, the increase in pres-
sure also increases the moisture content of the steam in the low-pressure end of the tur-
bine. The reheat cycle has been developed to take advantage of the increased
efficiency with higher pressures, and yet avoid excessive moisture in the low-pressure
stages of the turbine. This cycle is shown schematically and on a T-s diagram in Fig.
11. 7. The unique feature of this cycle is that the steam is expanded to some intermedi-
ate pressure in the turbine and is then reheated in the boiler, after which it expands in
the turbine to the exhaust pressure. It is evident from the T-s diagram that there is very
little gain in efficiency from reheating the steam, because the average temperature at
which heat is supplied is not greatly changed. The chief advantage is in decreasing to a
safe value the moisture content in the low-pressure stages of the turbine. If metals
could be found that would enable us to superheat the steam to 3', the simple Rankine
cycle would be more efficient than the reheat cycle, and there would be no need for the
reheat cycle.
Boiler.
P 2 , h 2 known; state fixed.
State 3 fixed (above).
h 3 - h 2 = 1407.6 - 71.5 = 1336.1 Btu/lbm
w«t _ 492.8 „ « Q0/
^- 1336T" 36 - 9/ °
394 B CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
Turbine
FIGURE 11.7 The
ideal reheat cycle.
Pump
s
EXAMPLE 11 .3 Consider a reheat cycle utilizing steam. Steam leaves the boiler and enters the turbine at
4 MPa, 400°C. After expansion in the turbine to 400 kPa, the steam is reheated to 400°C
and then expanded in the low-pressure turbine to 10 kPa. Determine the cycle efficiency.
For each control volume analyzed, the thermodynamic model is the steam tables,
the process is steady state, and changes in kinetic and potential energies are negligible.
For the high pressure turbine,
Control volume: High-pressure turbine.
Inlet state: P 3 , T 3 known; state fixed.
Exit state: P 4 known .
Analysis
The first law is
Solution
Substituting,
hi = 3213.6, ^ s 3 = 6.7690
s 4 =s 2 = 6.7690 = 1.7766 + x 4 5.1193, x 4 = 0.9752
h A = 604.7 + 0.9752(2133.8) = 2685.6
For the low-pressure turbine,
Control volume: Low-pressure turbine.
Inlet state: P 5 , T 5 known; state fixed.
Exit state: P 6 known.
The second law is
s 3 — s 4
THE REHEAT CYCLE B 395
Analysis
The first law is
= h 5 - h 6
For the second law,
*5 = *6
Solution
Upon substituting,
h 5 = 3273.4 s 5 = 7.8985
s 6 = s 5 = 7.8985 - 0.6493 + ,r 6 7.5 009, x s = 0.9664
h 6 = 191.8 + 0.9664(2392.8) = 2504.3
For the overall turbine, the total work output w t is the sum of w k - p and w } - p , so that
w ( ^ (A 3 " h) + <h s - A 6 )
= (3213.6 - 2685.6) + (3273.4 - 2504.3)
= 1297.1 kJ/kg
For the pump,
Control volume: Pump.
Inlet state: P t known, saturated liquid; state fixed.
Exit state: P 2 known.
Analysis
From the first law, we have
w p -h 2 -h x
and the second law,
s 2 = Si
Since s 2 = s u
h 2 -h x = ^vdP = v{P 2 -P l )
Solution
Substituting,
w p = v{P 2 - P x ) = (0.001 01)(4000 - 10) - 4.0 kJ/kg
h 2 = 191.8 + 4.0 - 195.8
396 E CHAPTERELEVEN power and refrigeration systems
Finally, for the boiler
Control volume:
Inlet states:
Exit states:
Boiler.
States 2 and 4 both known (above).
States 3 and 5 both known (as given).
Analysis
The first law gives
q H =(h 3 -h 2 ) + (h s -hJ
Solution
Substituting,
q H = (h 3 - h 2 ) + (h 5 ~ h A )
= (3213.6 - 195.8) + (3273.4 - 2685.6) = 3605.6 kj/kg
Therefore,
Wnrt = w t - w p = 1297.1 - 4.0 = 1293.1 kj/kg
By comparing this example with Example 1 1.2, we find that through reheating the gain
in efficiency is relatively small, but the moisture content of the vapor leaving the turbine
is decreased from 18.4 to 3.4%.
Another important variation from the Rankine cycle is the regenerative cycle, which uses
feed water heaters. The basic concepts of this cycle can be demonstrated by considering
the Rankine cycle without superheat as shown in Fig. 11.8. During the process between
states 2 and 2', the working fluid is heated while in the liquid phase, and the average tem-
perature of the working fluid is much lower than during the vaporization process 2'-3.
The process between states 2 and 2' causes the average temperature at which heat is sup-
plied in the Rankine cycle to be lower than in the Camot cycle 1 2'— 3-4— 1 Conse-
quently, the efficiency of the Rankine cycle is lower than that of the corresponding Carnot
cycle. In the regenerative cycle the working fluid enters the boiler at some state between 2
and 2'; consequently, the average temperature at which heat is supplied is higher.
Consider first an idealized regenerative cycle, as shown in Fig. 11.9. The unique
feature of this cycle compared to the Rankine cycle is that after leaving the pump, the liq-
uid circulates around the turbine casing, counterflow to the direction of vapor flow in the
turbine. Thus, it is possible to transfer to the liquid flowing around the turbine the heat
from the vapor as it flows through the turbine. Let us assume for the moment that this is a
reversible heat transfer; that is, at each point the temperature of the vapor is only infmites-
imally higher than the temperature of the liquid. In this instance line 4-5 on the Ts dia-
11.5 The Regenerative Cycle
THE REGENERATIVE CYCLE El 397
FIGURE 11.8
Temperature-entropy
diagram showing the
relationships between
Camot-cycle efficiency
and Rankine-cycle
efficiency.
gram of Fig. 1 1.9, which represents the states of the vapor flowing through the turbine, is
exactly parallel to line 1-2-3, which represents the pumping process (1-2) and the states
of the liquid flowing around the turbine. Consequently, areas 2-~3~b~a-2 and 5-A-d-c-5
are not only equal but congruous, and these areas, respectively, represent the heat trans-
ferred to the liquid and from the vapor. Heat is also transferred to the working fluid at
constant temperature in process 3-4, and area 3^4-^d-b~3 represents this heat transfer.
Heat is transferred from the working fluid in process 5-1, and area l-5-c-a-\ represents
this heat transfer. This area is exactly equal to area V~5'~d-b-V , which is the heat re-
jected in the related Carnot cycle l'-3-4-5'-F, Thus, the efficiency of this idealized re-
generative cycle is exactly equal to the efficiency of the Carnot cycle with the same heat
supply and heat rejection temperatures.
Quite obviously, this idealized regenerative cycle is impractical. First, it would be
impossible to effect the necessary heat transfer from the vapor in the turbine to the liquid
feedwater. Furthermore, the moisture content of the vapor leaving the turbine increases
considerably as a result of the heat transfer. The disadvantage of this has been noted pre-
viously. The practical regenerative cycle extracts some of the vapor after it has partially
expanded in the turbine and uses feedwater heaters, as shown in Fig. 11.10.
Steam enters the turbine at state 5. After expansion to state 6, some of the steam is
extracted and enters the feedwater heater. The steam that is not extracted is expanded in
FIGURE 11.9 The
ideal regenerative cycle.
Pump
398 ffl CHAPTER ELEVEN POWER. AND REFRIGERATION SYSTEMS
FIGURE 11.10
Regenerative cycle with
open feedwater heater.
Boiler
the turbine to state 7 and is then condensed in the condenser. This condensate is
pumped into the feedwater heater where it mixes with the steam extracted from the tur-
bine The proportion of steam extracted is just sufficient to cause the liquid leaving the
feedwater heater to be saturated at state 3. Note that the liquid has not been pumpec to
the boiler pressure, but only to the intermediate pressure corresponding to state 6. An-
other pump is required to pump the liquid leaving the feedwater heater to boiler pres-
sure. The significant point is that the average temperature at which heat is supplied has
been increased. . - -n,...,.
Consider a control volume around the open feedwater heater in Fig. 1 1 . 10. The con-
servation of mass requires:
m 2 + >"e = '"3
satisfied with the extraction fraction as
(11.2)
so
(11.3)
m 7 = (1 - x)m 5 = m x = m 2
The energy equation with no external heat transfer and no work becomes
m 2 h 2 + m^i 6 = m^hy
into which we substitute the mass flow rates (m 3 = »i 5 ) as
(1 - x)m$h 2 + Jewish = m s h 3 C 1 *- 4 )
We take state 3 as the limit of saturated liquid (we do not want to heat further as it would
move into the two-phase region and damage the pump P2) and then solve for x
(11.5)
x =
h 6 - h 2
This establishes the maximum extraction fraction we should take out at this extraction
pressure.
The regenerative Cycle H 399
This cycle is somewhat difficult to show on a T—s diagram because the masses of
steam flowing through the various cdmponents vary. The T—s diagram of Fig. 11.10 sim-
ply shows the state of the fluid at the various points.
Area 4-5-c-b-A in Fig. 11.10 represents the heat transferred per kilogram of work-
ing fluid. Process 7-1 is the heat rejection process, but since not all the steam passes
through the condenser, area 1—1—c—a-l represents the heat transfer per kilogram flowing
through the condenser, which does not represent the heat transfer per kilogram of working
fluid entering the turbine. Between states 6 and 7 only part of the steam is flowing through
the turbine. The example that follows illustrates the calculations for the regenerative cycle.
EXAMPLE 11,4 Consider a regenerative cycle using steam as the working fluid. Steam leaves the boiler
and enters the turbine at 4 MPa, 400°C. After expansion to 400 kPa, some of the steam is
extracted from the turbine for the purpose of heating the feedwater in an open feedwater
heater. The pressure in the feedwater heater is 400 kPa, and the water leaving it is satu-
rated liquid at 400 kPa. The steam not extracted expands to 10 kPa. Determine the cycle
efficiency.
The line diagram and T—s diagram for this cycle are shown in Fig, 11.10
As in previous examples, the model for each control volume is the steam tables,
the process is steady state, and kinetic and potential energy changes are negligible.
From Examples 1 1 .2 and 1 1 .3 we have the following properties:
h s = 3213.6 h s = 2685.6
A 7 = 2144.1 = 191.8
For the low-pressure pump,
Control volume: Low-pressure pump.
Inlet state: P x known, saturated liquid; state fixed.
Exit state: P 2 known.
Analysis
The first law is
Wpi = h 2 - h x
For the second law,
Therefore,
h~h x = j\dP = v(P 2 -Pd
Solution
Substituting,
w P i = v(P 2 ~ Pi) = (0-001 01)(400 - 10) = 0.4 kJ/kg
h 2 = A, + w p = 191.8 + 0.4 = 192.2
400 H Chapter eleven power and refrigeration Systems
For the turbine,
Confrol volume: Turbine.
Inlet state: P 5 , T 5 known; state fixed.
Exit state: P 6 known; P 7 known.
Analysis
From the first law,
" ( -(/>5 ' (i --'"i)(*6 ih) :
and the second law,
S $ - S 6 ~
Solution
From the second law, the values for h 6 and h 7 given previously were calculated in Exam
pies 11.2 and 11.3.
For the feedwater heater,
Control volume: Feedwater heater.
Inlet states: States 2 and 6 both known (as given).
Exit state: P 3 known, saturated liquid; state fixed.
Analysis
The first law gives
mi (h 6 ) + (1 - m x )h 2 = h 2
Solution
After substitution,
7^(2685.6) + (1 - m0(192.2) = 604.7
m x = 0.1654
We can now calculate the turbine work.
w, = (/r 5 -A 6 ) + (l-md(Ji 6 -kj)
= (3213.6 - 2685.6) + (1 - 0.1654)(2685.6 - 2144.1)
= 979.9 kJ/kg
For the high-pressure pump,
Control volume: High-pressure pump.
Inlet state: State 3 known (as given).
Exit state: F 4 known.
THE REGENERATIVE CYCLE ffl 401
Analysis
For the first law,
w p2 = h 4 -h 3
The second law is
S 4 — £3
Solution
Substituting,
w p2 = v(P 4 - P 3 ) = (0.001 084)(4000 - 400) = 3.9 kJ/kg
h 4 = h 3 + w p2 = 604.7 + 3.9 = 608.6
Therefore,
Wmi = w t ~ (1 - m{)w pl - w p2
= 979.9 - (1 - 0.1654)(0.4) - 3.9 = 975.7 kJ/kg
Finally, for the boiler,
Control volume: Boiler.
Inlet state: P 4 , h 4 known (as given); state fixed.
Exit state: State 5 known (as given).
Analysis
The first law gives
q H ^h s -h A
Solution
Substituting,
q H = h 5 - h 4 = 3213.6 - 608.6 - 2605.0 kJ/kg
. . W net 975.7 . ^ , 0/
77(11 " " 2605.0 " 5
Note the increase in efficiency over the efficiency of the Rankine cycle of Example 11.2.
Up to this point, the discussion and examples have tacitly assumed that the extrac-
tion steam and feedwater are mixed in the feedwater heater. Another much-used type of
feedwater heater, known as a closed heater, is one in which the steam and feedwater do
not mix. Rather, heat is transferred from the extracted steam as it condenses on the outside
of tubes while the feedwater flows through the tubes. In a closed heater, a schematic
sketch of which is shown in Fig. 11 ,1 1, the steam and feedwater may be at considerably
different pressures. The condensate may be pumped into the feedwater line, or it may be
removed through a trap to a lower-pressure heater or to the condenser. (A trap is a device
that permits liquid but not vapor to flow to a region of lower pressure.)
402 m CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
FIGURE 11.11
Schematic arrangement
for a closed feedwater
heater.
Extraction steam
Feedwater
Condensate to Sower pressure
heater or condenser
Let us analyze the closed feedwater heater in Fig. 11.11 when a trap with a dram to
the condenser is used. We assume we can heat the feedwater up to the temperature of the
condensing extraction flow, that is T 3 = T A = T ta as there is no drip pump. Conservation
of mass for the feedwater heater is
m 4 - m 3 = m 2 = m 5 \ m 6 = xm 5 = m 6a = m €c
Notice that the extraction flow is added to the condenser, so the flowrate at 2 is the same
as at state 5. The energy equation is
m 5 h 2 + xm 5 h 6 = m s h 3 + xm s h^ O 1 - 6 )
which we can solve for.r as
'6a
Open feedwater heaters have the advantage of being less expensive and having bet-
ter heat-transfer characteristics than closed feedwater heaters. They have the disadvantage
of requiring a pump to handle the feedwater between each heater.
In many power plants a number of stages of extraction are used, though only rarely
more than five. The number is, of course, determined by economics. It is evident tot
using a very large number of extraction stages and feedwater heaters allows the cycle effi-
ciency to approach that of the idealized regenerative cycle of Fig. 1 1 .9, where the feedwa-
ter enters the boiler as saturated liquid at the maximum pressure. In practice, however,
this could not be economically justified because the savings effected by the increase m ef-
ficiency would be more than, offset by the cost of additional equipment (feedwater heaters,
piping, and so forth). . ,
A typical arrangement of the main components in an actual power plant is shown m
Fig 1 1 12 Note that one open feedwater heater is a deaerating feedwater heater, this
heater has the dual purpose of heating and removing the air from the feedwater Unless
the air is removed, excessive corrosion occurs in the boiler. Note also that the condensate
from the high-pressure heater drains (through a trap) to the intermediate heater, and the
intermediate heater drains to the deaerating feedwater heater. The low-pressure heater
drains to the condenser.
Many actual power plants combine one reheat stage with a number of extraction
stages. The principles already considered are readily applied to such a cycle.
deviation of actual cycles from ideal cycles B 403
8.7 MPa
500°C
Boiler feed
pump
320,000 kg/h
i
JL
High-
pressure
heater
Intermediate-
pressure
heater
Trap
^ Trap — ' > — ->I
Deaerating
open feed-
water heater
Low-
pressure
heater
^ Booster
J pump
Trap
Condensate
pump
FIGURE 11,12 Arrangement of heaters in an actual power plant utilizing regenerative feedwater heaters.
11.6 Deviation of Actual Cycles
from Ideal Cycles
Before we leave the matter of vapor power cycles, a few comments are in order regarding
the ways in which an actual cycle deviates from an ideal cycle. The most important of
these losses are due to the turbine, the pump(s), the pipes, and the condenser. These losses
are discussed next.
Turbine Losses
Turbine losses, as described in Section 9.5, represent by far the largest discrepancy between
the performance of a real cycle and a corresponding ideal Rankine-cycle power plant. The
large positive turbine work is the principal number in the numerator of the cycle thermal effi-
ciency and is directly reduced by the factor of the isentropic turbine efficiency. Turbine losses
are primarily those associated with the flow of the working fluid through the turbine blades
and passages, with heat transfer to the surroundings also being a loss but of secondary impor-
tance. The turbine process might be represented as shown in Fig. 11.13 where state 4 S is the
state after an ideal isentropic turbine expansion and state 4 is the actual state leaving the tur-
bine following an irreversible process. The turbine governing procedures may also cause a
loss in the turbine, particularly if a throttling process is used to govern the turbine operation.
Pump Losses
The losses in the pump are similar to those of the turbine and are due primarily to the irre-
versibilities with the fluid flow. Pump efficiency was discussed in Section 9.5, and the
ideal exit state 2 S and real exit state 2 are shown in Fig. 1 1.13. Pump losses are much
smaller than those of the turbine, since the associated work is very much smaller.
404 M CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
T
FIGURE 11.13
Temperature-entropy
diagram showing effect of
turbine and pump
inefficiencies on cycle
performance.
Piping Losses
Pressure drops caused by frictional effects and heat transfer to the surroundings are the
most important piping losses. Consider, for example, the pipe connecting the turbine to
the boiler. If only frictional effects occur, states a and b in Fig. 11.14 would represent
the states of the steam leaving the boiler and entering the turbine, respectively. Note
that the frictional effects cause an increase in entropy. Heat transferred to the surround-
ings at constant pressure can be represented by process be. This effect decreases en-
tropy. Both the pressure drop and heat transfer decrease the availability of the steam
entering the turbine. The irreversibility of this process can be calculated by the methods
outlined in Chapter 10.
A similar loss is the pressure drop in the boiler. Because of this pressure drop, the
water entering the boiler must be pumped to a higher pressure than the desired steam pres-
sure leaving the boiler, and this requires additional pump work.
Condenser Losses
The losses in the condenser are relatively small. One of these minor losses is the cool-
ing below the saturation temperature of the liquid leaving the condenser. This repre-
sents a loss because additional heat transfer is necessary to bring the water to its
saturation temperature.
The influence of these losses on the cycle is illustrated in the following example,
which should be compared to Example 1 1 .2.
FIGURE 11.14
Temperature-entropy
diagram showing effect of
losses between boiler and
turbine.
DEVIATION OF ACTUAL CYCLES FROM IDEAL CYCLES El 405
EXAMPLE 11.5 A steam power plant operates on a cycle with pressures and temperatures as designated
in Fig. 11.15. The efficiency of the' turbine is 86%, and the efficiency of the pump is
80%. Determine the thermal efficiency of this cycle.
As in previous examples, for each control volume the model used is the steam ta-
bles, and each process is steady state with no changes in kinetic or potential energy. This
cycle is shown on the T-s diagram of Fig. 11.16.
Control volume:
Inlet state:
Exit state:
Turbine.
P 5 , T 5 known; state fixed.
P 6 known.
Analysis
From the first law, we have
The second law is
w, = h 5 - h 6
The efficiency is
lu ~ h A
h« - h< c A, - hf,
FIGURE 11.15
Schematic diagram for
Example 11.5.
Pump
FIGURE 11.16
Temperature-entropy
diagram for Example 1 1 .5.
406 H CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
Solution
From the steam tables, we get
h s -3169.1 kJ/kg, A' 5 - 6.7235
% = Ss = 6.7235 - 0.6493 + x 6i 7.5009, x 6i = 0.8098
h 6i = 191.8 4- 0.8098(2392.8) = 2129.5 kJ/kg
W( = Vl (h 5 - h 6 ) = 0.86(3169.1 - 2129.5) = 894.1 kJ/kg
For the pump, we have:
Control volume: Pump.
Met state: P h 2\ known; state fixed.
Exit state: P 2 known.
Analysis
The first law gives
The second law gives
The pump efficiency is
Since s 2j — s x ,
Therefore,
Solution
Substituting, we obtain
v p - Hi
si, - s x
h 2s - hi _ h ls - hi
h 2s - A L = v(P 2 - P,)
h ~ h v(_P 2 ~ P t )
Pl) _ (a ° 01 ° 09 ^ 5000 " 10) - 6 ? kFkr
w * " tj, 0.80
Therefore,
HVe. = w ( - w p = 894.1 - 6.3 = 887.8 kJ/kg
Finally, for the boiler:
Control volume: Boiler.
Inlet state: P 3 , T 3 known; state fixed.
Exit state: P 4 , T 4 known, state fixed.
Deviation of Actual Cycles from Ideal Cycles B 407
Analysis
The first law is
Solution
Substitution gives
q H - h A - A 3 = 3213.6 - 171.8 = 3041.8 kJ/kg
887.8
3041.8
= 29.2%
This result compares to the Rankine efficiency of 35.3% for the similar cycle of Exam-
ple 11.2.
EXAMPLE 11. 5E A steam power plant operates on a cycle with pressure and temperatures as designated in
Fig. 1 1. 15E. The efficiency of the turbine is 86%, and the efficiency of the pump is 80%.
Determine the thermal efficiency of this cycle.
As in previous examples, for each control volume the model used is the steam ta-
bles, and each process is steady state with no changes in kinetic or potential energy. This
cycle is shown on the Ts diagram of Fig, 11.16.
Control volume: Turbine.
Inlet state: P 5> T s known; state fixed.
Exit state: P 6 known.
Analysis
From the first law, we have
The second law is
w t = h 5 - h 6
s 6s ~ s 5
FIGURE 11.15E
Schematic diagram for
Example 11.5E.
©600 I bf/ in. 2
800 F
560 fbf/in. 2
© 760 F
-®760!bf/in 2
95 F
800!bf/in. 2
© Pump
1 lbf/in. 2
93 F
408 H Chapter Eleven Power and Refrigeration systems
The efficiency is
= W l - ^ ~ h 6
Solution
From the steam tables, we get
h s - 1386.8 s s = 1.6248
s fc = s s = 1.6248 = 1.9779 - (1 -x) b lM53
A & = 1 105.8 - 0.1912(1036.0) = 907.6
MV = U(^5 - A&) = 0.86(1386.8 - 907.6)
= 0.86(479.2) = 412.1 Btu/lbm
For the pump, we have:
Control volume: Pump.
Inlet state: P u T x known; state fixed.
Exit state: P 2 known.
Analysis
The first law gives
The second law gives
The pump efficiency is
Since s 2s — s u
Therefore,
w p = h 2 - h x
s 2s ~ s l
^ w p 1h - A,
h ls ~h l = v(P 2 -P l )
„ K -h x _ v(P 2 - P{)
Solution
Substituting, we obtain
i/(P 2 -Pi) 0.016 15(800 - 1)144
w, = = Q.8 X 778 = 3 -° Btu/lbm
Cog ene rati on □ 409
Therefore,
Wnrt = Wl - Wp - 412.1 - 3.0 = 409.1 Btu/lbm
Finally, for the boiler:
Control volume: Boiler.
Inlet state: P J} T 3 known; state fixed.
Exit state: P 4) T 4 known; state fixed.
Analysis
The first law is
1h = ] h ~ h
Solution
Substitution gives
q H = h 4 ~ k 3 = 1407.6 - 65.1 = 1342.5 Btu/lbm
This result compares to the Rankine efficiency of 36.9% for the similar cycle of Exam-
ple 11. 2E.
11.7 COGENERATION
There are many occasions in industrial settings where the need arises for a specific
source or supply of energy within the environment in which a steam power plant is
being used to generate electricity. In such cases, it is appropriate to consider supply-
ing this source of energy in the form of steam that has already been expanded through
the high-pressure section of the turbine in the power plant cycle, thereby eliminat-
ing the construction and use of a second boiler or other energy source. Such an
arrangement is shown in Fig, 1 1.17, in which the turbine is tapped at some intermedi-
ate pressure to furnish the necessary amount of process steam required for the particu-
lar energy need— perhaps to operate a special process in the plant, or in many cases
simply for the purpose of space heating the facilities. This type of application is
termed cogeneration. If the system is designed as a package with both the electrical
and the process steam requirements in mind, it is possible to achieve a substantial sav-
ings in capital cost of equipment and in the operating cost, through careful considera-
tion of all the requirements and optimization of the various parameters involved.
Specific examples of cogeneration systems are considered in the problems at the end
of the chapter.
410 H Chapter eleven power and Refrigeration Systems
11.8 Air-Stanbard Power cycles
In Section 11.1, we considered idealized four-process cycles, including both steady-
state-process and cylinder/piston boundary-movement cycles. The question of phase-
change cycles and single-phase cycles was also mentioned. We then proceeded to
examine the Rankine power plant cycle in detail, the idealized model of a phase-change
power cycle. However, many work-producing devices (engines) utilize a working fluid
that is always a gas. The spark-ignition automotive engine is a familiar example, as are
the diesel engine and the conventional gas turbine. In all these engines there is a change
in the composition of the working fluid, because during combustion it changes from air
and fuel to combustion products. For this reason these engines are called internal-
combustion engines. In contrast, the steam power plant may be called an external-
combustion engine because heat is transferred from the products of combustion to the
working fluid. External- combustion engines using a gaseous working fluid (usually air)
have been built. To date they have had only limited application, but use of the gas-tur-
bine cycle in conjunction with a nuclear reactor has been investigated extensively.
Other external-combustion engines are currently receiving serious attention in an effort
to combat air pollution.
Because the working fluid does not go through a complete thermodynamic cycle in
the engine (even though the engine operates in a mechanical cycle), the internal-
combustion engine operates on the so-called open cycle. However, for analyzing
internal-combustion engines, it is advantageous to devise closed cycles that closely ap-
proximate the open cycles. One such approach is the air-standard cycle, which is based on
the following assumptions:
1. A fixed mass of air is the working fluid throughout the entire cycle, and the air is al-
ways an ideal gas. Thus, there is no inlet process or exhaust process.
The Brayton Cycle H 411
2. The combustion process is replaced by a process transferring heat from an external
source.
3. The cycle is completed by heat transfer to the surroundings (in contrast to the ex-
haust and intake process of an actual engine).
4. All processes are internally reversible.
5. An additional assumption is often made that air has a constant specific heat, recog-
nizing that this is not the most accurate model.
The principal value of the air- standard cycle is to enable us to examine qualitatively
the influence of a number of variables on performance. The quantitative results obtained
from the air-standard cycle, such as efficiency and mean effective pressure, will differ
from those of the actual engine. Our emphasis, therefore, in our consideration of the air-
standard cycle will be primarily on the qualitative aspects.
The term mean effective pressure (mep), which is used in conjunction with recipro-
cating engines, is defined as the pressure that, if it acted on the piston during the entire
power stroke, would do an amount of work equal to that actually done on the piston. The
work for one cycle is found by multiplying this mean effective pressure by the area of
the piston (minus the area of the rod on the crank end of a double-acting engine) and by
the stroke.
11.9 THE BRAYTON CYCLE
In discussing idealized four-steady-state-process power cycles in Section 1 1.1, a cycle in-
volving two constant-pressure and two isentropic processes was examined, and the results
are shown in Fig. 11.2. This cycle used with a condensing working fluid is the Rankine
cycle, but when used with a single-phase, gaseous working fluid it is termed the Brayton
cycle. The air-standard Brayton cycle is the ideal cycle for the simple gas turbine. The
simple open-cycle gas turbine utilizing an internal-combustion process and the simple
closed-cycle gas turbine, which utilizes heat-transfer processes, are both shown schemati-
cally in Fig. 11.18. The air-standard Brayton cycle is shown on the P-v and T-s diagrams
ofFig. 11.19.
FIGURE 11.18 A gas
turbine operating on the
Brayton cycle, (a) Open
cycle, (b) Closed cycle.
Fuel
Combustion
chamber
Turbine : pO*
1
■ .".Heat
4
— <§— | —
I «g—
exchanger
Ql
412 iJ chapter Eleven power and refrigeration Systems
The efficiency of the air-standard Brayton cycle is found as follows:
^ Qh C p (T, - T 2 ) T 2 (T 3 /T 2 - 1)
"We note, however, that
Pa P,
P 1 _( TM k ~ l) _ ^3 _ ( TM k ~»
T A Ti" T 2 T { T 2 T 2
* ml -h l -wfc& (11 - 8)
The efficiency of the air-standard Brayton cycle is therefore a function of the
isentropic pressure ratio. The fact that efficiency increases with pressure ratio is evident
from the T—s diagram of Fig. 11.19 because increasing the pressure ratio changes the
cycle from 1-2-3-4-1 to 1—2'— 3'— 4—1. The latter cycle has a greater heat supply and
the same heat rejected as the original cycle; therefore, it has a greater efficiency. Note
that the latter cycle has a higher maximum temperature, T y> than the original cycle, T 2 .
In the actual gas turbine, the maximum temperature of the gas entering the turbine is
fixed by material considerations. Therefore, if we fix the temperature T 3 and increase
the pressure ratio, the resulting cycle is l-2'-3"-4"-l. This cycle would have a higher
efficiency than the original cycle, but the work per kilogram of working fluid is thereby
changed.
With the advent of nuclear reactors, the closed-cycle gas turbine has become more
important. Heat is transferred, either directly or via a second fluid, from the fuel in the nu-
clear reactor to the working fluid in the gas turbine. Heat is rejected from the working
fluid to the surroundings.
The actual gas-turbine engine differs from the ideal cycle primarily because of
irreversibilities in the compressor and turbine, and because of pressure drop in the
flow passages and combustion chamber (or in the heat exchanger of a closed-cycle tur-
bine). Thus, the state points in a simple open-cycle gas turbine might be as shown in
Fig. 11.20.
The brayton Cycle M 413
FIGURE 11,20 Effect
of inefficiencies on the
gas-turbine cycle.
The efficiencies of the compressor and turbine are defined in relation to isentropic
processes. With the states designated as in Fig. 11.20, the definitions of compressor and
turbine efficiencies are
A, - h.
h 4
h - h 4s
(11.9)
(11.10)
Another important feature of the Brayton cycle is the large amount of compressor work
(also called back work) compared to turbine work. Thus, the compressor might require from
40 to 80% of the output of the turbine. This is particularly important when the actual cycle is
considered because the effect of the losses is to require a larger amount of compression work
from a smaller amount of turbine work. Thus, the overall efficiency drops very rapidly with a
decrease in the efficiencies of the compressor and turbine. In fact, if these efficiencies drop
below about 60%, all the work of the turbine will be required to drive the compressor, and the
overall efficiency will be zero. This is in sharp contrast to the Rankine cycle, where only 1 or
2% of the turbine work is required to drive the pump. This demonstrates the inherent advan-
tage of the cycle utilizing a condensing working fluid, such that a much larger difference in
specific volume between the expansion and compression processes is utilized effectively.
EXAMPLE 11.6 In an air-standard Brayton cycle the air enters the compressor at 0.1 MPa and 15°C. The
pressure leaving the compressor is 1.0 MPa, and the maximum temperature in the cycle
is 1100°C. Determine
1. The pressure and temperature at each point in the cycle.
2. The compressor work, turbine work, and cycle efficiency.
For each of the control volumes analyzed, the model is ideal gas with constant
specific heat, at 300 K, and each process is steady state with no kinetic or potential en-
ergy changes. The diagram for this example is Fig. 11.19.
414 H chapter eleven power and refrigeration Systems
We consider the compressor, the turbine, and the high-temperature and low-
temperature heat exchangers in turn.
Control volume: Compressor.
Inlet state: P u T x known; state fixed.
Exit state: P 2 known.
Analysis
The first law gives
w c = h 2 - h x
(Note that the compressor work w c is here defined as work input to the compressor.) The
second law is
s 2 — s x
so that
Solution
Solving for T 2) we get
(k-iyk
= lO om = 1.932, T 2 = 556.8 K
Therefore,
w c = h 2 -h t = c p {T 2 ~ r,)
- 1.004(556.8 - 288.2) = 269.5 kJ/kg
Consider the turbine next.
Control volume: Turbine.
Inlet state: P 3 (= P 2 ) known, T 3 known, state fixed.
Exit state: P 4 (= P x ) known .
Analysis
The first law gives
w, = /t 3 — /r 4 "
The second law is
^3 = *4
so that
The brayton Cycle l;I 415
Solution
Solving for T 4 , we get
/ p\{k-\yk
\~J = lO om = 1.932, T 4 = 710.8 K
Therefore,
w t = h 3 -h 4 = C p (T 3 -T 4 )
= 1.004(1373.2 - 710.8) = 664.7 kJ/kg
w oet = w t - w c = 664.7 - 269.5 = 395.2 kJ/kg
Now we turn to the heat exchangers.
Control volume: High-temperature heat exchanger.
Inlet state: State 2 fixed (as given).
Exit state: State 3 fixed (as given).
Analysis
The first law is
9h = h ~ h 2 - C p (T 3 - T 2 )
Solution
Substitution gives
q H = h 3 - h 2 = C p (T 3 - T 2 ) = 1.004(1373.2 - 556.8) = 819.3 kJ/kg
Control volume: Low-temperature heat exchanger.
Inlet state: State 4 fixed (above).
Exit state: State 1 fixed (above).
Analysis
The first law is
gL = h A -h l = C p (T 4 -T l )
Solution
Upon substitution we have
q L = h 4 ~h i = C p (T 4 - r,) = 1.004(710.8 - 288.2) = 424.1 kJ/kg
Therefore,
^net 395.2 AO O0/
^ = ^r = 8i9i = 48 - 2%
This may be checked by using Eq. 1 1 .8.
416 @ Chapter Eleven power and Refrigeration Systems
EXAMPLE 11.7 Consider a gas turbine with air entering the compressor under the same conditions as in
Example 11.6 and leaving at a pressure of 1.0 MPa. The maximum temperature is
3 100°C. Assume a compressor efficiency of 80%, a turbine efficiency of 85%, and a
pressure drop between the compressor and turbine of 15 kPa. Determine the compressor
work, turbine work, and cycle efficiency.
As in the previous example, for each control volume the model is ideal gas with
constant specific heat, at 300 K, and each process is steady state with no kinetic or po-
tential energy changes. In this example the diagram is Fig. 11. 20.
We consider the compressor, the turbine and the high-temperature heat exchanger
in turn.
Control volume: Compressor.
Inlet state: P u T x known; state fixed.
Exit state: P 2 known.
Analysis
The first law for the real process is
w c = h 1 -h l
The second law for the ideal process is
so that
T 2i (PM-W
In addition,
h t ~ } h _ T 2j ~ Ti
Solution
Solving for T 2 ^ we get
fp 2 y-W T 2
-± = io°* 2S6 - 1 .932, T 2i = 556.8 K
P l j T x
The efficiency is
K ~ h = T 2i - T x _ 5 56i8 _ 288.2
Ve h-K T 2 -T, T 2 ~T {
Therefore,
= 0.80
T 2 - r, = 556,8 080 288 ' 2 = 335.8, T 2 = 624.0 K
w c ~h 2 -h l = : C P (T 2 - r t )
- 1.004(624.0 - 288.2) = 337.0 kJ/kg
The Brayton Cycle B 417
For the turbine, we have:
Control volume: Turbine.
Inlet state: P 3 (P 2 - drop) known, T 3 known; state fixed.
Exit state: P A known.
Analysis
The first law. for the real process is
w t = h - /i 4
The second law for the ideal process is
s 4, ~ s 3
so that
In addition,
Solution
Substituting numerical values, we obtain
P % = P 2 - pressure drop = 1.0 - 0.015 = 0.985 MPa
11) =^ = 9.85 0286 = 1.9236, T 4 = 713.9 K
Pa) U t
Vt
lh K T 2 T 4
T 3 -T 4 = 0.85(1373.2 - 713.9) = 560.4 K
T A = 812.8 K
w t = h~h A = c p (r 2 - r 4 )
= 1.004(1373.2 - 812.8) = 562.4 kJ/kg
w ait = w, - w e = 562.4 - 337.0 = 225.4 kJ/kg
Finally, for the heat exchanger;
Control volume: High-temperature heat exchanger.
Inlet state: State 2 fixed (as given).
Exit state: State 3 fixed (as given).
Analysis
The first law is
q H = h 3 - h 2
418 H Chapter Eleven power and Refrigeration Systems
Solution
Substituting, we have
q ll -h,-h 2 ---c p (;r l -T 7 )
- 1.004(1373.2 - 624.0) = 751.8 kJ/kg
so that
w net 225.4 _ on no/
^ = W = 75TS- 30 - 0%
The following comparisons can be made between Examples 1 1.6 and 1 1.7:
w c
Example 1L6 (Ideal)
Example 1 1 .7 (Actual)
269.5
337.0
664.7
562.4
395.2
225.4
819.3
751.8
48.2
30.0
As stated previously, the irreversibilities decrease the turbine work and increase the
compressor work. Since the net work is the difference between these two, it decreases
very rapidly as compressor and turbine efficiencies decrease. The development of com-
pressors and turbines of high efficiency is therefore an important aspect of the develop-
ment of gas turbines.
Note that in the ideal cycle (Example 11.6) about 41% of the turbine work is re-
quired to drive the compressor and 59% is delivered as net work. In the actual turbine
(Example 1 1.7) 60% of the turbine work is required to drive the compressor and 40% is
delivered as net work. Thus, if the net power of this unit is to be 10 000 kW, a 25 000 k\V
turbine and a 15 000 kW compressor are required. This result demonstrates that a gas tur-
bine has a high back-work ratio.
11.10 THE SIMPLE GAS-TURBINE
Cycle with a Regenerator
The efficiency of the gas-turbine cycle may be improved by introducing a regenerator.
The simple open-cycle gas-turbine cycle with a regenerator is shown in Fig. 1 1.21, and
the corresponding ideal air-standard cycle with a regenerator is shown on the P-v and T-s
diagrams. In cycle l-2-x-3-4-;>-l, the temperature of the exhaust gas leaving the turbine
in state 4 is higher than the temperature of the gas leaving the compressor. Therefore, heat
can be transferred from the exhaust gases to the high-pressure gases leaving the compres-
sor. If this is done in a counterflow heat exchanger (a regenerator), the temperature of the
high-pressure gas leaving the regenerator, T x , may, m the ideal case, have a temperature
equal to T 4 , the temperature of the gas leaving the turbine. Heat transfer from the external
source is necessary only to increase the temperature from T x to T 3 . Aiea x-3-d-b-x repre-
sents the heat transferred, and area y-\—a—c—y represents the heat rejected.
The influence of pressure ratio on the simple gas-turbine cycle with a regenerator is
shown by considering cycle l-2'-3'-4-l. In this cycle the temperature of the exhaust gas
the Simple Gas-Turbine Cycle with a regenerator B 419
leaving the turbine is just equal to the temperature of the gas leaving the compressor;
therefore, utilizing a regenerator is not possible. This can be shown more exactly be deter-
mining the efficiency of the ideal gas-turbine cycle with a regenerator.
The efficiency of this cycle with regeneration is found as follows, where the states
areas given in Fig, 11.21.
H'net _
% =
< }h =
C P (T 3
~T 4 )
But for an ideal regenerator, T 4 = T x , and therefore q H — w,. Consequently,
C P (T 2 - T,)
Cm - T 4 )
Uyr, - i) _ r j^/Ftf-'^-i]
Thus, for the ideal cycle with regeneration the thermal efficiency depends not only
on the pressure ratio but also on the ratio of the minimum to maximum temperature. We
note that, in contrast to the Brayton cycle, the efficiency decreases with an increase in
pressure ratio.
The effectiveness or efficiency of a regenerator is given by the regenerator effi-
ciency, which can best be defined by reference to Fig. 11.22. State x represents the high-
pressure gas leaving the regenerator. In the ideal regenerator there would be only an
infinitesimal temperature difference between the two streams, and the high-pressure gas
would leave the regenerator at temperature T' xi and T' x = T 4 . In an actual regenerator,
420 H CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
which must operate with a finite temperature difference T„ the actual temperature leaving
the regenerator is therefore less than T' x . The regenerator efficiency is defined by
lu - h-,
v (11.11)
If the specific heat is assumed to be constant, the regenerator efficiency is also given by
the relation
A higher efficiency can be achieved by using a regenerator with a greater heat-
transfer area. However, this also increases the pressure drop, which represents a loss, and
both the pressure drop and the regenerator efficiency must be considered in determining
which regenerator gives maximum thermal efficiency for the cycle. From an economic
point of view, the cost of the regenerator must be weighed against the savings that can be
effected by its use.
EXAMPLE 11.8 If an ideal regenerator is incorporated into the cycle of Example 11.6, determine the
thermal efficiency of the cycle.
The diagram for this example is Fig. 1 1 .22. Values are from Example 1 1 .6. There-
fore, for the analysis of the high-temperature heat exchanger (combustion chamber),
from the first law, we have
q H =h 3 - h x
so that the solution is
T x = T 4 = 710.8 K
q H = J h - h x = C P (T 3 - T s ) = 1.004(1373.2 - 710.8) = 664.7 kJ/kg
w wt = 395.2 kJ/kg (from Example 11.6)
GAS-TURBINE POWER CYCLE CONFIGURATIONS M 421
11.11 Gas°turbine Power cycle
Configurations
The Brayton cycle, being the idealized model for the gas-turbine power plant, has a re-
versible, adiabatic compressor and a reversible, adiabatic turbine. In the following exam-
ple, we consider the effect of replacing these components with reversible, isothermal
processes.
EXAMPLE 11.9 An air-standard power cycle has the same states as given in Example 11.6. In this
cycle, however, the compressor and turbine are both reversible, isothermal processes.
Calculate the compressor work and the turbine work, and compare with the results of
Example 11.6.
Control volumes: Compressor, turbine.
Analysis
For each reversible, isothermal process, from Eq. 9. 1 8,
compared with +664.7 kJ/kg in the adiabatic turbine.
It is found that the isothermal process would be preferable to the adiabatic process
in both the compressor and turbine. The resulting cycle, called the Ericsson cycle, consists
of two reversible, constant-pressure processes and two reversible, constant-temperature
processes. The reason that the actual gas turbine does not attempt to emulate this cycle
rather than the Brayton cycle is that the compressor and turbine processes are both high-
flow rate processes involving work-related devices in which it is not practical to attempt
to transfer large quantities of heat. As a consequence, the processes tend to be essentially
adiabatic, so that this becomes the process in the model cycle.
There is a modification of the Brayton/gas turbine cycle that tends to change its per-
formance in the direction of the Ericsson cycle. This modification is to use multiple stages
of compression with intercooling, and also multiple stages of expansion with reheat. Such
a cycle with two stages of compression and expansion, and also incorporating a regenera-
tor, is shown in Fig. 11.23. The air-standard cycle is given on the corresponding T-s dia-
gram. It may be shown that for this cycle the maximum efficiency is obtained if equal
pressure ratios are maintained across the two compressors and the two turbines. In this
Solution
For the compressor,
w = -0.287 X 288.2 X In 10 = -190.5 kJ/kg
compared with -269.5 kJ/kg in the adiabatic compressor.
For the turbine,
w = -0.287 X 1373.2 X In 0.1 = +907.5 kJ/kg
422 H CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
Regenerator
FIGURE 11.23 The
ideal gas-turbine cycle
utilizing intercooling,
reheat, and a regenerator.
ideal cycle, it is assumed that the temperature of the air leaving the intercooler, 2* 3 , is
equal to the temperature of the air entering the first stage of compression, T u and that the
temperature after reheating, T $ , is equal to the temperature entering the first turbine, r ? .
Furthermore, in the idea! cycle it is assumed that the temperature of the high-pressure air
leaving the regenerator, T s , is equal to the temperature of the low-pressure air leavmg the
turbine, T 9 . , t . ,
If a large number of stages of compression and expansion are used, it is evident mat
the Ericsson cycle is approached. This is shown in Fig. 11. 24. In practice, the economical
limit to the number of stages is usually two or three. The turbine and compressor losses
and pressure drops that have already been discussed would be involved in any actual unit
employing this cycle. ,
The turbines and the compressors using this cycle can be utilized m a variety ot
ways. Two possible arrangements for closed cycles are shown in Fig. 1 1.25. One ad-
vantage frequently sought in a given arrangement is ease of control of the unit under
various loads. Detailed discussion of this point, however, is beyond the scope of this
text.
Gas-Turbine Power Cycle Configurations W 423
FIGURE 11.24
Temperature-entropy
diagram that shows how
the gas-turbine cycle with
many stages approaches
the Ericsson cycle.
VvWVA-
FIGURE 11.25 Some
arrangements of
components that may be
utilized in stationary gas-
hirbine power plants.
424 H Chapter eleven power and refrigeration Systems
11.12 The air-Standard Cycle
for jet propulsion
The next air-standard power cycle we consider is utilized in jet propulsion. In this cycle
the work done by the turbine is just sufficient to drive the compressor. The gases are ex-
panded in the turbine to a pressure for which the turbine work is just equal to the com-
pressor work. The exhaust pressure of the turbine will then be greater than that of the
surroundings, and the gas can be expanded in a nozzle to the pressure of the surroundings.
Since the gases leave at a high velocity, the change in momentum that the gases undergo
gives a thrust to the aircraft in which the engine is installed. A jet engine was shown in
Fig. 1.11, and the air-standard cycle for this situation is shown in Fig. 1 1.26. The princi-
ples governing this cycle follow from the analysis of the Brayton cycle plus that for a re-
versible, adiabatic nozzle.
THE AIR-STANDARD CYCLE FOR JET PROPULSION M 425
EXAMPLE 11.10
Consider an ideal jet propulsion cycle in which air enters the compressor at 0.1 MPa and
15°C. The pressure leaving the compressor is 1.0 MPa, and the maximum temperature is
1 100°C. The air expands in the turbine to a pressure at which the turbine work is just equal
to the compressor work. On leaving the turbine, the ah expands in a nozzle to 0.1 MPa. The
process is reversible and adiabatic. Determine the velocity of the air leaving the nozzle.
The model used is ideal gas with constant specific heat, at 300 K, and each process
is steady state with no potential energy change. The only kinetic energy change occurs
in the nozzle. The diagram is shown in Fig. 1 1.26.
The compressor analysis is the same as in Example 11.6. From the results of that
solution, we have
Pi -0.1 MPa, r 3 — 288.2 K
P 2 = 1.0 MPa, T 2 = 556.8 K
w c = 269.5 kj/kg
The turbine analysis is also the same as in Example 11.6. Here, however,
P 3 = 1.0 MPa, T 3 - 1373.2 K
w e = w t = Cffi - T 4 ) = 269.5 kJ/kg
T 3 -T 4 =
so that
269.5
1.004
- 268.6, T 4 = 1104.6 K
1373.2
1104.6
= 1.2432
- 2.142, P 4 = 0.4668 MPa
Control volume: Nozzle.
Inlet state: State 4 fixed (above).
Exit state: P 5 known.
Analysis
The first law gives
VI
h = h 5 + ^
The second law is
s 4 = s s
Solution
Since P 5 is 0,1 MPa, from the second law we find that T s = 710.8 K. Then
- 2^(7; - r 5 )
Vf = 2 X 1000 X 1.004(1104.6 - 710.8)
V 5 = 889 m/s
426 B Chapter Eleven Power and Refrigeration Systems
11.13 RECIPROCATING ENGINE POWER CYCLES
In Section 1 1.1, we discussed power cycles incorporating either steady-state processes or
cylinder/piston boundary- work processes. In that section, it was noted that for the steady-
state process, there is no work in a constant-pressure process. Each of the steady-state
power cycles presented in subsequent sections of this chapter incorporated two constant-
pressure heat-transfer processes. It was further noted in Section 11,1 that in a boundary-
movement work process, there is no work in a constant-volume process. In the next three
sections, we will present ideal air-standard power cycles for cylinder/piston boundary-
movement work processes, each example of which includes either one or two constant-
volume heat-transfer processes.
Before we describe the reciprocating engine cycles, we want to present a few com-
mon definitions and terms. Car engines typically have 4, 6, or 8 cylinders, each with a di-
ameter called bore B. The piston is connected to a crankshaft, as shown in Figure 1 1.27,
and as it rotates changing the crank angle, 0, the piston moves up or down with a stroke.
S = 2i? ereni (11.12)
This gives a displacement for all cylinders as
which is the main characterization of the engine size
est volume is the compression ratio
r v =CR= V m JV^ (11.14)
--N cyi A cyi S (11.13)
The ratio of the largest to the small-
The otto Cycle m 427
and both of these characteristics are fixed with the engine geometry. The net specific work
in a complete cycle is used to define a mean effective pressure
w«t = §Pdv s?^^ - ymin ) (n.15)
or net work per cylinder per cycle
= »"fnet = P^ftiY^ ~ Knin) (11.16)
We now Use this to find the rate of work (power) for the whole engine as
n iv cyX mw nil ^ — J r nM . ff P disp ] ^ (1 1.17}
where RPM is revolutions per minute. This result should be corrected with a factor \ for a
four-stroke engine, where 2 revolutions are needed for a complete cycle to also accom-
plish the intake and exhaust strokes.
11.14 the Otto Cycle
The air-standard Otto cycle is an ideal cycle that approximates a spark-ignition internal-
combustion engine. This cycle is shown on the P-v and T-s diagrams of Fig. 11.28.
Process 1-2 is an isentropic compression of the air as the piston moves from crank-end
dead center to head-end dead center. Heat is then added at constant volume while the pis-
ton is momentarily at rest at head-end dead center. (This process corresponds to the igni-
tion of the fuel-air mixture by the spark and the subsequent burning in the actual engine.)
Process 3-4 is an isentropic expansion, and process 4-1 is the rejection of heat from the
air while the piston is at crank-end dead center.
The thermal efficiency of this cycle is found as follows, assuming constant specific
heat of air;
mC v (T, - r 2 )
* — oir^-Q,
_ « WJT, - 1)
r,(7yr 2 - n
i -
428 M Chapter eleven power and Refrigeration Systems
We note further that
Therefore,
and
where
T 2 T x
. f . Vi _ n
r v — compression ratio — yr — y~
It is important to note that the efficiency of the air-standard Otto cycle is a function only
of the compression ratio and that the efficiency is increased by increasing the compression
ratio. Figure 11.29 shows a plot of the air-standard cycle thermal efficiency versus com-
pression ratio. It is also true of an actual spark-ignition engine that the efficiency can be
increased by increasing the compression ratio. The trend toward higher compression ra-
tios is prompted by the effort to obtain higher thermal efficiency. In the actual engine
there is an increased tendency for the fuel to detonate as the compression ratio is in-
creased. After detonation the fuel burns rapidly, and strong pressure waves present in the
engine cylinder give rise to the so-called spark knock. Therefore, the maximum compres-
sion ratio that can be used is fixed by the fact that detonation must be avoided. Advances
in compression ratios over the years in actual engines were originally made possible by
developing fuels with better antiknock characteristics, primarily through the addition of
tetraethy! lead. More recently, however, nonleaded gasolines with good antiknock charac-
teristics have been developed in an effort to reduce atmospheric contamination.
Some of the most important ways in which the actual open-cycle spark-ignition en-
gine deviates from the air-standard cycle are as follows:
1. The specific heats of the actual gases increase with an increase in temperature. ,
2. The combustion process replaces the heat-transfer process at high temperature, and
combustion may be incomplete.
The Otto Cycle 9 429
3. Each mechanical cycle of the engine involves an inlet and an exhaust process and,
because of the pressure drop through the valves, a certain amount of work is re-
quired to charge the cylinder with air and exhaust the products of combustion.
4. There will be considerable heat transfer between the gases in the cylinder and the
cylinder walls.
5. There will be irreversibilities associated with pressure and temperature gradients.
EXAMPLE 11.11 The compression ratio in an air-standard Otto cycle is 10. At the beginning of the com-
pression stoke the pressure is 0.1 MPa and the temperature is 15°C. The heat transfer to
the air per cycle is 1800 kJ/kg air. Determine
1. The pressure and temperature at the end of each process of the cycle.
2. The thermal efficiency.
3. The mean effective pressure.
Control mass: Air inside cylinder.
Diagram: Fig. 11.29.
P t = 0.1 MPa, T x = 288.2 K.
Four processes known (Fig. 11.29). Also, r v = 10 and
q H = 1800kJ/kg.
Ideal gas, constant specific heat, value at 300 K.
State information:
Process information:
Model:
Analysis
The second law for compression process 1-2 is
H = *l
so that
r
The first law for heat addition process 2-3 is
to - 2?3 = ih ~u 2 = C V (T } - r 2 )
The second law for expansion process 3-4 is
so that
= *3
430 M CHAPTER ELEVEN POWER. AND REFRIGERATION SYSTEMS
In addition,
1 Wntf
Solution
Substitution yields the following:
= Q.287 X 288.2 = a82 7 m 3 /kg
y * 100
= ( IlV" 1 = 10 - 4 = 2.51 19, T 2 = 723.9 K
^ = = io u = 25.12, P 2 = 2.512 MPa
y = M|Z = 0.0827 m 3 /kg
' 10
rf3 = Q(r 3 -T 2 ) = i800kJ/kg
7,-^ = ^ = 2510^ 7/ 3 = 3234K
3 2 0.717
r * _ = 3234 = 4467 j>_ = 11.222 MPa
2^ " P 2 723.9 ' '
II = f Y" 1 = 10 ' 4 = 2.5 1 19, T A - 1287.5 K
^ = fiiY = 10 L4 = 25.12, P 4 = 0.4467 MPa
__ 1 _J_=l-J- = 0.602 = 60.2%
This can be checked by finding the heat rejected:
tfi = CJPi ~ r 4 ) = 0.717(288.2 - 1287.5) = -716.5 kJ/kg
_ ! _ 21^1 = 0.602 = 60.2%
™> 1800
Wnrt = 1800 -716.5 = 1083.5 kJ/kg = (v, - u 2 )mep
me ? = (0^27^827) = 1456 ^
This is a high value for mean effective pressure, largely because the two constant-
volume heat transfer processes keep the total volume change to a — (compared
with a Brayton cycle, for example). Thus, the Otto cycle is a good model to emulate m
the cylinder-piston internal-combustion engine. At the other extreme, a low ™n defec-
tive pressure means a large piston displacement for a given power output, which in turn
means high frictional losses in an actual engine.
The Diesel Cycle m 431
11.15 the Diesel Cycle
The air-standard diesel cycle is shown in Fig. 11. 30. This is the ideal cycle for the diesel
engine, which is also called the compression-ignition engine.
In this cycle the heat is transferred to the working fluid at constant pressure. This
process corresponds to the injection and burning of the fuel in the actual engine. Since the
gas is expanding during the heat addition in the air-standard cycle, the heat transfer must
be just sufficient to maintain constant pressure. When state 3 is reached, the heat addition
ceases and the gas undergoes an isentropic expansion, process 3^1, until the piston
reaches crank-end dead center. As in the air-standard Otto cycle, a constant-volume rejec-
tion of heat at crank-end dead center replaces the exhaust and intake processes of the ac-
tual engine.
The efficiency of the diesel cycle is given by the relation
„ - i _ 6* - i c ^ " *i ) _ i flgyr, - i)
'* Qh C p {T, - T 2 ) kT 2 {yT 2 - 1) CU ■ 1 y)
The isentropic compression ratio is greater than the isentropic expansion ratio in the
diesel cycle. In addition, for a given state before compression and a given compression
ratio (that is 3 given states 1 and 2), the cycle efficiency decreases as the maximum temper-
ature increases. This is evident from the T-s diagram because the constant-pressure and
constant- volume lines converge, and increasing the temperature from 3 to 3' requires a
large addition of heat (area 3-3'-e-6-3) and results in a relatively small increase in work
(area 3-3'-4'-4-3).
A number of comparisons may be made between the Otto cycle and the diesel
cycle, but here we will note only two. Consider Otto cycle 1— 2— 3 W — 4— 1 and diesel cycle
1-2-3-4-1, which have the same state at the beginning of the compression stroke and the
same piston displacement and compression ratio. From the T-s diagram we see that the
Otto cycle has the higher efficiency. In practice, however, the diesel engine can operate
on a higher compression ratio than the spark-ignition engine. The reason is that in the
spark-ignition engine an air-fuel mixture is compressed, and detonation (spark knock) be-
comes a serious problem if too high a compression ratio is used. This problem does not
exist in the diesel engine because only air is compressed during the compression stroke.
432 9 CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
Therefore, we might compare an Otto cycle with a diesel cycle and in each case select
a compression ratio that might be achieved in practice. Such a comparison can be made by
considering Otto cycle 1—2'— 3— 4—1 and diesel cycle 1-2-3-4-1. The maximum pressure and
temperature are the same for both cycles, which means that the Otto cycle has a lower com-
pression ratio than the diesel cycle. It is evident from the T~s diagram that in this case the
diesel cycle has the higher efficiency. Thus, the conclusions drawn from a comparison of
these two cycles must always be related to the basis on which the comparison has been made.
The actual compression-ignition open cycle differs from the air-standard diesel
cycle in much the same way that the spark-ignition open cycle differs from the air-
standard Otto cycle.
EXAMPLE 11.12
An air-standard diesel cycle has a compression ratio of 20, and the heat transferred to
the working fluid per cycle is 1800 kJ/kg. At the beginning of the compression process,
the pressure is 0. 1 MPa and the temperature is 15°C. Determine
1. The pressure and temperature at each point in the cycle.
2. The thermal efficiency.
3. The mean effective pressure.
Control mass:
Diagram:
State information:
Process information:
Model:
Air inside cylinder.
Fig. 11.30.
J 9 ! = 0.1 MPa, T x = 288.2 K.
Four processes known (Fig. 11.30). Also,^ 20 and
q H = 1800kJ/kg.
Ideal gas, constant specific heat, value at 300 K.
Analysis
The second law for compression process 1-2 is
so that
V
Pi
The first law for heat addition.process 2-3 is
%h = ill = C P ( T 3 ~ T 2)
and the second law for expansion process 3-4 is
so that
The Stirling Cycle El 433
In addition.
Solution
Substitution gives
• _ 0.287 X 288.2
= 0.827 m 3 /kg
2 ~~ 20 " ~^20~~ = °- 04135 m &&
T ( V \ k ~~ l
^ = Ml - 20 - 4 = 3.3 145, T 2 = 955.2 K
Y X = \Y 2 ) = 20 w = 66.29, P 2 = 6.629 MPa
= = q,^ - r 2 ) = 1800 kJ/kg
T 2 -T 2 = - 1793 K, r 3 = 2748 K
V % T 3 2748
= ~ = ~™ = 2.8769, y 3 = 0.118 96 m 3 /kg
i = (tr = (oifkr = 2,i?i9) 74 = 1265 K
q L = A q x = C V (T X - T 4 ) = 0.717(288.2 - 1265) = -700.4 kJ/kg
HVt = 1800 - 700.4 = 1099.6 kJ/kg
„ _ w«t „ 1099.6 „ ^ 10/
m ,„ - W «t 1099.6 1 /fAA 1 T>
me P " «T=^ " 0.827 - 0.04135 = 1400kPa
11.16 The Stirling Cycle
The final air-standard power cycle to be discussed is the Stirling cycle, which is shown on
the P~v and T-s diagrams of Fig. 11.31. Heat is transferred to the working fluid during
the constant-volume process 2-3 and also during the isothermal expansion process 3-4.
Heat is rejected during the constant- volume process 4-1 and also during the isothermal
434 M CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
FIGURE 11.31 The
air-standard Stirling cycle.
compression process 1-2. Thus, this cycle is the same as the Otto cycle, with the adiabatic
processes of that cycle replaced with isothermal processes. Since the Stirling cycle in-
cludes two constant-volume heat-transfer processes, keeping the total volume change dur-
ing the cycle to a minimum, it is a good candidate for a cylinder/piston boundary work
application; it should have a high mean effective pressure.
Stirling-cycle engines have been developed in recent years as external-combustion en-
gines with regeneration. The significance of regeneration is noted from the ideal case shown
in Fig. 1 1 .3 1. Note that the heat transfer to the gas between states 2 and 3, area 2-3-&-0-2,
is exactly equal to the heat transfer from the gas between states 4 and 1, area \-4-d-c-l.
Thus, in the ideal cycle, all external heat supplied Q H takes place in the isothermal expan-
sion process 3-4, and all external heat rejection Q L takes place in the isothermal compres-
sion process 1-2. Since all heat is supplied and rejected isothermaHy, the efficiency of this
cycle equals the efficiency of a Carnot cycle operating between the same temperatures. The
same conclusions would be drawn in the case of an Ericsson cycle, which was discussed
briefly in Section 11.11, if that cycle were to include a regenerator as well.
11.17 Introduction to
Refrigeration Systems
In Section 1 U, we discussed cyclic heat engines consisting of four separate processes, ei-
ther steady-state or cylinder-piston boundary-movement work devices. We further al-
lowed for a working fluid that changes phase or for one that remains in a single phase
throughout the cycle. We then considered a power system comprised of four reversible
steady-state processes, two of which were constant-pressure heat-transfer processes, for
simplicity of equipment requirements, since these two processes involve no work. It was
further assumed that the other two work-involved processes were adiabatic and therefore
isentropic. The resulting power cycle appeared as in Fig. 1 1.2.
We now consider the basic ideal refrigeration system cycle in exactly the same terms
as those described earlier, except that each process is the reverse of that in the power cycle.
The result is the ideal cycle shown in Fig. 1 1.32. Note that if the entire cycle takes place in-
side the two-phase liquid-vapor dome, the resulting cycle is, as with the power cycle, the
Camot cycle, since the two constant-pressure processes are also isothermal. Otherwise, this
cycle is not a Carnot cycle. It is also noted, as before, that the net work input to the cycle is
equal to the area enclosed by the process lines 1-2-3-4-1, independently of whether.the
individual processes are steady state or cylinder/piston boundary movement.
The Vapor-Compression Refrigeration Cycle M 435
P
3
P
2
\
4
FIGURE 11.32 Four-
process refrigeration
cycle.
v
In the next section, we make one modification to this idealized basic refrigera-
tion system cycle in presenting and applying the model of refrigeration and heat pump
systems.
In this section, we consider the ideal refrigeration cycle for a working substance that
changes phase during the cycle, in a manner equivalent to that done with the Rankine
power cycle in Section 1 1.2. In doing so, we note that state 3 in Fig. 1 1.32 is saturated liq-
uid at the condenser temperature and state 1 is saturated vapor at the evaporator tempera-
ture. This means that the isentropic expansion process from 3-4 will be in the two-phase
region, and the substance there will be mostly liquid. As a consequence, there will be very
little work output from this process, such that it is not worth the cost of including this
piece of equipment in the system. We therefore replace the turbine with a throttling de-
vice, usually a valve or a length of small-diameter tubing, by which the working fluid is
throttled from the high-pressure to the low-pressure side. The resulting cycle become the
ideal model for a vapor-compression refrigeration system, which is shown in Fig. 11.33.
Saturated vapor at low pressure enters the compressor and undergoes a reversible adia-
batic compression, process 1-2. Heat is then rejected at constant pressure in process 2-3,
and the working fluid exits the condenser as saturated liquid. An adiabatic throttling
process, 3^, follows, and the working fluid is then evaporated at constant pressure,
process 4—1, to complete the cycle.
The similarity of this cycle to the reverse of the Rankine cycle has already been
noted. We also note the difference between this cycle and the ideal Carnot cycle, in which
the working fluid always remains inside the two-phase region, l'-2'— 3— 4'-l'. It is much
more expedient to have a compressor handle only vapor than a mixture of liquid and
vapor, as would be required in process 1 '-2' of the Carnot cycle. It is virtually impossible
to compress, at a reasonable rate, a mixture such as that represented by state 1 ' and still
maintain equilibrium between liquid and vapor. The other difference, that of replacing the
turbine by the throttling process, has already been discussed.
11.18 The Vapor- Compression
Eefrigeration Cycle
436 H ChapterEleven power and Refrigeration Systems
FIGURE 11.33 The
ideal vapor-compression
refrigeration cycle.
©
Expansion
vaive or
capillary tube
©-
Compressor
Work
4' 4
r r
The system described in Fig. 11,33 can be used for either of two purposes. The
first use is as a refrigeration system, in which case it is desired to maintain a space at a
low temperature 7\ relative to the ambient temperature T 3 . (In a real system, it would
be necessary to allow a finite temperature difference in both the evaporator and con-
denser to provide a finite rate of heat transfer in each.) Thus, the reason for building
the system in this case is the quantity q L . The measure of performance of a refrigera-
tion system is given in terms of the coefficient of performance, which was defined
in Chapter 7 as
j6=|r (11-20)
The second use of this system described in Fig. 1 1.33 is as a heat pump system, in
which case it is desired to maintain a space at a temperature T 3 above that of the ambient
(or other source) 7\, In this case, the reason for building the system is the quantity q H , and
the coefficient of performance for the heat pump, is now
)3'=f (11-21)
Refrigeration systems and heat pump systems are, of course, different in terms of
design variables, but the analysis of the two is the same. When we discuss refrigerators in
this and the following two sections, it should be kept in mind that the same comments
generally apply to heat pump systems as well.
EXAMPLE 11.13 Consider an ideal refrigeration cycle that uses R-134a as the working fluid. The temper-
ature of the refrigerant in the evaporator is -20°C, and in the condenser it is 40°C. The
refrigerant is circulated at the rate of 0.03 kg/s. Determine the coefficient of perfor-
mance and the capacity of the plant in rate of refrigeration.
The diagram for this example is as shown in Fig. 1 1.33. For each control volume
analyzed, the thermodynamic model is as exhibited in the R-134a tables. Each process is
steady state, with no changes in kinetic or potential energy.
Conti'ol volume: Compressor.
Inlet state: T x known, saturated vapor; state fixed.
Exit state: P 2 known (saturation pressure at T 3 ).
The Vapor-Compression Refrigeration Cycle B 437
Analysis
The first and second laws are
w c = h 2 - ky
s 2 =s x
Solution
At r 3 = 40°G,
P g = P 2 = 1017 kPa
From the R-134a tables, we get
h x = 386.1 kJ/kg, s x = 1.7395 kJ/kg K
Therefore,
s 2 =Si = 1.7395 kJ/kgK
so that
T 2 = 47.7°C and h 2 = 428.4 kJ/kg
w c = h 2 - A, - 428.4 - 386.1 = 42.3 kJ/kg
Control volume: Expansion valve.
Inlet state: T 3 known, saturated liquid; state fixed.
Exit state: T 4 known.
Analysis
The first law is
h 3 = h A
Solution
Numerically, we have
h 4 = h = 256.5 kJ/kg
Conti'ol volume: Evaporator.
Inlet state: State 4 known (as given).
Exit state: State 1 known (as given).
Analysis
The first law is
q L = h x ~ h 4
Solution
Substituting, we have
qi = h l - h 4 = 386.1 - 256.5 = 129.6 kJ/kg
438 B Chapter Eleven power and Refrigeration Systems
Therefore,
Refrigeration capacity = 129.6 X 0.03 - 3.89 kW
11.19 WORKING FLUIDS FOR VAPOR-
COMPRESSION REFRIGERATION Systems
A much larger number of different working fluids (refrigerants) are utilized in vapor-
compression refrigeration systems than in vapor power cycles. Ammonia and sulfur diox-
ide were important in the early days of vapor-compression refrigeration, but both are
highly toxic and therefore dangerous substances. For many years now, the principal re-
frigerants have been the halogenated hydrocarbons, which are marketed under the trade
names of Freon and Genatron. For example, dichlorodifluoromethane (CCI 2 F 2 ) is known
as Freon-12 and Genatron- 12, and therefore as refrigerant- 12 or R-12. This group of sub-
stances, known commonly as chlorofluorocarbons or CFCs, are chemically very stable at
ambient temperature, especially those lacking any hydrogen atoms. This characteristic is
necessary for a refrigerant working fluid. This same characteristic, however, has devastat-
ing consequences if the gas, having leaked from an appliance into the atmosphere, spends
many years slowly diffusing upward into the stratosphere. There it is broken down, re-
leasing chlorine, which destroys the protective ozone layer of the stratosphere. It is there-
fore of ovenvhelrrring importance to us all to eliminate completely the widely used but
life-threatening CFCs, particularly R-l 1 and R-12, and to develop suitable and acceptable
replacements. The CFCs containing hydrogen (often termed HCFCs), such as R-22, have
shorter atmospheric lifetimes and therefore are not as likely to reach the stratosphere be-
fore being broken up and rendered harmless. The most desirable fluids, called HFCs, con-
tain no chlorine atoms at all.
There are two important considerations when selecting refrigerant working fluids:
the temperature at which refrigeration is needed and the type of equipment to be used.
As the refrigerant undergoes a change of phase during the heat-transfer process, the
pressure of the refrigerant will be the saturation pressure during the heat supply and heat
rejection processes. Low pressures mean large specific volumes and correspondingly
large equipment. High pressures mean smaller equipment, but it must be designed to
withstand higher pressure. In particular, the pressures should be well below the critical
pressure. For extremely low temperature applications a binary fluid system may be used
by cascading two separate systems.
The type of compressor used has a particular bearing on the refrigerant. Reciprocat-
ing compressors are best adapted to low specific volumes, which means higher pressures,
whereas centrifugal compressors are most suitable for low pressures and high specific
volumes.
It is also important that the refrigerants used in domestic appliances be nontoxic.
Other beneficial characteristics, in addition to being environmentally acceptable, are iras-
cibility with compressor oil, dielectric strength, stability, and low cost. Refrigerants, how-
DEVIATION OF THE ACTUAL VAPOR- COM PRESS ION REFRIGERATION CYCLE
a 439
ever, have an "unfortunate tendency to cause corrosion. For given temperatures during
evaporation and condensation, not all refrigerants have the same coefficient of perfor-
mance for the ideal cycle. It is, of course, desirable to use the refrigerant with the highest
coefficient of performance, other factors permitting.
11,20 Deviation of the Actual
vapor-compression refrigeration
Cycle from the ideal cycle
The actual refrigeration cycle deviates from the ideal cycle primarily because of pressure
drops associated with fluid flow and heat transfer to or from the surroundings. The actual
cycle might approach the one shown in Fig. 1 1 .34.
The vapor entering the compressor will probably be superheated. During the
compression process, there are irreversibilities and heat transfer either to or from the
surroundings, depending on the temperature of the refrigerant and the surroundings.
Therefore, the entropy might increase or decrease during this process, for the irre-
versibility and the heat transferred to the refrigerant cause an increase in entropy, and
the heat transferred from the refrigerant causes a decrease in entropy. These possibili-
ties are represented by the two dashed lines 1-2 and 1-2'. The pressure of the liquid
leaving the condenser will be less than the pressure of the vapor entering, and the tem-
perature of the refrigerant in the condenser will be somewhat higher than that of the
surroundings to which heat is being transferred. Usually, the temperature of the liquid
leaving the condenser is lower than the saturation temperature. It might drop some-
what more in the piping between the condenser and expansion valve. This represents a
gain, however, because as a result of this heat transfer the refrigerant enters the evapo-
rator with a lower enthalpy, which permits more heat to be transferred to the refriger-
ant in the evaporator.
There is some drop in pressure as the refrigerant flows through the evaporator. It
may be slightly superheated as it leaves the evaporator, and through heat transferred from
the surroundings its temperature will increase in the piping between the evaporator and
the compressor. This heat transfer represents a loss because it increases the work of the
compressor, since the fluid entering it has an increased specific volume.
440 H Chapter Eleven Power and Refrigeration systems
EXAMPLE 11.14 A refrigeration cycle utilizes R-12 as the working fluid. The following are the properties
at various points of the cycle designated in Fig. 1 1 .34.
Pi
= 125 kPa,
- -io°c
Pi
= 1.2 MPa,
T 2
= I00°C
P3
= 1.19 MPa,
T 3
= 80°C
Pa
- 1.16 MPa,
T 4
= 45°C
Ps
- 1.15 MPa,
Ts
-40°C
P6
= P 7 = 140 kPa,
x 6
= *7
P*
= 130 kPa,
= ~20°C
The heat transfer from R-12 during the compression process is 4 kJ/kg. Deteimine
the coefficient of performance of this cycle.
For each control volume, the R-12 tables are the model. Each process is steady
state with no changes in kinetic or potential energy.
As before, we break the process down into stages, treating the compressor, the
throttling value and line, and the evaporator in turn.
Control volume: Compressor.
Inlet state: P u T { known; state fixed.
Exit state: P 2s T 2 known; state fixed.
Analysis
From the first law, we have
q + h,
W c
Solution
From the R-12 tables, we read
ft, = 185.16 kJ/kg, h 2 = 245.52 kJ/kg
Therefore,
w c = 245.52 - 185.16 - (-4) = 64.36 kJ/kg
Control volume: Throttling valve plus line.
Inlet state: P 5$ T 5 known; state fixed.
Exit state: P 7 = P 6 known, x 7 = ,r 6 .
Analysis
The first law is
h 5 = h 6
= h 2 + w
= -w = h z - h x - q
The Ammonia Absorption refrigeration Cycle M 441
Since x 7 = x 6i it follows that h 7 = h 6 .
Solution
Numerically, we obtain
A s = h 6 = h 7 = 74.53
Control volume: Evaporator.
Inlet state: P 1) h 7 known (above).
Exit state: P g , T $ known; state fixed.
Analysis
The first law is
q L = h s - h n
Solution
Substitution gives
<7z = h - Ay = 179.12 - 74.53 = 104.59 kJ/kg
Therefore,
11.21 The Ammonia Absorption
Refrigeration cycle
The ammonia absorption refrigeration cycle differs from the vapor-compression cycle in
the manner in which compression is achieved. In the absorption cycle the low-pressure
ammonia vapor is absorbed in water, and the liquid solution is pumped to a high pressure
by a liquid pump. Figure 11.35 shows a schematic arrangement of the essential elements
of such a system.
The low-pressure amrnonia vapor leaving the evaporator enters the absorber
where it is absorbed in the weak ammonia solution. This process takes place at a tem-
perature slightly higher than that of the surroundings. Heat must be transferred to the
surroundings during this process. The strong ammonia solution is then pumped through
a heat exchanger to the generator where a higher pressure and temperature are main-
tained. Under these conditions, ammonia vapor is driven from the solution as heat is
transferred from a high- temperature source. The ammonia vapor goes to the condenser
where it is condensed, as in a vapor-compression system, and then to the expansion
valve and evaporator. The weak ammonia solution is returned to the absorber through
the heat exchanger.
The distinctive feature of the absorption system is that very little work input is
required because the pumping process involves a liquid. This follows from the fact
442 B Chapter Eleven Power and refrigeration Systems
FIGURE 11.35 The
ammonia-absorption
refrigeration cycle.
Off (from =;
high-temperature
source)
Strong
ammonia
solution
High-pressure ammonia vapor
Generator
Weak
r ammonia
solution
H (to surroundings)
Condenser
Liquid i r
ammonia
Heat
exchanger
Low-pressure ammonia vapor
-sj — —
Expansion
valve
<8>
Absorber
Pump
Evaporator
Q L ' (\o surroundings)
Ql
{from cold box)
that for a reversible steady-state process with negligible changes in kinetic and poten-
tial energy, the work is equal to - / v dP and the specific volume of the liquid is much
less than the specific volume of the vapor. However, a relatively high-temperature
source of heat must be available (100° to 200°C). There is more equipment in an ab-
sorption system than in a vapor-compression system, and it can usually be economi-
cally justified only when a suitable source of heat is available that would otherwise be
wasted. In recent years, the absorption cycle has been given increased attention in con-
nection with alternative energy sources, for example, solar energy or supplies of geo-
thermal energy.
This cycle brings out the important principle that since the shaft work in a re-
versible steady-state process with negligible changes in kinetic and potential energy is
-J v dP, a compression process should take place with the smallest possible specific
volume.
11.22 The Air-Standard
refrigeration cycle
If we consider the original ideal four-process refrigeration cycle of Fig. 11.32 with a non-
condensing (gaseous) working fluid, then the work output during the isentropic expansion
process is not negligibly small, as was the case with a condensing working fluid. There-
fore, we retain the turbine in the four-steady-state process ideal air-standard refrigeration
cycle shown in Fig. 11.36. This cycle is seen to be the reverse Brayton cycle, and it is
the ammonia absorption refrigeration Cycle ffl 443
used in practice in the liquefaction of air and other gases and also in certain special situa-
tions that require refrigeration, such as aircraft cooling systems. After compression from
states 1 to 2, the air is cooled as heat is transferred to the surroundings at temperature T .
The air is then expanded in process 3-4 to the pressure entering the compressor, and the
temperature drops to 7 4 in the expander. Heat may then be transferred to the air until tem-
perature T L is reached. The work for this cycle is represented by area 1-2—3-4—1, and the
refrigeration effect is represented by area 4-l-b-a-4. The coefficient of performance is
the ratio of these two areas.
In practice, this cycle has been used to cool aircraft in an open cycle, A simplified
form is shown in Fig. 1 1 .37. Upon leaving the expander, the cool air is blown directly into
the cabin, thus providing the cooling effect where needed.
When counterflow heat exchangers are incorporated, very low temperatures can be
obtained. This is essentially the cycle used in low-pressure air liquefaction plants and in
other liquefaction devices such as the Collins helium liquefier. The ideal cycle is as
shown in Fig. 11.38. Because the expander operates at very low temperature, the de-
signer is faced with unique problems in providing lubrication and choosing materials.
444 ffl CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
Heat exchanger
FIGURE 11.38 The
air-refrigeration cycle
utilizing a heat exchanger.
=£4 Compressor
EXAMPLE 11.15 Consider the simple air-standard refrigeration cycle of Fig. 1 1 .36. Air enters the compressor
at 0.1 MPa and -20°C and leaves at 0.5 MPa. Air enters the expander at 15°C. Determine
1. The coefficient of performance for this cycle.
2. The rate at which air must enter the compressor to provide 1 kW of refrigeration.
For each control volume in this example, the model is ideal gas with constant specific
heat, at 300 K, and each process is steady state with no kinetic or potential energy
changes. The diagram for this example is Fig. 1 1.36.
Control volume: Compressor.
Met state: P u 7\ known; state fixed.
Exit state: P 2 known.
Analysis *
The first law is
w c - h 2 A,
(Here w c designates work into the compressor.)
The second law gives
s l =s 2
so that
2 toY*- 1 "*
Solution
Substituting, we obtain
= 1.004(401.2 - 253.2) - 148.5 kj/kg
_ <r0.2SS _
= 1.5845,
401.2 K
THE AMMONIA ABSORPTION REFRIGERATION CYCLE H 445
Control volume: Expander. ,
Inlet state: P 3 (= P 7 ) known, T 3 known; state fixed.
Exit state: P 4 (= P^) known.
Analysis
From the first law, we have
The second law gives
so that
Solution
Therefore,
w t — h 3 — h A
s 3 = s 4
_ 1 0.286 _
_ , _ , - 1.5845, T 4 = 181.9 K
J- 4 \"4j
w, = h 3 - h A = 1.004(288.2 - 181.9) = 106.7 kJ/kg
Control volume: High-temperature heat exchanger.
Inlet state: State 2 known (as given).
Exit state: State 3 known (as given).
Analysis
The first law is
Solution
Substitution gives
q H - h 2
q H = h 2 - h 3 (heat rejected)
h 3 = CJT 2 - T 3 ) = 1.004(401.2 - 288.2) = 113.4 kJ/kg
Control volume: Low-temperature heat exchanger.
Inlet state: State 4 known (as given).
Exit state: State 1 known (as given).
Analysis
The first law is
446 m Chapter Eleven power and Refrigeration Systems
Solution
Substituting, we obtain
q L = h y - h = C p (Ti ~ t a) = 1.004(253.2 - 181.9) = 71.6 kJ/kg
Therefore,
w«t = w c - w t = 148.5 - 106.7 = 41.8 kJ/kg
. ( h. _ 71.6 .... , 71 o
^- w IiCt 41.8 ' L713 ; /;
To provide 1 kW of refrigeration capacity, we have
"-|-7T6 =a014ks/s : }>:^:(:
11.23 Combined-Cycle Power
and refrigeration systems
In many situations it is desirable to combine two cycles in series, either power systems or
refrigeration systems, to take advantage of a very wide temperature range or to utilize what
would otherwise be waste heat to improve efficiency. One combined power cycle, shown
in Fig. 1 1 .39 as a simple steam cycle with a mercury-topping cycle, is often referred to as a
binary cycle. The advantage of this combined system is that mercury has a very low vapor
pressure relative to that for water. Therefore, it is possible for an isothermal boiling process
in the mercury to take place at a high temperature, much higher than the critical tempera-
ture of water, but still at a moderate pressure. The mercury condenser then provides an
isothermal heat source as input to the steam boiler, such that the two cycles can be closely
matched by proper selection of the cycle variables, with the resulting combined cycle then
Combined-Cycle Power and refrigeration Systems
m 447
FIGURE 11.40
Combined
Brayton/Rankine cycle
power system.
P 5 = p 4 © v /y//////////////////// .
AA/VW
/ V V V V V Y/
— ^— WVVv — $~ = °
-©
Qcond
having a high thermal efficiency. Saturation pressures and temperatures for a typical
mercury-water binary cycle are shown in the T~s diagram of Fig. II. 39.
A different type of combined cycle that has seen considerable attention is to use the
"waste heat" exhaust from a Brayton cycle gas-turbine engine (or another combustion en-
gine such as a diesel engine) as the heat source for a steam or other vapor power cycle, in
which case the vapor cycle acts as a bottoming cycle for the gas engine, in order to im-
prove the overall thermal efficiency of the combined power system. Such a system, utiliz-
ing a gas turbine and a steam Rankine cycle, is shown in Fig. 1 1 .40. In such a combination,
there is a natural mismatch using the cooling of a noncondensing gas as the energy source
to effect an isothermal boiling process plus superheating the vapor, and careful design is
required to avoid a pinch point, a condition at which the gas has cooled to the vapor boiling
temperature without having provided sufficient energy to complete the boiling process,
j One way to take advantage of the cooling exhaust gas in the Brayton-cycle portion
I of the combined system is to utilize a mixture as the working fluid in the Rankine cycle.
! An example of this type of application is the Kalina cycle, which uses ammonia-water
I mixtures as the working fluid in the Rankine-type cycle. Such a cycle can be made very
efficient, inasmuch as the temperature differences between the two fluid streams can be
controlled through careful design of the combined system.
Combined cycles are used in refrigeration systems in cases in which there is a very
large temperature difference between the ambient surroundings and the refrigerated space.
Such a refrigeration system is often called a cascade system, an example of which is shown
in Fig. 1 1 ,4 1 . In this case, the refrigerant R-22 is used in the refrigeration system rejecting
heat to the ambient surroundings, while its evaporator picks up the heat rejected in the low-
temperature system condenser, the low temperature working fluid in this case being R-23,
L
448 m CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
FIGURE 11.41
Combined-cycle cascade
refrigeration system.
Sat. liquid R-22
40°C
<X> Valve
v//////////////////////.
Sat.vaporR-22 f _ 5 ^ WWV
Insulated heat
exchanger
Sat. liquid R-23
-10°C
Valve
Sat. vapor R-23
-80°C
I Cold space
whose thermodynamic properties are suited to work as a refrigerant in this low-temperafure
range. As with the other combined-cycle systems, the working fluids and design variables
must be considered very carefully to optimize the performance of each unit.
We have described only a few combined-cycle systems here, as examples of the
types of applications that can be dealt with and the resulting improvement in overall per-
formance that can occur. Obviously, many other combinations of power and refrigeration
systems are possible. Some of these are discussed in the problems at the end of the chapter.
SUMMARY a numrjer f standard power-producing cycles and refrigeration cycles are presented.
First we cover a number of stationary and mobile power-producing heat engines. The
Rankine cycle and its variations represent a steam power plant, which produces most of
the world production of electricity. The heat input can come from combustion of fossil
fuels, a nuclear reactor, solar radiation, or any other heat source that can generate a tem-
perature high enough to boil water at a high pressure. In low- or very high-temperature
applications, substances other than water can be used. Modifications to the basic cycle
such as reheat, closed, and open feedwater heaters are covered together with applications
where the electricity is cogenerated with a base demand for process steam.
A Brayton cycle is a gas turbine producing electricity and, with a modification, a jet
engine producing thrust. This is a high-power, low-mass, and low-volume device and is
used where space and weight are at a premium cost. A high back-work ratio makes this
Key concepts and formulas H 449
cycle sensitive to the compressor efficiency. A number of variations and configurations
for the Brayton cycle with regenerators and intercoolers are shown.
Piston/cylinder devices are shown for the Otto and Diesel cycles modeling the gaso-
line and diesel engines, which can be two- or four-stroke engines. Cold air properties are
used to show the influence of compression ratio on the thermal efficiency, and the mean
effective pressure is used to relate the engine size to total power output. We give a short
mention of the Stirling cycle as an example of an external combustion engine.
Standard refrigeration systems are covered by the vapor-compression refrigeration
cycle. This applies to household refrigerators, air conditioners, and heat pumps as well as
commercial units to lower temperature ranges. As a special cycle, we mention the ammo-
nia absorption cycle and cover the air-standard refrigeration cycle in detail.
The chapter is completed with a short description of combined cycle applications. This
covers stacked or cascade systems for large temperature spans and combinations of different
kinds of cycles where one can be added as a topping cycle or a bottoming cycle. Often a
Rankine cycle uses exhaust energy from a Brayton cycle in larger stationary applications.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
• Apply the general laws to control volumes with several devices forming a complete
system.
• Have a knowledge of how many common power-producing devices work.
• Have a knowledge of how simple refrigerators and heat pumps work.
• Know that most cycle devices do not operate in Camot cycles.
• Know that real devices have lower efficiencies/COP than ideal cycles.
• Have a sense of the most influential parameters for each type of cycle.
• Have an idea about the importance of the component efficiency for the overall cycle
efficiency or COP.
• Know that most real cycles have modifications to the basic cycle setup.
• Know that many of these devices affect our environment
• Know the principle of combining different cycles.
Key concepts
and formulas
Rankine Cycle
Open feedwater heater
Closed feedwater heater
Deaerating FWH
Cogeneration
Brayton Cycle
Compression ratio
Regenerator
Intercooler
Jet engine
Thrust
Propulsive power
Feedwater mixed with extraction steam, exit as saturated
liquid
Feedwater heated by extraction steam, no mixing
Open feedwater heater operating at P atm to vent gas out
Turbine power is cogenerated with a desired steam supply
Pressure ratio r p = Pug/Pi^,
Dual fluid heat exchanger, uses exhaust flow energy
Cooler between compressor stages, reduces work input
No shaftwork out, kinetic energy generated in exit nozzle
F = m(V e ~ Vf) momentum equation
» f =/ r V lliKBft = i«(V < -V / )V,
450 M Chapter Eleven power and Refrigeration Systems
Piston Cylinder Power Cycles
Compression ratio
Displacement (1 cyl.)
Stroke
Mean effective pressure
Power by 1 cylinder
Volume ratio /-„ = CR = V m3J JV min
Vmn ~V mln = mtiu^ ~ U min ) = SA cyl
5 = 2 i? crank , piston travel in compression or expansion.
W — mw.
RPM
60
(times | for four-stroke cycle)
Refrigeration Cycle
Coefficient of performance
_ Ql _ %l
COP - 0REF - -r- -
Wr
Combined Cycles
Topping, bottoming cycle
Cascade system
The high- and low-temperature cycles
Stacked refrigeration cycles
Concept-Study guide problems
11.1 Is a steam power plant nmrring in a Carnot cycle? 11,12
Name the four processes.
11.2 Consider a Rankine cycle without superheat. How 11.13
many single properties are needed to determine
the cycle? Repeat the answer for a cycle with su- n.14
perheat.
11.3 Which component determines the high pressure in
a Rankine cycle? What determines the low pres- jx.is
sure?
11.4 Mention two benefits of a reheat cycle.
11.5 What is the difference between an open and a 11,16
closed feedwater heater?
11.6 Can the energy removed in a power plant con-
denser be useful? 11.17
11.7 In a cogenerating power plant, what is cogener-
ated?
11.8 Why is the back- work ratio much higher in the.
Brayton cycle than in the Rankine cycle? 11.18
11.9 The Brayton cycle has the same four processes as
the Rankine cycle, but the T-s and P-v diagrams
look very different; why is that? 11.19
11.10 Is it always possible to add a regenerator to the
Brayton cycle? What happens when the pressure
ratio is increased? 11.20
11.11 Why would you use an intercooler between com-
pressor stages?
The jet engine does not produce shaftwork; how
is power produced?
How is the compression in the Otto cycle differ-
ent from that in the Brayton cycle?
Does the inlet state (P b 7\) have any influence on
the Otto-cycle efficiency? How about the power
produced by a real car engine?
How many parameters do you need to know to
completely describe the Otto cycle? How about
the Diesel cycle?
The exhaust and inlet flow processes are not in-
cluded in the Otto or Diesel cycles. How do these
necessary processes affect the cycle performance?
A refrigerator in my 20°C kitchen uses R-12, and
I want to make ice cubes at — 5°C. What is the
minimum high P and the maximum low P it can
use?
How many parameters are needed to completely
determine a standard vapor-compression refriger-
ation cycle?
Why would one consider a combined cycle sys-
tem for a power plant? for a heat pump or refrig-
erator?
Since any heat transfer is driven by a temperature
difference, how does that affect all the real cycles
relative to the ideal cycles?
Homework Problems H 451
Homework problems
Rankine Cycles, Power Plants
Simple Cycles
11.21 A steam power plant, as shown in Fig. 11.3, oper-
ating in a Ranldne cycle has saturated vapor at 3
MPa leaving the boiler. The turbine exhausts to
the condenser operating at 10 kPa. Find the spe-
cific work and heat transfer in each of the ideal
components and the cycle efficiency.
11.22 Consider a solar-energy-powered ideal Rankine
cycle that uses water as the working fluid. Satu-
rated vapor leaves the solar collector at 175°Cj
and the condenser pressure is 1 kPa. Determine
the thermal efficiency of this cycle.
11.23 A utility runs a Rankine cycle with a water boiler
at 3MPa } and the cycle has the highest and lowest
temperatures of 450°C and 45°C, respectively.
Find the plant efficiency and the efficiency of a
Carnot cycle with the same temperatures.
11.24 A Rankine cycle uses ammonia as the working
substance and powered by solar energy. It heats
the ammonia to 140°C at 5000 kPa in the
boiler/superheater. The condenser is water
cooled, and the exit is kept at 25°C. Find (T, P,
andx if applicable) for all four states in the cycle.
11.25 A steam power plant operating in an ideal Rankine
cycle has a high pressure of 5 MPa and a low pres-
sure of 15 kPa. The turbine exhaust state should
have a quality of at least 95%, and the turbine power
generated should be 7.5 MW. Find the necessary
boiler exit temperature and the total mass flow rate.
11.26 A supply of geothermal hot water is to be used as
the energy source in an ideal Rankine cycle, with
R-134a as the cycle working fluid. Saturated
vapor R-134a leaves the boiler at a temperature of
85°C, and the condenser temperature is 40°C. Cal-
culate the thermal efficiency of this cycle.
11.27 Do Problem 1 1 .26 with R-22 as the working fluid.
11.28 Do Problem 11.26 with ammonia as the working
fluid.
11.29 Consider the boiler in Problem 11.26 where the
geothermal hot water brings the R-134a to satu-
rated vapor. Assume a counterflowing heat ex-
changer arrangement. The geothermal water
temperature should be equal to or greater than the
R-134a temperature at any location inside the heat
exchanger. The point with the smallest tempera-
ture difference between the source and the work-
ing fluid is called the pinch point shown in Fig.
PI 1.29. If 2 kg/s of geothermal water is available
at 95°C, what is the maximum power output of
this cycle for R-134a as the working fluid? {Hint
split the heat exchanger C.V. into two so that the
pinch point with AT — 0,T— 85°C appears.)
Pinch
point
R-134a-
Liquid
D
Boiler
3
heater
[ — >
— I— s-
C
B
-«-+—
-« — I
R-134a
85° C
H 2
95° C
FIGURE P11.29
11.30 Do the previous problem with R-22 as the work-
ing fluid,
11.31 Consider the ammonia Rankine-cycle power plant
shown in Fig. PI 1.31, a plant that was designed to
operate in a location where the ocean water tem-
perature is 25°C near the surface and 5°C at some
greater depth. The mass flow rate of the working
fluid is 1000 kg/s.
a. Determine the turbine power output and the
pump power input for the cycle.
23 °c
Insulated heat
exchanger
©—
Purnprh
-MWv-f
25 C C
Saturated
liquid NH 3 @-
T 3 = 10 C C
NH-, CYCLE
' //////////////////////A
Surface
H,0
Saturated
-© vapor NH 3
= 20=C
5°C deep H 2
FIGURE P11.31
AW/v-l
© P 2 = P 3
Insulated heat
exchanger
7 e C
452 B CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
b. Determine the mass flow rate of water through
each heat exchanger.
c. What is the thermal efficiency of this power
plant?
11.32 A smaller power plant produces 25 kg/s steam at
3 MPa, 600°C, in the boiler. It cools the con-
denser with ocean water coming in at 12°C and
returned at 15°C so the condenser exit is at 45°C.
Find the net power output and the required mass
flow rate of ocean water.
11.33 The power plant in Problem 11.21 is modified to
have a superheater section following the boiler so
that the steam leaves the superheater at 3 MPa
and 400°C. Find the specific work and heat trans-
fer in each of the ideal components and the cycle
efficiency,
11.34 A steam power plant has a steam generator exit at
4 MPa and 500°C and a condenser exit tempera-
ture of 45°C. Assume all components are ideal
and find the cycle efficiency and the specific work
and heat transfer in the components.
11.35 Consider an ideal Rankine cycle using water with
a high-pressure side of the cycle at a supercritical
pressure. Such a cycle has a potential advantage
of minimizing local temperature differences be-
tween the fluids in the steam generator, such as
the instance in which the high-temperafure energy
source is the hot exhaust gas from a gas-turbine
engine. Calculate the thermal efficiency of the
cycle if the state entering the turbine is 30 MPa,
550°C, and the condenser pressure is 5 kPa. What
is the steam quality at the turbine exit?
Reheat Cycles
11.36 A smaller power plant produces steam at 3 MPa,
600°C, in the boiler. It keeps the condenser at
45°C by transfer of 10 MW out as heat transfer.
The first turbine section expands to 500 kPa, and
then flow is reheated followed by the expansion
in the low-pressure turbine. Find the reheat tem-
perature so that the turbine output is saturated
vapor. For this reheat, find the total turbine power
output and the boiler heat transfer.
11.37 Consider an ideal steam reheat cycle as shown in
Fig. 11.7, where steam enters the high-pressure
turbine at 3 MPa and 400°C and then expands to
0.8 MPa. It is then reheated at constant pressure
0.8 MPa to 400°C and expands to 10 kPa in the
low-pressure turbine. Calculate the thermal effi-
ciency and the moisture content of the steam leav-
ing the low-pressure turbine.
11.38 A smaller power plant produces 25 kg/s steam at
3 MPa, 600°C, in the boiler. It cools the con-
denser with ocean water so that the condenser exit
is at 45°C, There is a reheat done at 500 kPa up to
400°C, and then expansion takes place in the low-
pressure turbine. Find the net power output and
the total heat transfer in the boiler.
11.39 The reheat pressure affects the operating variables
and thus turbine performance. Repeat Problem
11.37 twice, using 0.6 and 1.0 MPa for the reheat
pressure.
11.40 The effect of a number of reheat stages on the
ideal steam reheat cycle is to be studied. Repeat
Problem 11.37 using two reheat stages, one stage
at 1.2 MPa and the second at 0.2 MPa, instead of
the single reheat stage at 0.8 MPa.
Open Feedwater Heaters
11.41 An open feedwater heater in a regenerative steam
power cycle receives 20 kg/s of water at 100°C
and 2 MPa. The extraction steam from the turbine
enters the heater at 2 MPa and 275°C, and all the
feedwater leaves as saturated liquid. What is the
required mass flow rate of the extraction steam?
11.42 A power plant with one open feedwater heater has
a condenser temperature of 45°C, a maximum
pressure of 5 MPa, and boiler exit temperature of
900°C. Extraction steam at 1 MPa to the feedwa-
ter heater is mixed with the feedwater line so that
the exit is saturated liquid into the second pump.
Find the fraction of extraction steam flow and the
two specific pump work inputs.
11.43 A Rankine cycle operating with ammonia is
heated by some low-temperature source so that
the highest T is 120°C at a pressure of 5000 kPa.
Its low pressure is 1003 kPa, and it operates with
one open feedwater heater at 2033 kPa. The total
flow rate is 5 kg/s. Find the extraction flow rate to
the feedwater heater assuming its outlet state is
saturated liquid at 2033 kPa. Find the total power
to the two pumps.
11.44 A steam power plant operates with a boiler output
of 20 kg/s steam at 2 MPa and 600°C. The con-
denser operates at 50°C, dumping energy to a
river that has an average temperature of 20°C.
Homework Problems H 453
There is one open feedwater heater with extrac-
tion from the turbine at 600 kPa, and its exit is
saturated liquid. Find the mass flow rate of the ex-
traction flow. If the river water should not be
heated more than 5°C, how much water should be
pumped from the river to the heat exchanger
(condenser)?
11.45 Consider an ideal steam regenerative cycle in
which steam enters the turbine at 3 MPa and
400°C and exhausts to the condenser at 10 kPa.
Steam is extracted from the turbine at 0.8 MPa for
an open feedwater heater. The feedwater leaves
the heater as saturated liquid. The appropriate
pumps are used for the water leaving the con-
denser and the feedwater heater. Calculate the
thermal efficiency of the cycle and the net work
per kilogram of steam.
11.46 In one type of nuclear power plant, heat is trans-
ferred in the nuclear reactor to liquid sodium. The
liquid sodium is then pumped through a heat ex-
changer where heat is transferred to boiling water.
Saturated vapor steam at 5 MPa exits this heat ex-
changer and is then superheated to 600°C in an
external gas-fired superheater. The steam enters
the turbine, which has one (open-type) feedwater
extraction at 0.4 MPa. The condenser pressure is
7.5 kPa. Determine the heat transfer in the reactor
and in the superheater to produce a net power out-
put of 1 MW.
11.47 A steam power plant has high and low pressures
of 20 MPa and 10 kPa, and one open feedwater
heater operating at 1 MPa with the exit as satu-
rated liquid. The maximum temperature is 800°C,
and the turbine has a total power output of 5 MW.
Find the fraction of the flow for extraction to the
feedwater and the total condenser heat transfer
rate.
Closed Feedwater Heaters
11.48 A closed feedwater heater in a regenerative steam
power cycle, as shown in Fig. 11.11, heats 20 kg/s
of water from 100°C and 20 MPa to 250°C and 20
MPa. The extraction steam from the turbine en-
ters the heater at 4 MPa and 275°C and leaves as
saturated liquid. What is the required mass flow
rate of the extraction steam?
11.49 A power plant with one closed feedwater heater
has a condenser temperature of 45°C, a maximum
pressure of 5 MPa, and boiler exit temperature of
• 900 CI C. Extraction steam at 1 MPa to the feedwater
heater condenses and is pumped up to the 5 MPa
feedwater line where all the water goes to the
boiler at 200°C. Find the fraction of extraction
steam flow and the two specific pump work inputs.
11.50 Repeat Problem 11.45, but assume a closed in-
stead of an open feedwater heater. A single pump
is used to pump the water leaving the condenser
up to the boiler pressure of 3 MPa. Condensate
from the feedwater heater is drained through a
trap to the condenser.
11.51 Do Problem 11.47 with a closed feedwater heater
instead of an open heater and a drip pump to add
the extraction flow to the feedwater line at 20 MPa.
Assume the temperature is 175°C after the drip
pump flow is added to the line. One main pump
brings the water to 20 MPa from the condenser.
11.52 Assume the power plant in Problem 1 1 .43 has one
closed feedwater heater (FWH) instead of the
open FWH. The extraction flow out of the FWH
is saturated liquid at 2033 kPa being dumped into
the condenser and the feedwater is heated to
50°C. Find the extraction flow rate and the total
turbine power output.
Nonideal Cycles
11.53 Steam enters the turbine of a power plant at 5
MPa and 400°C and exhausts to the condenser at
10 kPa. The turbine produces a power output of
20 000 kW with an isentropic efficiency of 85%.
What is the mass flow rate of steam around the
cycle and the rate of heat rejection in the con-
denser? Find the thermal efficiency of the power
plant. How does this compare with a Carnot
cycle?
11.54 A steam power plant has a high pressure of 5 MPa
and maintains 50°C in the condenser. The boiler
exit temperature is 600°C. All the components are
ideal except the turbine, which has an actual exit
state of saturated vapor at 50°C. Find the cycle ef-
ficiency with the actual turbine and the turbine
isentropic efficiency.
11.55 A steam power cycle has a high pressure of 3
MPa and a condenser exit temperature of 45°C.
The turbine efficiency is 85%, and other cycle
components are ideal. If the boiler superheats to
800°C, find the cycle thermal efficiency.
454 ffl CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
11.56 A steam power plant operates with a high pres-
sure of 5 MPa and has a boiler exit temperature of
600°C receiving heat from a 700°C source. The
ambient air at 20°C provides cooling for the con-
denser so it can maintain 45°C inside. All the ;
components are ideal except for the turbine,
which has an exit state with a quality of 97%.
Find the work and heat transfer in all components
per kg of water and the turbine isentropic effi-
ciency. Find the rate of entropy generation per kg
of water in the boiler/heat source setup.
11.57 For the steam power plant described in Problem
11.21, assume the isentropic efficiencies of the
turbine and pump are 85% and 80%, respectively.
Find the component specific work and heat trans-
fers and the cycle efficiency.
11.58 A small steam power plant has a boiler exit of
3 MPa and 400° C, while it maintains 50 kPa in
the condenser. All the components are ideal ex-
cept the turbine, which has an isentropic effi-
ciency of 80%, and it should deliver a shaft power
of 9.0 MW to an electric generator. Find the spe-
cific turbine work, the needed flow rate of steam,
and the cycle efficiency.
11.59 Repeat Problem 11.47 assuming the turbine has
an isentropic efficiency of 85%.
1 1.60 Steam leaves a power plant steam generator at 3 .5
MPa, 400°C, and enters the turbine at 3.4 MPa,
375°C. The isentropic turbine efficiency is 88%,
and the turbine exhaust pressure is 10 kPa, Con-
densate leaves the condenser and enters the pump
at 35°C, 10 kPa. The isentropic pump efficiency is
80%, and the discharge pressure is 3.7 MPa. The
feedwater enters the steam generator at 3.6 MPa,
30°C. Calculate the thermal efficiency of the
cycle and the entropy generation for the process
in the line between the steam generator exit and
the turbine inlet, assuming an ambient tempera-
ture of 25°C.
11.61 In a particular reheat-cycle power plant, steam en-
ters the high-pressure turbine at 5 MPa, 450°C,
and expands to 0.5 MPa, after which it is reheated
to 450°C. The steam is then expanded through the
low-pressure turbine to 7.5 kPa. Liquid water
leaves the condenser at 30°C, is pumped to
5 MPa, and then returned to the steam generator.
Each turbine is adiabatic with an isentropic effi-
ciency of 87% and a pump efficiency of 82%. If
the total power output of the turbines is 10 MW,
determine the mass flow rate of steam, the pump
power input, and the thermal efficiency of the
power plant.
i 1.62 A supercritical steam power plant has a high pres-
sure of 30 MPa and an exit condenser temperature
of 50°C. The maximum temperature in the boiler
is 1000°C, and the turbine exhaust is saturated
vapor. There is one open feedwater heater receiv-
ing extraction from the turbine at 1 MPa, and its
exit is saturated liquid flowing to pump 2. The
isentropic efficiency for the first section and the
overall turbine are both 88.5%. Find the ratio of
the extraction mass flow to total flow into turbine.
What is the boiler inlet temperature with and
without the feedwater heater?
Cogeneration
11.63 A cogenerating steam power plant, as in Fig.
11.17, operates with a boiler output of 25 kg/s
steam at 7 MPa and 500°C. The condenser oper-
ates at 7.5 kPa, and the process heat is extracted at
5 kg/s from the turbine at 500 kPa, state 6, and
after use is returned as saturated liquid at 100
kPa, state 8. Assume ail components are ideal and
find the temperature after pump 1, the total tur-
bine output, and the total process heat transfer.
11.64 A 10-kg/s steady supply of saturated-vapor steam
at 500 kPa is required for drying a wood pulp
slurry in a paper mill (see Fig. PI 1.64). It is de-
cided to supply this steam by cogeneration; that
is, the steam supply will be the exhaust from a
steam turbine. Water at 20°C and 100 kPa is
pumped to a pressure of 5 MPa and then fed to a
steam generator with an exit at 400°C. What is the
additional heat-transfer rate to the steam genera-
tor beyond what would have been required to pro-
duce only the desired steam supply? What is the
difference in net power?
Water
FIGURE PI 1.64
Homework Problems H 455
11.65 In a cogenerating steam power plant the turbine
receives steam from a high-pressure steam drum
and a low-pressure steam drum, as shown in Fig.
Pi 1.65. The condenser is made as two closed
heat exchangers used to heat water running in
a separate loop for district heating. The high-
temperature heater adds 30 MW, and the low-
temperature heater adds 31 MW to the district
heating water flow. Find the power cogenerated
by the turbine and the temperature in the return
line to the deaerator.
5 kg/s
95°C
To district
heating
50 kPa, 14 kg/s
30 MW
31 MW
60°C
415 kg/s
- - 27 kg/s to
deaerator
FIGURE P11.65
11.66 A boiler delivers steam at 10 MPa, 550°C to a two-
stage turbine as shown in Fig. 11.17. After the first
stage, 25% of the steam is extracted at 1.4 MPa for
a process application and returned at 1 MPa, 90°C,
to the feedwater line. The remainder of the steam
continues through the low-pressure turbine stage,
which exhausts to the condenser at 10 kPa. One
pump brings the feedwater to 1 MPa, and a second
pump brings it to 10 MPa. Assume the first and
second stages in the steam turbine have isentropic
efficiencies of 85% and 80%, respectively, and that
both pumps are ideal. If the process application re-
quires 5 MW of power, how much power can then
be cogenerated by the turbine?
11.67 A smaller power plant produces 25 kg/s steam at
3 MPa, 600°C, in the boiler. It cools the con-
denser to an exit of 45°C and the cycle is shown
in Fig. PI 1.67. There is an extraction done at 500
kPa to an open feedwater hater, and in addition a
steam supply of 5 kg/s is taken out and not re-
turned. The missing 5 kg/s water is added to the
feedwater heater from a 20°C, 500 kPa source.
Find the needed extraction flow rate to cover both
the feedwater heater and the steam supply. Find
the total turbine power output.
FIGURE PI 1.67
Brayton Cycles, Gas Turbines
11.68 Consider an ideal air-standard Brayton cycle in
which the air into the compressor is at 100 kPa
and 20°C, and the pressure ratio across the com-
pressor is 12:1. The maximum temperature in the
cycle is 1100°C, and the airflow rate is 10 kg/s.
Assume constant specific heat for the air (from
Table A.5). Determine the compressor work, the
turbine work, and the thermal efficiency of the
cycle.
11.69 Repeat Problem 11.68, but assume variable spe-
cific heat for the air (Table A.7).
11.70 A Brayton-cycle inlet is at 300 K and 100 kPa,
and the combustion adds 670 kJ/kg. The maxi-
mum temperature is 1200 K due to material con-
siderations. What is the maximum allowed
compression ratio? For this ratio, calculate the net
work and cycle efficiency assuming variable spe-
cific heat for the air (Table A.7).
11.71 A large stationary Brayton-cycle gas turbine
power plant delivers a power output of 100 MW
to an electric generator. The minimum tempera-
456 H Chapter eleven power and refrigeration Systems
ture in the cycle is 300 K, and the maximum tem-
perature is 1600 K. The minimum pressure in the
cycle is 100 kPa, and the compressor pressure
ratio is 14 to 1. Calculate the power output of the
turbine. What fraction of the turbine output is re-
quired to drive the compressor? What is the ther-
mal efficiency of the cycle?
11.72 A Brayton cycle produces 14 MW with an inlet
state of 17°C, 100 kPa, and a compression ratio
of 16:1. The heat added in the combustion is
960 kJ/kg. What is the highest temperature and
the mass flow rate of air, assuming cold air
properties?
11.73 Do the previous problem with properties from
Table A.7.1 instead of cold air properties.
Regenerators, Intercoolers, and Nonideal Cycles
11.74 An ideal regenerator is incorporated into the ideal
air-standard Brayton cycle of Problem 11.68.
Find the thermal efficiency of the cycle with this
modification.
11.75 The gas-turbine cycle shown in Fig. P11.75 is
used as an automotive engine. In the first rur- _
bine, the gas expands to pressure P 5 , just low
enough for this turbine to drive the compressor.
The gas is then expanded through the second
turbine connected to the drive wheels. The data
for the engine are shown in the figure and as-
sume that all processes are ideal. Determine the
intermediate pressure P 5 , the net specific work
output of the engine, and the mass flow rate
through the engine. Find also the air tempera-
ture entering the burner T 3 , and the thermal effi-
ciency of the engine.
11.76 Repeat Problem 11.71, but include a regenerator
with 75% efficiency in the cycle.
11.77 A two-stage air compressor has an intercooier be-
tween the two stages, as shown in Fig. PI 1.77.
The inlet state is 100 kPa, 290 K, and the final exit
pressure is 1.6 MPa. Assume that the constant-
pressure intercooier cools the air to the inlet tem-
perature, r 3 = T v It can be shown that the optimal
pressure is P 2 = (PiP^ a > for minimum total com-
pressor work. Find the specific compressor works
and the intercooier heat transfer for the optimal P 2 .
Compressor 1
FIGURE P11.77
11,78 A two-stage compressor in a gas turbine brings
atmospheric air at 100 kPa and 17°C to 500 kPa,
and then cools it in an intercooier to 27°C at con-
stant P. The second stage brings the air to 1000
kPa. Assume that both stages are adiabatic and re-
versible. Find the combined specific work to the
compressor stages. Compare that to the specific
work for the case of no intercooier (i.e., one com-
pressor from 100 to 1000 kPa).
FIGURE P11.75
Homework problems EI 457
11.79 A gas turbine with air as the working fluid has
two ideal turbine sections, as shown in Fig.
Pi 1.79, the first of which drives the ideal com-
pressor, with the second producing the power out-
put. The compressor input is at 290 K, 100 kPa,
and the exit is at 450 kPa. A fraction of flow, x,
bypasses the burner, and the rest (1 ~ x) goes
through the burner where 1200 kJ/kg is added by
combustion. The two flows then mix before enter-
ing the first turbine and continue through the sec-
ond turbine, with exhaust at 100 kPa. If the
mixing should result in a temperature of 1000 KL
into the first turbine, find the fraction x. Find the
required pressure and temperature into the second
turbine and its specific power output.
FIGURE PI 1.79
11.80 Repeat Problem 11.71, but assume that the com-
pressor has an isentropic efficiency of 85% and
the turbine an isentropic efficiency of 88%.
11.81 Repeat Problem 11.77 when the intercooler
brings the air to T 3 = 320 K. The corrected for-
mula for the optimal pressure is P 2 -
[P l P 4 (T 3 /T i y'^} m . See Problem 9.184, where n
is the exponent in the assumed polytropic process.
11.82 Consider an ideal gas-turbine cycle with a pres-
sure ratio across the compressor of 12 to 1. The
compressor inlet is at 300 K and 100 kPa, and the
cycle has a maximum temperature of 1600 K. An
ideal regenerator is also incorporated in the cycle.
Find the thermal efficiency of the cycle using cold
air (298 K) properties. If the compression ration is
raised, T 4 — T 2 goes down. At what compression
ratio is T 2 = T 4 so the regenerator cannot be used?
11.83 A gas turbine has two stages of compression,
with an intercooler between the stages (see Fig.
PI 1.77). Air enters the first stage at 100 kPa and
300 K. The pressure ratio across each compres-
• sor stage is 5 to 1, and each stage has an isen-
tropic efficiency of 82%. Air exits the
intercooler at 330 K. Calculate the exit tempera-
ture from each compressor stage and the total
specific work required.
11.84 Repeat the questions in Problem II. 75 when we
assume that friction causes pressure drops in the
burner and on both sides of the regenerator. In
each case, the pressure drop is estimated to be 2%
of the inlet pressure to that component of the sys-
tem, so Pj, = 588 kPa, P 4 = 0.98 P 3 , and P 6 -
102 kPa.
Ericsson Cycles
11.85 Consider an ideal air-standard Ericsson cycle that
has an ideal regenerator, as shown in Fig. PI 1.85,
The high pressure is 1 MPa, and the cycle effi-
ciency is 70%. Heat is rejected in the cycle at a
temperature of 300 K, and the cycle pressure at
the beginning of the isothermal compression
process is 100 kPa. Determine the high tempera-
ture, the compressor work, and the turbine work
per kilogram of air.
Regenerator
=>W H
Ql
FIGURE P11.8S
11.86 An air-standard Ericsson cycle has an ideal regen-
erator. Heat is supplied at 1000°C, and heat is re-
jected at 20°C. Pressure at the beginning of the
isothermal compression process is 70 kPa. The
heat added is 600 kJ/kg. Find the compressor
work, the turbine work, and the cycle efficiency.
Jet Engine Cycles
11.87 Consider an ideal air-standard cycle for a gas-
turbine, jet propulsion unit, such as that shown in
458 @ Chapter eleven power and refrigeration systems
Fig. 1 1.26. The pressure and temperature entering
the compressor are 90 kPa and 290 K. The pres-
sure ratio across the compressor is 14 to 1, and
the turbine inlet temperature is 1500 K. When the
air leaves the turbine, it enters the nozzle and ex-
pands to 90 kPa. Determine the pressure at the
nozzle inlet and the velocity of the air leaving the
nozzle.
11.88 The turbine section in a jet engine receives gas
(assumed to be air) at 1200 K and 800 kPa with
an ambient atmosphere at 80 kPa. The turbine is
followed by a nozzle open to the atmosphere,
and all the turbine work drives a compressor re-
ceiving air at 85 kPa and 270 K with the same
flow rate. Find the turbine exit pressure so that
the nozzle has an exit velocity of 800 m/s. To
what pressure can the compressor bring the in-
coming air?
kPa and the ambient temperature is - 1 8°C. The
velocity of the aircraft is 280 m/s, the pressure
ratio across the compressor is 14:1, and the
cycle maximum temperature is 1450 K. Assume
that the inlet flow goes through a diffuser to
zero relative velocity at state a, Fig. 11.26. Find
the temperature and pressure at state a and the
velocity (relative to the aircraft) of the air leav-
ing the engine at 55 kPa.
11.92 An afterburner in a jet engine adds fuel after the
turbine, thus raising the pressure and temperature
via the energy of combustion. Assume a standard
condition of 800 K and 250 kPa after the turbine
into the nozzle that exhausts at 95 kPa. Assume
the afterburner adds 450 kJ/kg to that state with a
rise in pressure for the same specific volume, and
neglect any upstream effects on the turbine. Find
the nozzle exit velocity before and after the after-
burner is turned on.
Fuel in
Turbojet engine
FIGURE P11.88
11.89 The turbine in a jet engine receives air at 1250 K
and 1.5 MPa. It exhausts to a nozzle at 250 kPa,
which in turn exhausts to the atmosphere at 100
kPa. The isentropic efficiency of the turbine is
85%, and the nozzle efficiency is 95%. Find the
nozzle inlet temperature and the nozzle exit ve-
locity. Assume negligible kinetic energy out of
the turbine.
11.90 Consider an air-standard jet engine cycle operat-
ing in a 280-K, 100-kPa environment. The com-
pressor requires a shaft power input of 4000 kW.
Air enters the turbine state 3 at 1600 K and 2
MPa, at the rate of 9 kg/s, and the isentropic effi-
ciency of the turbine is 85%. Determine the pres-
sure and temperature entering the nozzle.
11.91 A jet aircraft is flying at an altitude of 4900 m,
where the ambient pressure is approximately 55
Combuslors Fuel-spray bars
Diffuser Gas generator Afterburner Adjustable
duct nozzle
FIGURE P11.92
Otto Cycles
11.93 Air flows into a gasoline engine at 95 kPa and
300 K. The air is then compressed with a volu-
metric compression ratio of 8 : 1 . The combus-
tion process releases 1300 kJ/kg of energy as the
fuel burns. Find the temperature and pressure
after combustion using cold air properties.
11.94 A gasoline engine has a volumetric compression
ratio of 9. The state before compression is
290 K, 90 kPa, and the peak cycle temperature is
1800 K. Find the pressure after expansion, the
cycle net work, and the cycle efficiency using
properties from Table A.5.
11.95 To approximate an actual spark-ignition engine,
consider an air-standard Otto cycle that has a
Homework Problems H 459
heat addition of 1800 kj/kg of air, a compression
ratio of 7, and a pressure and temperature at the
beginning of the compression process of 90 kPa
and 10°C. Assuming constant specific heat, with
the value from Table A.5, determine the maxi-
mum pressure and temperature of the cycle, the
thermal efficiency of the cycle, and the mean ef-
fective pressure.
FIGURE PI 1,95
11.96 A gasoline engine has a volumetric compression
ratio of 8 and before compression has air at
280 K and 85 kPa. The combustion generates a
peak pressure of 6500 kPa. Find the peak tem-
perature, the energy added by the combustion
process, and the exhaust temperature.
11.97 A gasoline engine has a volumetric compression
ratio of 10 and before compression has air at
290 K, 85 kPa, in the cylinder. The combustion
peak pressure is 6000 kPa. Assume cold air
properties. What is the highest temperature in
the cycle? Find the temperature at the beginning
of the exhaust (heat rejection) and the overall
cycle efficiency.
11.98 A four-stroke gasoline engine has a compression
ratio of 10:1 with 4 cylinders of total displace-
ment 2.3 L. The inlet state is 280 K, 70 kPa, and
the engine is running at 2100 RPM with the fuel
adding 1800 kj/kg in the combustion process.
What is the net work in the cycle, and how much
power is produced?
FIGURE P11.98
11,99 A gasoline engine takes air in at 290 K and 90
kPa and then compresses it. The combustion
adds 1000 kJ/kg to the air, after which the tem-
perature is 2050 K. Use the cold atr properties
(i.e., constant heat capacities at 300 K) and find
the compression ratio, the compression specific
work, and the highest pressure in the cycle.
11.100 Answer the same three questions for the previ-
ous problem, but use variable heat capacities
(use Table A.7).
11.101 When methanol produced from coal is consid-
ered as an alternative fuel to gasoline for auto-
motive engines, it is recognized that the engine
can be designed with a higher compression ratio,
say 10 instead of 7, but that the energy release
with combustion for a stoichiometric mixture
with air is slightly smaller, about 1700 kJ/kg.
Repeat Problem 1 1.95 using these values.
11.102 A gasoline engine receives air at 10°C, 100 kPa,
having a compression ratio of 9 : 1 by volume.
The heat addition by combustion gives the high-
est temperature as 2500 K. Use cold air proper-
ties to find the highest cycle pressure, the
specific energy added by combustion, and the
mean effective pressure.
11.103 Repeat Problem 11.95, but assume variable spe-
cific heat. The ideal-gas air tables, Table A.7, are
recommended for this calculation (and the spe-
cific heat from Fig. 5.10 at high temperature).
11.104 It is found experimentally that the power stroke
expansion in an internal combustion engine can
be approximated, with a polytropic process with
a value of the polytropic exponent n somewhat
larger than the specific heat ratio k. Repeat Prob-
lem 11.95, but assume that the expansion
process is reversible and polytropic (instead of
460 ffl CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
the isentropic expansion in the Otto cycle) with
n equal to 1.50.
11.105 In the Otto cycle, all the heat transfer q H occurs
at constant volume. It is more realistic to assume
that part of q n occurs after the piston has started
its downward motion in the expansion stroke.
Therefore, consider a cycle identical to the Otto
cycle, except that the first two-thirds of the total
q H occurs at constant volume and the last one-
third occurs at constant pressure. Assume that
the total q H h 2100 kJ/kg, that the state at the be-
ginning of the compression process is 90 kPa,
20"C } and that the compression ratio is 9. Calcu-
late the maximum pressure and temperature and
the thermal efficiency of this cycle. Compare the
results with those of a conventional Otto cycle
having the same given variables-
Diesel Cycles
11.106 A diesel engine has a state before compression
of 95 kPa, 290 K, a peak pressure of 6000 kPa,
and a maximum temperature of 2400 K. Find the
volumetric compression ratio and the thermal ef-
ficiency.
11.107 A diesel engine has a bore of 0.1 m, a stroke of
0. 1 1 m, and a compression ratio of 19 : 1 running
at 2000 RPM (revolutions per minute). Each
cycle takes two revolutions and has a mean ef-
fective pressure of 1400 kPa. With a total of 6
cylinders, find the engine power in kW and
horsepower, hp.
FIGURE P11.107
11.108 A diesel engine has a compression ratio of 20 : 1
with an inlet of 95 kPa and 290 K, state 1, with
volume 0.5 L. The maximum cycle temperature
is 1800 K. Find the maximum pressure, the net
specific work, and the thermal efficiency.
11.109 At the beginning of compression in a diesel cycle
T = 300 K. and P = 200 kPa; after combustion
(heat addition) is complete T = 1500 K and P ~
7.0 MPa. Find the compression ratio, the thennal
efficiency, and the mean effective pressure.
11.110 Do Problem 11.106 but use the properties from
Table A.7 and not the cold air properties.
11.111 A diesel engine has air before compression at 280
K and 85 kPa. The highest temperature is 2200 K,
and the highest pressure is 6 MPa. Find the volu-
metric compression ratio and the mean effective
pressure using cold air properties at 300 K.
11.112 Consider an ideal air-standard diesel cycle in
which the state before the compression process is
95 kPa, 290 K, and the compression ratio is 20.
Find the maximum temperature (by iteration) in
the cycle to have a thermal efficiency of 60%.
Stirling and Carnot Cycles
11.113 Consider an ideal Stirling-cycle engine in which
the state at the beginning of the isothermal com-
pression process is 100 kPa, 25°C, the compres-
sion ratio is 6, and the maximum temperature in
the cycle is 1100°C. Calculate the maximum
cycle pressure and the thermal efficiency of the
cycle with and without regenerators.
11.114 An air-standard Stirling cycle uses helium as the
working fluid. The isothermal compression
brings helium from 100 kPa, 37°C to 600 kPa.
The expansion takes place at 1200 K, and there
is no regenerator. Find the work and heat trans-
fer in all of the four processes per kg helium and
the thermal cycle efficiency.
11.115 Consider an ideal air-standard Stirling cycle
with an ideal regenerator. The minimum pres-
sure and temperature in the cycle are 100 kPa,
25°C, the compression ratio is 1 0, and the maxi-
mum temperature in the cycle is 1000°C. Ana-
lyze each of the four processes in this cycle for
work and heat transfer, and determine the over-
all performance of the engine.
11.116 The air-standard Camot cycle was not shown in
the text; show the T-s diagram for this cycle. In
Homework Problems H 461
an air-standard Carnot cycle, the low tempera-
ture is 280 K and the efficiency is 60%. If the
pressure before compression and after heat re-
jection is 100 kPa, find the high temperature and
the pressure just before heat addition.
11.117 Air in a piston/cylinder setup goes through a
Carnot cycle in which T L = 26.8°C and the total
cycle efficiency is 17 = 2/3. Find T H , the specific
work, and the volume ratio in the adiabatic ex-
pansion for constant C p , C„.
11.118 Do the previous Problem, 11.117, using Table
A.7.I.
Refrigeration Cycles
11.119 A refrigerator with R-12 as the working fluid has a
minimum temperature of - 10°C and a maximum
pressure of 1 MPa. Assume an ideal refrigeration
cycle as in Fig. 11. 33. Find the specific heat trans-
fer from the cold space and that to the hot space,
and determine the coefficient of performance.
11.120 Consider an ideal refrigeration cycle that has a
condenser temperature of 45°C and an evapora-
tor temperature of - 1 5°C. Determine the coeffi-
cient of performance of this refrigerator for the
working fluids R-12 and R-22.
11.121 The environmentally safe refrigerant R-134a is
one of the replacements for R-12 in refrigeration
systems. Repeat Problem 11.120 using R-134a
and compare the result with that for R-12.
11.122 A refrigerator using R-22 is powered by a small
natural gas-fired heat engine with a thermal effi-
ciency of 25%, as shown in Fig. PI 1.122. The
R-22 condenses at 40°C, it evaporates at -20°C,
and the cycle is standard. Find the two specific
heat transfers in the refrigeration cycle. What is
the overall coefficient of performance as QrfQ x 1
L
I
\
Ql
/ \ / Cold space \ FIGURE P11.122
11.123 A refrigerator in a meat warehouse must keep a
low temperature of -15°C. It uses R-12 as the
refrigerant, which must remove 5 kW from the
cold space. Assume that the outside temperature
is 20°C. Find the flow rate of the R-12 needed,
assuming a standard vapor compression refriger-
ation cycle with a condenser at 20°C.
11.124 A refrigerator with R-12 as the working fluid has
a minimum temperature of -10°C and a maxi-
mum pressure of 1 MPa. The actual adiabatic
compressor exit temperature is 60°C. Assume no
pressure loss in the heat exchangers. Find the
specific heat transfer from the cold space and that
to the hot space, the coefficient of performance,
and the isentropic efficiency of the compressor.
11.125 Consider an ideal heat pump that has a con-
denser temperature of 50°C and an evaporator
temperature of 0°C. Determine the coefficient of
performance of this heat pump for the working
fluids R-12, R-22, and ammonia.
11.126 The air conditioner in a car uses R-134a, and the
compressor power input is 1.5 kW, bringing the
R-134a from 201.7 kPa to 1200 kPa by com-
pression. The cold space is a heat exchanger that
cools 30°C atmospheric air from the outside
down to 10°C and blows it into the car. What is
the mass flow rate of the R-134a, and what is the
low-temperature heat-transfer rate? What is the
mass flow rate of air at 10°C?
11.127 A refrigerator using R-134a is located in a 20°C
room. Consider the cycle to be ideal, except that
the compressor is neither adiabatic nor re-
versible. Saturated vapor at -20°C enters the
compressor, and the R-134a exits the compres-
sor at 50°C. The condenser temperature is 40°C.
The mass flow rate of refrigerant around the
cycle is 0.2 kg/s, and the coefficient of perfor-
mance is measured and found to be 2.3. Find the
power input to the compressor and the rate of
entropy generation in the compressor process.
11.128 A refrigerator has a steady flow of R-22 as satu-
rated vapor at -20°C into the adiabatic com-
pressor that brings it to 1000 kPa. After the
compressor, the temperature is measured to be
60°C. Find the actual compressor work and the
actual cycle coefficient of performance.
11.129 A small heat pump unit is used to heat water for
a hot-water supply. Assume that the unit uses
462 U CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
R-22 and operates on the ideal refrigeration cycle.
The evaporator temperature is 15 C, and the con-
denser temperature is 60°C. If the amonnt of hot
water needed is 0.1 kg/s, determine the amount of
energy saved by using the heat pump instead of
directly heating the water from 15 to 60°C.
11.130 The refrigerant R-22 is used as the working fluid
in a conventional heat pump cycle. Saturated
vapor enters the compressor of this unit at 10°C;
its exit temperature from the compressor is mea-
sured and found to be 85°C. If the compressor
exit is 2 MPa, what is the compressor isentropic
efficiency and the cycle COP?
11.131 A refrigerator in a laboratory uses R-22 as the
working substance. The high pressure is 1200
kPa, the low pressure is 201 kPa, and the com-
pressor is reversible. It should remove 500 W
from a specimen currently at -20°C (not equal
to T L in the cycle) that is inside the refrigerated
space. Find the cycle COP and the electrical
power required.
11.132 Consider the previous problem and find the two
rates of entropy generation in the process and
where they occur.
11.133 In an actual refrigeration cycle using R-12 as the
working fluid, the refrigerant flow rate is 0.05
kg/s. Vapor enters the compressor at 150 kPa
and - 10°C and leaves at 1.2 MPa and 75°C. The
power input to the nonadiabatic compressor is
measured and found to be 2.4 kW. The refriger-
ant enters the expansion valve at 1.15 MPa and
40°C and leaves the evaporator at 175 kPa and
- 15°C. Determine the entropy generation in the
compression process, the refrigeration capacity,
and the coefficient of performance for this cycle.
Ammonia Absorption Cycles
11.134 Consider a small ammonia absorption refrigera-
tion cycle that is powered by solar energy and is
to be used as an air conditioner. Saturated vapor
ammonia leaves the generator at 50°C, and satu-
rated vapor leaves the evaporator at 10°C. If 7000
kJ of heat is required in the generator (solar col-
lector) per kilogram of ammonia vapor generated,
determine the overall performance of this system.
11.135 The performance of an ammonia absorption
cycle refrigerator is to be compared with that of
a similar vapor-compression system. Consider
an absorption system having an evaporator tem-
perature of - 10°C and a condenser temperature
of 50°C. The generator temperature in this sys-
tem is 150°C. In this cycle 0.42 kJ is transferred
to the ammonia in the evaporator for each kilo-
joule transferred from the high-temperature
source to the ammonia solution in the generator.
To make the comparison, assume that a reser-
voir is available at 150°C and that heat is trans-
ferred from this reservoir to a reversible engine
that rejects heat to the surroundings at 25°C.
This work is then used to drive an ideal vapor-
compression system with ammonia as the refrig-
erant. Compare the amount of refrigeration that
can be achieved per kilojoule from the high tem-
perature source with the 0.42 kJ that can be
achieved in the absorption system.
Air-Standard Refrigeration Cycles
11.136 The formula for the coefficient of performance
when we use cold air properties is not given in
the text. Derive the expression for COP as func-
tion of the compression ratio similar to how the
Brayton-cycle efficiency was found.
11.137 A heat exchanger is incorporated into an ideal
air-standard refrigeration cycle, as shown in Fig.
PI 1.137. It may be assumed that both the com-
pression and the expansion are reversible adia-
batic processes in this ideal case. Determine the
coefficient of performance for the cycle.
FIGURE P11.137
11.138 Repeat Problem 11.137, but assume that helium
is the cycle working fluid instead of air. Discuss
the significance of the results.
HOMEWORK PROBLEMS K 463
0,139 Repeat Problem 11.137, but assume an isen-
tropic efficiency of 75% for botb the compressor
and the expander.
Combined Cycles
11.140 A binary system power plant uses mercury for
the high-temperature cycle and water for the
low-temperature cycle, as shown in Fig. 11.39.
The temperatures and pressures are shown in the
corresponding T-s diagram. The maximum tem-
perature in the steam cycle is where the steam
leaves the superheater at point 4 where it is
500°C. Determine the ratio of the mass flow rate
of mercury to the mass flow rate of water in the
beat exchanger that condenses mercury and
boils the water and the thermal efficiency of this
ideal cycle.
The following saturation properties for mer-
cury are known:
T
Sp kJ/
s g , kJ/
MPa
°C
kJ/kg
kJ/kg
kg-K
kg-K
0.04
309
42.21
335.64
0.1034
0.6073
1.60
562
75.37
364.04
0.1498
0.4954
11.141 A Rankine steam power plant should operate
with a high pressure of 3 MPa, a low pressure of
10 kPa, and a boiler exit temperature of 500°C.
The available high-temperature source is the ex-
haust of 175 kg/s air at 600°C from a gas tur-
bine. If the boiler operates as a counterflowing
heat exchanger where the temperature difference
at the pinch point is 20°C, find the maximum
water mass flow rate possible and the air exit
temperature.
11.142 A simple Rankine cycle with R-22 as the
working fluid is to be used as a bottoming
cycle for an electrical-generating facility dri-
ven by the exhaust gas from a diesel engine as
the high-temperature energy source in the
R-22 boiler. Diesel inlet conditions are 100
kPa, 20°C, the compression ratio is 20, and the
maximum temperature in the cycle is 2800°C.
Saturated vapor R-22 leaves the bottoming
cycle boiler at 110°C, and the condenser tem-
perature is 30°C. The power output of the
diesel engine is 1 MW. Assuming ideal cycles
■ throughout, determine
a. The flow rate required in the diesel engine.
b. The power output of the bottoming cycle, as-
suming that the diesel exhaust is cooled to
200°C in the R-22 boiler.
1 1 . 143 A cascade system is composed of two ideal refrig-
eration cycles, as shown in Fig. 11.41. The high-
temperature cycle uses R-22. Saturated liquid
leaves the condenser at 40°C ) and saturated vapor
leaves the heat exchanger at -20°C. The low-
temperature cycle uses a different refrigerant, R-23.
Saturated vapor leaves the evaporator at -80°C
with h = 330 kJ/kg, and saturated liquid leaves
the heat exchanger at - 10°C with h= 185 kJ/kg.
R-23 out of the compressor has h = 405 kJ/kg.
Calculate the ratio of the mass flow rates through
the two cycles an the COP of the total system.
11.144 Consider an ideal dual-loop heat-powered re-
frigeration cycle using R-12 as the working
fluid, as shown in Fig. PI 1.144. Saturated vapor
at 105°C leaves the boiler and expands in the
turbine to the condenser pressure. Saturated
vapor at -15°C leaves the evaporator and is
compressed to the condenser pressure. The ratio
of the flows through the two loops is such that
the turbine produces just enough power to drive
the compressor. The two exiting streams mix to-
gether and enter the condenser. Saturated liquid
Expansion
valve
FIGURE P11.144
464 M CHAPTER. ELEVEN POWER AND REFRIGERATION SYSTEMS
leaving the condenser at 45°C is then separated
into two streams in the necessary proportions.
Determine the ratio of mass flow rate through
the power loop to that through the refrigeration
loop. Find also the performance of the cycle, in
terms of the ratio QJQu-
11.145 For a cryogenic experiment, heat should be re-
moved from a space at 75 K to a reservoir at 180
K. A heat primp is designed to use nitrogen and
methane in a cascade arrangement (see Fig.
11.41), where the high temperature of the nitro-
gen condensation is at 10 K higher than the low-
temperature evaporation of the methane. The
two other phase changes take place at the listed
reservoir temperatures. Find the saturation tem-
peratures in the heat exchanger between the two
cycles that give the best coefficient of perfor-
mance for the overall system.
Availability or Exergy Concepts
11.146 Find the flows and fluxes of exergy in the con-
denser of Problem 11.32. Use those to determine
the second-law efficiency.
11.147 Find the availability of the water at all four
states in the Rankine cycle described in Problem
11.33. Assume that the high-temperature source
is 500°C and the low-temperature reservoir is at
25°C. Determine the flow of availability in or
out of the reservoirs per kilogram of steam flow-
ing in the cycle. What is the overall cycle sec-
ond-law efficiency?
11.148 Find the flows of exergy into and out of the
feedwater heater in Problem 1 1.43.
11.149 Find the availability of the water at all the states in
the steam power plant described in Problem 1 1.57.
Assume the heat source in the boiler is at 600°C
and the low-temperature reservoir is at 25°C. Give
the second-law efficiency of all the components.-
11.150 Consider the Brayton cycle in Problem 11.72.
Find all the flows and fluxes of exergy and find the
overall cycle second-law efficiency. Assume the
heat transfers are internally reversible processes,
and we then neglect any external irreversibility.
11.151 For Problem 11.141, determine the change of
availability of the water flow and that of the air-
flow. "Use these to determine a second-law effi-
ciency for the boiler heat exchanger.
Review Problems
11.152 A simple steam power plant is said to have the
four states as listed: (1) 20°C, 100 kPa, (2) 25°C,
1 MPa, (3) 1000°C, 1 MPa, (4) 250°C, 100 kPa,
with an energy source at H00°C, and it rejects
energy to a 0°C ambient. Is this cycle possible?
Are any of the devices impossible?
11.153 Do Problem 11.31 with R-134a as the working
fluid in the Rankine cycle.
11.154 An ideal steam power plant is designed to operate
on the combined reheat and regenerative cycle
and to produce a net power output of 10 MW.
Steam enters the high-pressure turbine at 8 MPa,
550°C, and is expanded to 0.6 MPa, at which
pressure some of the steam is fed to an open feed-
water heater, and the remainder is reheated to
550°C. The reheated steam is then expanded in
the low-pressure turbine to 10 kPa. Determine the
steam flow rte to the high-pressure turbine and
the power required to drive each pump.
11.155 Steam enters the turbine of a power plant at 5 MPa
and 400°C and exhausts to the condenser at
10 kPa. The turbine produces a power output of
20 000 kW with an isentropic efficiency of 85%.
What is the mass flow rate of steam around the
cycle and the rate of heat rejection in the con-
denser? Find the thermal efficiency of the power
plant. How does this compare with a Carnot cycle?
11.156 Consider an ideal steam reheat cycle as shown in
Fig. 11.7, where steam enters the high-pressure
turbine at 4 MPa and 450°C with a mass flow
rate of 20 kg/s. After expansion to 400 kPa, it is
reheated to T 5 flowing through the low-pressure
turbine out to the condenser operating at 10 kPa.
Find r s so the turbine exit quality is at least 95%.
For this reheat temperature find also the thermal
efficiency of the cycle and the net power output.
11.157 In one type of nuclear power plant, heat is trans-
ferred in the nuclear reactor to liquid sodium.
The liquid sodium is then pumped through a
heat exchanger where heat is transferred to boil-
ing water. Saturated vapor steam at 5 MPa exits
this heat exchanger and is then superheated to
600°C in an external gas-fired superheater. The
steam enters the turbine, which has one (open-
type) feedwater extraction at 0.4 MPa. The isen-
tropic turbine efficiency is 87%, and the
condenser pressure is 7.5 kPa. Determine the
HoiiEwoRK Problems H 465
heat transfer in the reactor and in the superheater
to produce a net power output of 1 MW,
11.158 An industrial application has the following
steam requirement: one 10-kg/s stream at a pres-
sure of 0.5 MPa and one 5-kg/s stream at 1.4
MPa (both saturated or slightly superheated
vapor). These are obtained by cogeneration,
whereby a high-pressure boiler supplies steam at
10 MPa and 500°C to a turbine. The required
amount is withdrawn at 1.4 MPa, and the re-
mainder is expanded in the low-pressure end of
the turbine to 0.5 MPa, providing the second re-
quired steam flow. Assuming both turbine sec-
tions have an isentropic efficiency of 85%.
a. Determine the power output of the turbine
and the heat-transfer rate in the boiler.
b. Compute the rates needed if the steam were
generated is a low-pressure boiler without co-
generation. Assume that for each, 20°C liquid
water is pumped to the required pressure and
fed to a boiler.
11.159 Repeat Problem 11.75, but
assume that the com-
pressor has an efficiency of 82%, that both tur-
bines have efficiencies of 87%, and that the
regenerator efficiency is 70%.
11.160 Consider a gas-turbine cycle with two stages of
compression and two stages of expansion. The
pressure ratio across each compressor stage and
each turbine stage is 8 to 1. The pressure at the
entrance to the first compressor is 100 kPa, the
temperature entering each compressor is 20°C,
and the temperature entering each turbine is
1100°C. A regenerator is also incorporated into
the cycle and it has an efficiency of 70%. Deter-
mine the compressor work, the turbine work,
and the thermal efficiency of the cycle.
11.161 A gas-turbine cycle has two stages of compres-
sion, with an intercooler between the stages. Air
enters the first stage at 100 kPa, 300 K. The
pressure ratio across each compressor stage is 5
to 1 , and each stage has an isentropic efficiency
of 82%. Air exits the intercooler at 330 K. The
maximum cycle temperature is 1500 K, and the
cycle has a single-turbine stage with an isen-
tropic efficiency of 86%. The cycle also includes
a regenerator with an efficiency of 80%. Calcu-
late the temperature at the exit of each compres-
sor stage, the second-law efficiency of the
turbine, and the cycle thermal efficiency.
11.162 A gasoline engine has a volumetric compression
ratio of 9. The state before compression is 290
K, 90 kPa, and the peak cycle temperature is
1800 K. Find the pressure after expansion, the
cycle net work, and the cycle efficiency using
properties from Table A.7.
11.163 The effect of a number of open feedwater heaters
on the thermal efficiency of an ideal cycle is to
be studied. Steam leaves the steam generator at
20 MPa, 600°C, and the cycle has a condenser
pressure of 10 kPa. Determine the thermal, effi-
ciency for each of the following cases. A: No
feedwater heater. B: One feedwater heater oper-
ating at 1 MPa. C: Two feedwater heaters, one
operating at 3 MPa and the other at 0.2 MPa.
11.164 The power plant shown in Fig. 1 1.40 combines a
gas-turbine cycle and a steam-turbine cycle. The
following data are known for the gas-turbine
cycle. Air enters the compressor at 100 kPa,
Regenerator
Compressor 1
FIGURE P11.161
466 m CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
25°C, the compressor pressure ratio is 14, and
the isentropic compressor efficiency is 87%; the
heater input rate is 60 MW; the turbine inlet
temperature is 1250°C, the exhaust pressure is
100 kPa, and the isentropic turbine efficiency is
87%; the cycle exhaust temperature from the
heat exchanger is 200°C. The following data are
known for the steam-turbine cycle. The pump
inlet state is saturated liquid at 10 kPa, the pump
exit pressure is 12.5 MPa, and the isentropic
pump efficiency is 85%; turbine inlet tempera-
ture is 500°C, and the isentropic turbine effi-
ciency is 87%. Determine
a. The mass flow rate of air in the gas-turbine
cycle.
b. The mass flow rate of water in the steam
cycle.
c. The overall thermal efficiency of the com-
bined cycle.
11.165 One means of improving the performance of a
refrigeration system that operates over a wide
temperature range is to use a two-stage compres-
sor. Consider an ideal refrigeration system of
this type that uses R-12 as the working fluid, as
shown in Fig. P11.165. Saturated liquid leaves
the condenser at 40°C and is throttled to -20°C.
The liquid and vapor at this temperature are sep-
arated, and the liquid is throttled to the evapora-
tor temperature, -70°C. Vapor leaving the
evaporator is compressed to the saturation pres-
sure corresponding to -20°C, after which it is
mixed with the vapor leaving the flash chamber.
It may be assumed that both the flash chamber
and the mixing chamber are well insulated to
prevent heat transfer from the ambient. Vapor
leaving the mixing chamber is compressed in the
second stage of the compressor to the saturation
pressure corresponding to the condenser temper-
ature, 40°C. Determine the following:
a. The coefficient of performance of the system.
b. The coefficient of performance of a simple
ideal refrigeration cycle operating over the
same condenser and evaporator ranges as
those of the two-stage compressor unit stud-
ied in this problem.
11,166 A jet ejector, a device with no moving parts,
functions as the equivalent of a coupled turbine-
compressor unit (see Problems 9.82 and 9.90).
Thus, the turbine-compressor in the dual-loop
-W.
Secondary
FIGURE P11.165
FIGURE PI 1.166
English Unit Problems B 467
cycle of Fig. PI 1.144 could be replaced by a jet
ejector. The primary stream of the jet ejector en-
ters from the boiler, the secondary stream enters
from the evaporator, and the discharge flows to
the condenser. Alternatively, a jet ejector may be
used with water as the working fluid. The pur-
pose of the device is to chill water, usually for an
air-conditioning system. In this application the
physical setup is as shown in Fig. PI 1.166.
Using the data given on the diagram, evaluate the
performance of this cycle in terms of the ratio
a. Assume an ideal cycle.
b. Assume an ejector efficiency of 20% (see
Problem 9.128).
English Unit Problems
Rankine Cycles
11.167E A steam power plant, as shown in Fig. 11.3,
operating in a Rankine cycle has saturated
vapor at 600 lbf/in. 2 leaving the boiler. The tur-
bine exhausts to the condenser operating at
2.225 lbf/in. 2 . Find the specific work and heat
transfer in each of the ideal components and
the cycle efficiency.
11.168E Consider a solar-energy-powered ideal Rank-
ine cycle that uses water as the working fluid.
Saturated vapor leaves the solar collector at
350 F, and the condenser pressure is 0.95
lbf/in. 2 . Determine the thermal efficiency of
this cycle.
11.169E A Rankine cycle uses ammonia as the working
substance and powered by solar energy. It
heats the ammonia to 320 F at 800 psia in the
boiler/superheater. The condenser is water
cooled, and the exit is kept at 70 F. Find (7*, P,
and x if applicable) for all four states in the
cycle.
11.170E A supply of geo thermal hot water is to be used
as the energy source in an ideal Rankine cycle,
with R-134a as the cycle working fluid. Satu-
rated vapor R-134a leaves the boiler at a tem-
perature of 180 F, and the condenser
temperature is 100 F. Calculate the thermal ef-
ficiency of this cycle.
11.171E Do Problem 1 1.170 with R-22 as the working
fluid.
11.172E A smaller power plant produces 50 lbm/s
steam at 400 psia, 1 100 F, in the boiler. It cools
the condenser with ocean water coming in at
55 F and returned at 60 F so that the condenser
exit is at 1 10 F. Find the net power output and
the required mass flow rate of ocean water.
11.173E The power plant in Problem 11. 167 is modified
to have a superheater section following the
boiler so the steam leaves the superheater at
600 lbf/in. 2 , 700 F. Find the specific work and
heat transfer in each of the ideal components
and the cycle efficiency.
11.174E Consider a simple ideal Rankine cycle using
water at a supercritical pressure. Such a cycle
has a potential advantage of minimizing local
temperature differences between the fluids in
the steam generator, such as the instance in
which the high-temperature energy source is
the hot exhaust gas from a gas-turbine engine.
Calculate the thermal efficiency of the cycle
if the state entering the turbine is 8000
lbf/in. 5 , 1300 F, and the condenser pressure is
0.95 lbf/in. 2 . What is the steam quality at the
turbine exit?
11.175E Consider an ideal steam reheat cycle in which
the steam enters the high-pressure turbine at
600 lbf/in. 2 , 700 F, and then expands to 150
lbf/in. 2 . It is then reheated to 700 F and ex-
pands to 2.225 lbf/in. 2 in the low-pressure tur-
bine. Calculate the thermal efficiency of the
cycle and the moisture content of the steam
leaving the low-pressure turbine.
11.176E Consider an ideal steam regenerative cycle in
which steam enters the turbine at 600 lbf/in. 2 ,
700 F, and exhausts to the condenser at 2.225
lbf/in. 2 . Steam is extracted from the turbine at
150 lbf/in. 2 for an open feedwater heater. The
feedwater leaves the heater as saturated liquid.
The appropriate pumps are used for the water
leaving the condenser and the feedwater heater.
Calculate the thermal efficiency of the cycle
and the net work per pound-mass of steam.
468 a
CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
11.177E A closed feedwater heater in a regenerative
steam power cycle heats 40 lbm/s of water
from 200 F, 2000 lbf/in. 2 to 450 F, 2000
lbf/in. 2 . The extraction steam from the turbine
enters the heater at 500 lbf/in. 2 , 550 F and
leaves as saturated liquid. What is the required
mass flow rate of the extraction steam?
11.178E A steam power cycle has a high pressure of
600 lbf/in. 2 and a condenser exit temperature
of 110 F. The turbine efficiency is 85%, and
other cycle components are ideal. If the boiler
superheats to 1400 F, find the cycle thermal ef-
ficiency.
11.179E The steam power cycle in Problem 1 1.167 has
an isentropic efficiency of the turbine of 85%
and that for the pump of 80%. Find the cycle
efficiency and the specific work and heat trans-
fer in the components.
11.180E Steam leaves a power plant steam generator at
500 lbf/in. 2 , 650 F, and enters the turbine at
490 lbf/in. 2 , 625 F. The isentropic turbine effi-
ciency is 88%, and the turbine exhaust pres-
sure is 1.7 lb/in. 2 . Condensate leaves the
condenser and enters the pump at 110 F, 1.7
lbf/in. 2 . The isentropic pump efficiency is 80%,
and the discharge pressure is 520 lbf/in. 2 . The
feedwater enters the steam generator at 510
lbf/in. 2 , 100 F. Calculate the thermal efficiency
of the cycle and the entropy generation of the
flow in the line between the steam generator
exit and the turbine inlet, assuming an ambient
temperature of 77 F.
11.181E A boiler delivers steam at 1500 lbf/in. 2 , 1000
F, to a two-stage turbine, as shown in Fig.
1 1 .21 . After the first stage, 25% of the steam is
extracted at 200 lbf/in. 2 for a process applica-
tion and returned at 150 lbf/in. 2 , 190 F, to the
feedwater line. The remainder of the steam
continues through the low-pressure turbine
stage, which exhausts to the condenser at
2 lbf/in. 2 . One pump brings the feedwater to
150 lbf/in. 2 and a second pump brings it to
1500 lbf/in. 2 . Assume the first and second
stages in the steam turbine have isentropic effi-
ciencies of 85% and 80% and that both pumps
are ideal. If the process application requires
5000 Btu/s of power, how much power can
then be cogenerated by the turbine?
Brayton Cycles
11.182E A large stationary Brayton-cycle gas-turbine
power plant delivers a power output of 100 000
hp to an electric generator. The minimum tem-
perature in the cycle is 540 R, and the maxi-
mum temperature is 2900 R. The minimum
pressure in the cycle is 1 atm, and the compres-
sor pressure ratio is 14 to 1. Calculate the
power output of the turbine, the fraction of the
turbine output required to drive the compres-
sor, and the thermal efficiency of the cycle.
11.183E A Brayton cycle produces 14 000 Btu/s with an
inlet state of 60 F, 14.5 psia, and a compres-
sion ratio of 16: 1. The heat added in the com-
bustion is 400 Btu/lbm. What is the highest
temperature and the mass flow rate of air as-
suming cold air properties.
11.184E Do the previous problem rising properties from
Table F.5.
11.185E An ideal regenerator is incorporated into the
ideal air-standard Brayton cycle of Problem
11.182. Calculate the cycle thermal efficiency
with this modification.
11.186E An air-standard Ericsson cycle has an ideal re-
generator, as shown in Fig. PI 1.85. Heat is
supplied at 1800 F, and heat is rejected at 68 F.
Pressure at the beginning of the isothermal
compression process is 10 lbf/in. 2 . The heat
added is 275 But/lbm. Find the compressor
work, the turbine work, and the cycle effi-
ciency.
11.187E The turbine in a jet engine receives air at 2200
R, 220 lbf/in 2 . It exhausts to a nozzle at 35
lbf/in. 2 , which in turn exhausts to the atmos-
phere at 14.7 lbf/in. 2 . The isentropic efficiency
of the turbine is 85%, and the nozzle efficiency
is 95%. Find the nozzle inlet temperature and
the nozzle exit velocity. Assume negligible ki-
netic energy out of the turbine.
Otto, Diesel, Stirling, and Carnot Cycles
11.188E Air flows into a gasoline engine at 14 lbf/in. 2 ,
540 R. The air is then compressed with a volu-
metric compression ratio of 8 : 1 . In the com-
bustion process 560 Btu/lbm of energy is
released as the fuel burns. Find the temperature
and pressure after combustion.
English Unit Problems H 469
11.189E To approximate an actual spark-ignition en-
gine consider an air-standard Otto cycle that
has a heat addition of 800 Btu/lbm of air, a
compression ratio of 7, and a pressure and
temperature at the beginning of the compres-
sion process of 13 lbf/in. 2 , 50 F. Assuming
constant specific heat, with the value from
Table F,4, determine the maximum pressure
and temperature of the cycle, the thermal effi-
ciency of the cycle, and the mean effective
pressure.
11.190E A gasoline engine has a volumetric compres-
sion ratio of 10 and before compression has air
at 520 R, 12.2 psia, in the cylinder. The com-
bustion peak pressure is 900 psia. Assume cold
air properties. What is the highest temperature
in the cycle? Find the temperature at the begin-
ning of the exhaust (heat rejection) and the
overall cycle efficiency.
11, 19 IE A four- stroke gasoline engine has a compres-
sion ratio of 10 : 1 with 4 cylinders of total dis-
placement 75 in. 3 . The inlet state is 500 R, 10
psia, and the engine is running at 2100 RPM,
with the fuel adding 750 Btu/lbm in the com-
bustion process. What is the net work in the
cycle, and how much power is produced?
11.192E It is found experimentally that the power
stroke expansion in an internal combustion en-
gine can be approximated with a polytropic
process with a value of the polytropic exponent
n somewhat larger than the specific heat ratio
k. Repeat Problem 1 1 . 1 89 but assume the ex-
pansion process is reversible and polytropic
(instead of the isentropic expansion in the Otto
cycle) with n equal to 1.50.
1U93E In the Otto cycle, all the heat transfer q H occurs
at constant volume. It is more realistic to as-
sume that part of q H occurs after the piston has
started its downwards motion in the expansion
stroke. Therefore, consider a cycle identical to
the Otto cycle, except that the first two-thirds
of the total q H occurs at constant volume and
the last one-third occurs at constant pressure.
Assume the total q H is 700 Btu/lbm, that the
state at the beginning of the compression
process is 13 lbf/in. 2 , 68 F, and that the com-
pression ratio is 9. Calculate the maximum
pressure and temperature and the thermal effi-
ciency of this cycle. Compare the results with
those of a conventional Otto cycle having the
same given variables.
11.194E A diesel engine has a bore of 4 in., a stroke of
4.3 in., and a compression ratio of 19:1 run-
ning at 2000 RPM (revolutions per minute).
Each cycle takes two revolutions and has a
mean effective pressure of 200 lbf/in. 2 . With a
total of six cylinders find the engine power in
Btu/s and horsepower, hp.
11.195E At the beginning of compression in a diesel
cycle T = 540 R, P = 30 lbf/in. 2 , and the state
after combustion (heat addition) is 2600 R and
1000 lbf/in. 2 . Find the compression ratio, the
thermal efficiency, and the mean effective
pressure.
11.196E Consider an ideal air-standard diesel cycle
where the state before the compression process
is 14 lbf/in. 2 , 63 F, and the compression ratio is
20. Find the maximum temperature (by itera-
tion) in the cycle to have a thermal efficiency
of 60%.
11.197E Consider an ideal Stirling-cycle engine in
which the pressure and temperature at the be-
ginning of the isothermal compression process
are 14.7 lbf/in. 2 , 80 F, the compression ratio is
6, and the maximum temperature in the cycle
is 2000 F. Calculate the maximum pressure in
the cycle and the thermal efficiency of the
cycle with and without regenerators.
11.198E An ideal air- standard Stirling cycle uses he-
lium as working fluid. The isothermal com-
pression brings the helium from 1 5 lbf/in. 2 , 70
F to 90 lbf/in. 2 . The expansion takes place at
2100 R, and there is no regenerator. Find the
work and heat transfer in all four processes per
lbm helium and the cycle efficiency.
11.199E The air- standard Carnot cycle was not shown
in the text; show the T-s diagram for this
cycle. In an air-standard Carnot cycle, the low
temperature is 500 R, and the efficiency is
60%. If the pressure before compression and
after heat rejection is 14.7 lbf/in. 2 , find the high
temperature and the pressure just before heat
addition.
11.200E Air in a piston/cylinder goes through a Camot
cycle in which T L — 80.3 F and the total cycle
470 H CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
efficiency is tj = 2/3. Find T H , the specific
work and volume ratio in the adiabatic expan-
sion for constant C p , C„.
11.201E Do the previous Problem, 11.200E, using
Table F.5.
Refrigeration Cycles
11.202E A car air-conditioner (refrigerator) in 70 F
ambient uses R-134a, and I want to have
cold air at 20 F produced. What is the mini-
mum high P and the maximum low P it can
use?
11.203E Consider an ideal refrigeration cycle that has a
condenser temperature of 1 10 F and an evapo-
rator temperature of 5 F. Determine the coeffi-
cient of performance of this refrigerator for the
working fluids R-12 and R-22.
11.204E The environmentally safe refrigerant R-134a
is one of the replacements for R-12 in refrig-
eration systems. Repeat Problem 1 1 .203 using
R-134a and compare the result with that for
R-12.
11.205E Consider an ideal heat pump that has a con-
denser temperature of 120 F and an evapo-
rator temperature of 30 F. Determine the
coefficient of performance of this heat pump
for the working fluids R-12, R-22, and
ammonia.
11.206E The refrigerant R-22 is used as the working
fluid in a conventional heat pump cycle. Satu-
rated vapor enters the compressor of this unit
at 50 F; its exit temperature from the com-
pressor is measured and found to be 185 F. If
the compressor exit is 300 psia, what is the
isentropic efficiency of the compressor and
the coefficient of performance of the heat
pump?
ll,207EAn air-standard refrigeration cycle has a
heat exchanger included as shown in Fig.
PI 1.137. The low pressure is 14.7 psia and
the high pressure is 200 psia. Tempera-
ture into the compressor is 60 F (T x and T 3
in Fig. 11.38) and T 4 = T 6 = -60 F. Deter-
mine the coefficient of performance of the
cycle.
Availability and Combined Cycles
11.208E Find the flows and fluxes of exergy in the con-
denser of Problem 11.172E. Use those to deter-
mine the second-law efficiency.
11, 209 E (Adv.) Find the availability of the water at
all four states in the Rankine cycle de-
scribed in Problem 11.173E. Assume the
high-temperature source is 900 F and the low-
temperature reservoir is at 65 F. Determine
the flow of availability in or out of the reser-
voirs per pound-mass of steam flowing in the
cycle. What is the overall cycle second-law
efficiency?
11.210E Find the flows of exergy into and out of the
feedwater heater in Problem 1 1.176E.
11.211 E Consider the Brayton cycle in Problem
11.183E. Find all the flows and fluxes of ex-
ergy and find the overall cycle second-law effi-
ciency. Assume the heat transfers are internally
reversible processes and we then neglect any
external irreversibility.
11.212E Consider an ideal dual-loop heat-powered re-
frigeration cycle using R-12 as the working
fluid, as shown in Fig. P.l 1.144. Saturated
vapor at 220 F leaves the boiler and expands in
the turbine to the condenser pressure. Saturated
vapor at F leaves the evaporator and is com-
pressed to the condenser pressure. The ratio of
the flows through the two loops is such that the
turbine produces just enough power to drive
the compressor. The two exiting streams mix
together and enter the condenser. Saturated liq-
uid leaving the condenser at 1 10 F is then sep-
arated into two streams in the necessary
proportions. Determine the ratio of mass flow
rate through the power loop to that through the
refrigeration loop. Find also the performance
of the cycle, in terms of the ratio QJQh-
Review Problems
11.213E Consider an ideal combined reheat and regener-
ative cycle in which steam enters the high-
pressure turbine at 500 lbf/in. 2 , 700 F, and is ex-
tracted to an open feedwater heater at 120
lbf/in. 2 with exit as saturated liquid. The remain-
der of the steam is reheated to 700 F at this pres-
Computer, Design, and Open-Ended Problems M 471
sure, 120 lbf/in 2 , and is fed to the low-pressure
turbine. The condenser pressure is 2 lbf/in. 2 .
Calculate the thermal efficiency of the cycle and
the net work per pound-mass of steam.
11.214E In one type of nuclear power plant, heat is
transferred in the nuclear reactor to liquid
sodium. The liquid sodium is then pumped
through a heat exchanger where heat is trans-
ferred to boiling water. Saturated vapor steam
at 700 lbf/in. 2 exits this heat exchanger and is
then superheated to 1 100 F in an external gas-
fired superheater. The steam enters the turbine,
which has one (open-type) feedwater extrac-
tion at 60 lbf/in. 2 . The isentropic turbine effi-
ciency is 87%, and the condenser pressure is 1
lbf/in. 2 . Determine the heat transfer in the reac-
tor and in the superheater to produce a net
power output of 1000 Btu/s.
11.215E Consider an ideal gas-turbine cycle with two
stages of compression and two stages of ex-
pansion. The pressure ratio across each com-
pressor stage and each turbine stage is 8 to 1 .
The pressure at the entrance to the first com-
pressor is 14 lbf/in. 2 , the temperature entering
each compressor is 70 F, and the temperature
entering each turbine is 2000 F. An ideal re-
generator is also incorporated into the cycle.
Determine the compressor work, the turbine
work, and the thermal efficiency of the cycle.
11.216E Repeat Problem 11.215E, but assume that each
compressor stage and each turbine stage has an
isentropic efficiency of 85%, Also assume that
the regenerator has an efficiency of 70%.
11.217E Consider a small ammonia absorption refriger-
ation cycle that is powered by solar energy and
is to be used as an air conditioner. Saturated
vapor ammonia leaves the generator at 120 F,
and saturated vapor leaves the evaporator at 50
F. If 3000 Btu of heat is required in the genera-
tor (solar collector) per pound-mass of ammo-
nia vapor generated, determine the overall
performance of this system.
Computer, Design, and Open-Ended Problems
11.218 The effect of turbine exhaust pressure on the
performance of the ideal steam Rankine cycle
given in Problem 11. 33 is to be studied. Calcu-
late the thermal efficiency of the cycle and the
moisture content of the steam leaving the turbine
for turbine exhaust pressures of 5, 10, 50, and
100 kPa. Plot the thermal efficiency versus tur-
bine exhaust pressure for the specified turbine
inlet pressure and temperature.
11.219 The effect of turbine inlet pressure on the perfor-
mance of the ideal steam Rankine cycle given in
Problem 11.33 is to be studied. Calculate the
thermal efficiency of the cycle and the moisture
content of the steam leaving the turbine for tur-
bine inlet pressures of 1, 3.5, 6, and 10 MPa.
Plot the thermal efficiency versus turbine inlet
pressure for the specified turbine inlet tempera-
ture and exhaust pressure.
11.220 The effect of turbine inlet temperature on the
performance of the ideal steam Rankine cycle
given in Problem 1 1.33 is to be studied. Calcu-
late the thermal efficiency of the cycle and the
moisture content of the steam leaving the turbine
for turbine inlet temperatures of 400°, 500°,
800°C, and saturated vapor (at 3.0 MPa). Plot
the thermal efficiency versus turbine inlet tem-
perature for the specified turbine inlet pressure
and exhaust pressure.
11.221 Write a program to solve the following problem.
The effects of varying parameters on the perfor-
mance of an air-standard Brayton cycle are to be
determined. Consider a compressor inlet condi-
tion of 100 kPa, 20°C, and assume constant spe-
cific heat. The thermal efficiency of the cycle and
the net specific work output should be determined
for the combinations of the following variables.
a. Compressor pressure ratio of 6, 9, 12, and 15.
b. Maximum cycle temperature of 900, 1100,
1300, and 1500°C.
c. Compressor and turbine isentropic efficien-
cies each 100, 90, 80, and 70%.
11.222 The effect of adding a regenerator to the gas-
turbine cycle in the previous two problems is to
472 H CHAPTER ELEVEN POWER AND REFRIGERATION SYSTEMS
I
be studied. Repeat one of these problems by in-
cluding a regenerator with, various values of the
regenerator efficiency.
1 1.223 Write a program to simulate the Otto cycle using
nitrogen as the working fluid. Use the variable
specific heat as given in Table A.6. The begin-
ning of compression has a state of 100 kPa t
20°C. Determine the net specific work output
and the cycle thermal efficiency for various
combinations of compression ratio and maxi-
mum cycle temperature. Compare the result
with those found when constant specific heat is
assumed.
11.224 A power plant is built to provide district heating
of buildings that requires 90°C liquid water at
150 kPa. The district heating water is returned at
50°C, 100 kPa, in a closed loop in an amount
such that 20 MW of power is delivered. This hot
water is produced from a steam power cycle
with a boiler making steam at 5 MPa, 600°C, de-
livered to the steam turbine. The steam cycle
could have its condenser operate at 90°C provid-
ing the power to the district heating. It could
also be done with extraction of steam from the
turbine. Suggest a system and evaluate its per-
formance in terms of the cogenerated amount of
turbine work.
11.225 The effect of evaporator temperature on the co-
efficient of performance of a heat pump is to be
studied. Consider an ideal cycle with R-22 as the
working fluid and a condenser temperature of
40°C. Plot a curve for the coefficient of perfor-
mance versus the evaporator temperature for
temperatures from + 15 to — 25°C.
11.226 A hospital requires 2 kg/s steam at 200°C, 125
kPa, for sterilization purposes, and space heating
requires 15 kg/s hot water at 90°C, 100 kPa.
Both of these requirements are provided by the
hospital's steam power plant. Discuss some
arrangement that will accomplish this.
11.227 Investigate the maximum power out of a steam
power plant with operating conditions as in
Problem 11.33. The energy source is 100 kg/s
combustion products (air) at 125 kPa, 1200 K.
Make sure the air temperature is higher than the
water temperature throughout the boiler.
11.228 In Problem 11.141, a steam cycle was powered
by the exhaust from a gas turbine. With a single
water flow and airflow heat exchanger, the air is
leaving with a relatively high temperature. Ana-
lyze how some more of the energy in the air can
be used before the air is flowing out to the chim-
ney. Can it be used in a feedwater heater?
Gas Mixtures
Up to this point in our development of thermodynamics, we have considered primarily
pure substances. A large number of thermodynamic problems involve mixtures of differ-
ent pure substances. Sometimes these mixtures are referred to as solutions, particularly in
the liquid and solid phases.
In this chapter we shall turn our attention to various thermodynamic considerations
of gas mixtures. We begin with a consideration of a rather simple problem: mixtures of
ideal gases. This leads to a consideration of a simplified but very useful model of certain
mixtures, such as air and water vapor, which may involve a condensed (solid or liquid)
phase of one of the components.
12.1 GENERAL CONSIDERATIONS AND
MIXTURES OF IDEAL GASES
Let us consider a general mixture of A'" components, each a pure substance, so the total
mass and the total number of moles are
'"tot ~ m \ + m 2 + ' ' ' + m N = 2) m i
«tot = "1 + «2 + ' • ' + n N = 2 »/
The mixture is usually described by a mass fraction (concentration)
or a mole fraction for each component as
* = £ • W
which are related through the molecular weight, M h as m t ~ nftfj. We may then convert
from a mole basis to a mass basis as
and from a mass basis to a mole basis as
= A = milM ~' = m,/(M»Jtot) = C,fMt
473
474 B CHAPTER TWELVE GAS MIXTURES
The molecular weight for the mixture becomes
= « £L=; s«^ =2 m (12<5)
ml * "tot "tot AW t
that is also the denominator in Eq. 12.3.
EXAMPLE 12.1 A mole-basis analysis of a gaseous mixture yields the following results:
C0 2 12.0%
2 4.0
N 2 82.0
CO 2.0
Determine the analysis on a mass basis and the molecular weight for the mixture.
Control mass: Gas mixture.
State: Composition known.
TABLE 12.1
Constituent
Percent
by Mole
Mole
Fraction
Molecular
Weight
Mass kg
per kmoi
of Mixture
Analysis
on Mass Basis,
Percent
co 2
12
0.12
X
44.0
= 5.28
5.28 - 17.55
30.08
o 2
4
0.04
X
32.0
= 1.28
L2S - 4.26
30.08
N 2
82
0.82
X
28.0
= 22.96
22.96 _ ?6 33
30.08
CO
2
0.02
X
28.0
0.56
30.08
0.56 _ 1.86
30.08 100.00
TABLE 12,2
kmol
Mass
Molecular
per kg
Mole
Mole
Constituent
Fraction
Weight
of Mixture
Fraction
Percent
co 2
0.1755
+ 44.0
= 0.003 99
0.120
12.0
o 2
0.0426
^ 32.0
= 0.00133
0.040
4.0
N 2
0.7633
+ 28.0
= 0.027 26
0.820
82.0
CO
0.0186
+ 28.0
= 0.000 66
0.020
2.0
0.033 24
1.000
100.0
General Considerations and Mixtures of Ideal Gases M 475
Solution
It is convenient to set up and solve this problem as shown in Table 12.1. The mass-basis
analysis is found using Eq. 12.3, as shown in the table. It is also noted that during this
calculation, the molecular weight of the mixture is found to be 30.08.
If the analysis has been given on a mass basis, and the mole fractions or percent-
ages are desired, the procedure shown in Table 12.2 is followed, using Eq. 12.4.
Consider a mixture of two gases (not necessarily ideal gases) such as shown in Fig.
12,1. What properties can we experimentally measure for such a mixture? Certainly we
can measure the pressure, temperature, volume, and mass of the mixture. We can also ex-
perimentally measure the composition of the mixture, and thus determine the mole and
mass fractions.
Suppose that this mixture undergoes a process or a chemical reaction and we wish
to perform a thermodynamic analysis of this process or reaction. What type of thermody-
namic data would we use in performing such an analysis? One possibility would be to
have tables of thermodynamic properties of mixtures. However, the number of different
mixtures that is possible, both as regards the substances involved and the relative amounts
of each, is such that we would need a library full of tables of thermodynamic properties to
handle all possible situations. It would be much simpler if we could determine the ther-
modynamic properties of a mixture from the properties of the pure components. This is in
essence the approach that is used in dealing with ideal gases and certain other simplified
models of mixtures.
One exception to this procedure is the case where a particular mixture is encoun-
tered very frequently, the most familiar being air. Tables and charts of the thermodynamic
properties of air are available. However, even in this case it is necessary to define the
composition of the "air" for which the tables are given, because the composition of the at-
mosphere varies with altitude, with the number of pollutants, and with other variables at a
given location. The composition of air on which air tables are usually based is as follows:
Component
% on Mole Basis
Nitrogen
78.10
Oxygen
20.95
Argon
0.92
C0 2 & trace elements
0.03
FIGURE 12.1 A
mixture of two gases.
Temperature = T
Pressure = P
Gases A + B :
.. - Volume V
■
476 M CHAPTER TWELVE GAS MIXTURES
In this chapter we focus on mixtures of ideal gases. We assume that each compo-
nent is uninfluenced by the presence of the other components and that each component
can be treated as an ideal gas. In the case of a real gaseous mixture at high pressure, this
assumption would probably not be accurate because of the nature of the interaction be-
tween the molecules of the different components. In this text, we will consider only a sin-
gle model in analyzing gas mixtures, namely, the Dal ton model.
Dalton Model
For the Dalton model of gas mixtures, the properties of each component of the mixture
are considered as though each component exists separately and independently at the tem-
perature and volume of the mixture, as shown in Fig. 12.2. We further assume that both
the gas mixture and the separated components behave according to the ideal gas model,
Eqs. 3.3-3.6, In general, we would prefer to analyze gas mixture behavior on a mass
basis. However, in this particular case it is more convenient to use a mole basis, since the
gas constant is then the universal gas constant for each component and also for the mix-
ture. Thus, we may write for the mixture (Fig. 12,1)
PV = nRT
n = n A + n B (12.6)
and for the components (Fig. 12.2)
On substituting, we have
P A V = n A RT
P B V = n B RT (12.7)
n = n A + n B
RT RT RT
IV^ P JL + P JL ( 12.8)
P-P a +Pb O 2 - 9 )
where P A and P a are referred to as partial pressures. Thus, for a mixture of ideal gases, the
pressure is the sum of the partial pressures of the individual components, where, using
Eqs. 12.6 and 12.7,
FIGURE 12,2 The
Dalton model.
Temperature = T
Pressure = P A
Temperature = T
Pressure = P B
b
Gas B
Volume V,
General considerations and Mixtures of ideal gases H 477
That is, each partial pressure is the product of that component's mole fraction and the
mixture pressure.
In determining the internal energy, enthalpy, and entropy of a mixture of ideal
gases, the Dalton model proves useful because the assumption is made that each con-
stituent behaves as though it occupies the entire volume by itself. Thus, the internal en-
ergy, enthalpy, and entropy can be evaluated as the sum of the respective properties of the
constituent gases at the condition at which the component exists in the mixture. Since for
ideal gases the internal energy and enthalpy are functions only of temperature, it follows
that for a mixture of components A and B, on a mass basis,
U — mn = m A a A + m B u B
= m{c A u A + c B a B ) (12.11)
H -- mh — m A h A + m B h B
- m{c A h A + c s h B ) (12.12)
In Eqs. 12.11 and 12.12, the quantities u A) u B , h A , and h 3 are the ideal-gas properties of the
components at the temperature of the mixture. For a process involvirig a change of tem-
perature, the changes in these values are evaluated by one of the three models discussed in
Section 5 .7 — involving either the ideal-gas Tables A.7 or the specific heats of the compo-
nents. In a similar manner to Eqs. 12.11 and 12.12, the mixture energy and enthalpy could
be expressed as the sums of the component mole fractions and properties per mole.
The ideal-gas mixture equation of state on a mass basis is
(12.13)
where
Alternatively,
^ = - (tv = s <«*) = m ™ t ii14 )
= Wl ( m A R <i + m B R B)
= c A R A + c B R B (12.15)
The entropy of an ideal-gas mixture is expressed as
S — ms = in A s A + msS B
= ™(p A s A + CbS b ) (12.16)
It must be emphasized that the component entropies in Eq. 12.16 must each be evaluated
at the mixture temperature and the corresponding partial pressure of the component in the
mixture, using Eq. 12.10 in terms of the mole fraction.
To evaluate Eq. 12.16 using the ideal-gas entropy expression 8.24, it is necessary to
use one of the specific heat models discussed in Section 8.10. The simplest model is con-
stant specific heat, Eq. 8.25, using an arbitrary reference state T 0> P 0l s 0i for each compo-
nent /' in the mixture at T and P,
Sl - s 0i + C p0i In if I - R t In I ^- ] (12.17)
478 M CHAPTER TWELVE GAS MIXTURES
Consider a process with constant-mixture composition between state 1 and state 2
and let us calculate the entropy change for component / with Eq. 12. 17.
"l r2
In —
In-p-
= + C p0i In
n"
-i? f ln
= C, o; ln^-tf f ln-^
We observe here that this expression is very similar to Eq. 8.25 and that the reference val-
ues %, T , P D all cancel out, as does the mole fraction.
An alternative model is to use the s° T function defined in Eq. 8.27, in which case
each component entropy in Eq. 12.16 is expressed as
«-4.-* ta $) (12 ' 18)
The mixture entropy could also be expressed as the sum of component properties on a
mole basis.
EXAMPLE 12.2 Let a mass m A of ideal gas A at a given pressure and temperature, P and T, be mixed with
m B of ideal gas B at the same P and T, such that the final ideal-gas mixture is also at P
and T. Determine the change in entropy for this process.
Control mass: All gas (A and B).
Initial states: P, T known for A and B.
Final state: P, T of mixture known.
Analysis and Solution
The mixture entropy if given by Eq. 12.16. Therefore, the change of entropy can be
grouped into changes for A and for B, with each change expressed by Eq. 8.24. Since
there is no temperature change for either component, this reduces to
ASU = m A (o - R A In + m B (o - R 3 In ^)
- ~m A R A \viy A - m s R B \ny B
which can also be written in the form
ASU = ~n A Rlny A - n B Rhiy B
The result of Example 12.2 can readily be generalized to account for the mixing of
any number of components at the same temperature and pressure. The result is
k
The interesting thing about this equation is that the increase in entropy depends only on
the number of moles of component gases and is independent of the composition of the
General considerations and Mixtures of Ideal Gases H 479
gas. For example, when 1 mol of oxygen and 1 mol of nitrogen are mixed, the increase in
entropy is the same as when 1 mol of hydrogen and 1 mol of nitrogen are mixed. But we
also know that if 1 mol of nitrogen is "mixed" with another mole of nitrogen, there is no
increase in entropy. The question that arises is, how dissimilar must the gases be in order
to have an increase in entropy? The answer lies in our ability to distinguish between the
two gases (based on their different molecular weights). The entropy increases whenever
we can distinguish between the gases being mixed. When we cannot distinguish between
the gases, there is no increase in entropy.
One special case that arises frequently involves an ideal -gas mixture undergoing a
process in which there is no change in composition. Let us also assume that the constant
specific heat model is reasonable. For this case, from Eq. 12.1 1 on a unit mass basis, the
internal energy change is
h 2 - ui = c A C vQA (T 2 - Tj) + c B C^ B {T 2 - T x )
~ Qomix (?2 ~ ^i) (12.20)
where
Qo mix ~ C A Q'O A + c B^vOB (12.21)
Similarly, from Eq. 12.12, the enthalpy change is
h 2 ~hi = c A C^ A {T 2 - I*,) + CjC^Ti ~ Tj)
-C^^-T,) (12.22)
where
CpOmix = C ACpOA + c bCj>ob (12.23)
The entropy change for a single component was calculated from Eq. 12.17, so we
substitute this result into Eq. 12.16 to evaluate the change as
S 2 ~Si = c A (s 2 - Si) A + c B {s 2 - s{) B
- c aC{&a In y - c A R A In ~ + c B C p0 B In ~ - c B R B In y
= C^m^-^ln^ (12.24)
The last expression was using Eq. 12.15 for the mixture gas constant andEq. 12.23 for the
mixture heat capacity. We see that Eqs. 12,20, 12.22, and 12.24 are the same as those for
the pure substance, Eqs. 5.20, 5.29, and 8.25. So we can treat a mixture similarly to a pure
substance once the mixture properties are found from the composition and the component
properties in Eqs. 12.15, 12.21, and 12.23.
This also implies that all the polytropic processes in a mixture can be treated simi-
larly to the way it is done for a pure substance (recall Sections 8.10 and 8.1 1). Specifically
the isentropic process where s is constant leads to the power relation between temperature
and pressure from Eq. 12.24. This is similar to Eq. 8.29, provided we use the mixture heat
capacity and gas constant. The ratio of specific heats becomes
C
-71 mix
mix f~> f~i p
W mix ^/>mix -"-n
and the relation can then also be written as in Eq. 8.32.
480 ■ CHAPTER TWELVE GAS MIXTURES
So far we have looked at mixtures of ideal gases as a natural extension to the de-
scription of processes involving pure substances. The treatment of mixtures for nonideal
(real) gases and multiphase states is important for many technical applications, for in-
stance, in the chemical process industry. It does require a more extensive study of the
properties and general equations of state, so we will defer this subject to Chapter 13.
12.2 A Simplified model oe a Mixture
Involving Gases and a Vapor
Let us now consider a simplification, which is often a reasonable one, of the problem in-
volving a mixture of ideal gases that is in contact with a solid or liquid phase of one of
the components. The most familiar example is a mixture of air and water vapor in con-
tact with liquid water or ice, such as is encountered in air conditioning or in drying. We
are all familiar with the condensation of water from the atmosphere when it cools on a
summer day.
This problem and a number of similar problems can be analyzed quite simply and
with considerable accuracy if the following assumptions are made:
1. The solid or liquid phase contains no dissolved gases.
2. The gaseous phase can be treated as a mixture of ideal gases.
3. When the mixture and the condensed phase are at a given pressure and temperature,
the equilibrium between the condensed phase and its vapor is not influenced by the
presence of the other component. This means that when equilibrium is achieved, the
partial pressure of the vapor will be equal to the saturation pressure corresponding
to the temperature of the mixture.
Since this approach is used extensively and with considerable accuracy, let us give
some attention to the terms that have been defined and the type of problems for which
this approach is valid and relevant. In our discussion we will refer to this as a gas-vapor
mixture.
The dew point of a gas-vapor mixture is the temperature at which the vapor con-
denses or solidifies when it is cooled at constant pressure. This is shown on the T-s dia-
gram for the vapor shown in Fig. 12.3. Suppose that the temperature of the gaseous
mixture and the partial pressure of the vapor in the mixture are such that the vapor is im-
v = constant ■
FIGURE 12.3
Temperature-entropy
diagram to show
definition of the dew
point.
A' Simplified Model of a Mixture Involving Gases and a Vapor H 481
tially supherheated at state I. If the mixture is cooled at constant pressure, the partial pres-
sure of the vapor remains constant until point 2 is reached, and then condensation begins.
The temperature at state 2 is the dew-point temperature. Lines 1-3 on the diagram indi-
cate that if the mixture is cooled at constant volume the condensation begins at point 3,
which is slightly lower than the dew-point temperature.
If the vapor is at the saturation pressure and temperature, the mixture is referred to
as a saturated mixture, and for an air-water vapor mixture, the term saturated air is used.
The relative humidity <f> is defined as the ratio of the mole fraction of the vapor in
the mixture to the mole fraction of vapor in a saturated mixture at the same temperature
and total pressure. Since the vapor is considered an ideal gas, the definition reduces to the
ratio of the partial pressure of the vapor as it exists in the mixture, P v , to the saturation
pressure of the vapor at the same temperature, P g :
4>-p-
In terms of the numbers on the T-s diagram of Fig. 12.3, the relative humidity (f>
would be
*-£
Since we are considering the vapor to be an ideal gas, the relative humidity can also be
defined in terms of specific volume or density:
< 12 - 25 >
The humidity ratio co of an air-water vapor mixture is defined as the ratio of the
mass of water vapor m u to the mass of dry air m a . The term dry air is used to emphasize
that this refers only to air and not to the water vapor. The term specific humidity is used
synonymously with humidity ratio.
» = £ (12.26)
This defmition is identical for any other gas-vapor mixture, and the subscript a
refers to the gas, exclusive of the vapor. Since we consider both the vapor and the mixture
to be ideal gases, a very useful expression for the humidity ratio in terms of partial pres-
sures and molecular weights can be developed. Writing
_P V_ P V VM U _P a V _ P a VM a
m " R V T r T ' m ° R a T Jt
we have
_P V V/B V T_RJ> V _M V P V
P a ¥IR a T RuPc MJ> a
For an air-water vapor mixture, this reduces to
(12.27)
co = 0.622 j- (12.28)
The degree of saturation is defined as the ratio of the actual humidity ratio to the hu-
midity ratio of a saturated mixture at the same temperature and total pressure.
482 Ll Chapter Twelve gas mixtures
FIGURE 12.4
Temperature-entropy
diagram to show the
cooling of a gas-vapor
mixture at a constant
pressure.
An expression for the relation between the relative humidity <p and the humidity
ratio (o can be found by solving Eqs. 12.25 and 12.28 for P u and equating them. The re-
sulting relation for an air-water vapor mixture is
Af = aP " (12.29)
9 0.622 P g
A few words should also be said about the nature of the process that occurs when a
gas— vapor mixture is cooled at constant pressure. Suppose that the vapor is initially su-
perheated at state 1 in Fig. 12.4. As the mixture is cooled at constant pressure, the partial
pressure of the vapor remains constant until the dew point is reached at point 2, where the
vapor in the mixture is saturated. The initial condensate is at state 4 and is in equilibrium
with the vapor at state 2. As the temperature is lowered further, more of the vapor con-
denses, which lowers the partial pressure of the vapor in the mixture. The vapor that re-
mains in the mixture is always saturated, and the liquid or solid is in equilibrium with it.
For example, when the temperature is reduced to T 3> the vapor in the mixture is at state 3,
and its partial pressure is the saturation pressure corresponding to T v The liquid in equi-
librium with it is at state 5.
EXAMPLE 12.3 Consider 100 m 3 of an air-water vapor mixture at 0.1 MPa, 35°C, and 70% relative hu-
midity. Calculate the humidity ratio, dew point, mass of air, and mass of vapor.
Control ttiass: Mixture.
State: P } T, <p known; state fixed.
Analysis and Solution
From Eq. 12.25 and the steam tables, we have
4> - 0.70 - ~;
' g
P u ■■- 0.70(5.628) = 3.94 kPa .
The dew point is the saturation temperature corresponding to this pressure, which
is 28.6°C.
The partial pressure of the air is
P a = P - P v = 100 - 3.94 - 96.06 kPa
A" SIMPLIFIED MODEL OF A MIXTURE INVOLVING GASES AND A VAPOR H 483
The humidity ratio can be calculated -from Eq. 12.28:
a> = 0.622 X ^ = 0.622 X -^1 = .0255
The mass of air is
„. _ 96.06 X 100 _. nnri _
a R a T 0.287 X 308.2 1U8 - bK S
The mass of the vapor can be calculated by using the humidity ratio or by using
the ideal-gas equation of state:
m v (om„ - 0.0255(108.6) - 2.77 kg
_ 3.94 X 100
" 0.4615 X 308.2 Z - //K §
EXAMPLE 12.3E Consider 2000 ft 3 of an air-water vapor mixture at 14.7 Ibf/in. 2 , 90 F, 70% relative hu-
midity. Calculate the humidity ratio, dew point, mass of air, and mass of vapor.
Control mass: Mixture.
State; P, T, <f> known; state fixed.
Analysis and Solution
From Eq. 12.25 and the steam tables,
r s
P v = 0.70(0.6988) = 0.4892 lbf/in 2
The dew point is the saturation temperature corresponding to this pressure, which
is 78.9 F.
The partial pressure of the air is
P a =P-P D = 14.70 - 0.49 = 14.21 lbf/in 2
The humidity ratio can be calculated from Eq. 12.28.
The mass of air is
to = 0.622 X ^ = 0.622 X = 0.02135
m - P ° V - H.21 X 144 X 2000 t , Qfil .
m ° ™ RJ ~ 53.34 X 550 = 139,6 lbm
The mass of the vapor can be calculated by using the humidity ratio or by using
the ideal-gas equation of state.
m v = a>m a = 0.02135(139.6) - 2.98 lbm
JM - 0-4892 X 144 X 2000 n QQ „
85.7 X 550 = 2-»81hm
1
484 a CHAPTER TWELVE GAS MIXTURES
EXAMPLE 12.4 Calculate the amount of water vapor condensed if the mixture of Example 12.3 is cooled
to 5°C in a constant-pressure process.
Control mass: Mixture.
Initial state: Known (Example 12.3).
Final state: T known.
Process: Constant pressure.
Analysis
At the final temperature, 5°C, the mixture is saturated, since this is below the dew-point
temperature. Therefore,
and
w 2 - 0.622
From the conservation of mass, it follows that the amount of water condensed is equal to
the difference between the initial and final mass of water vapor, or
Mass of vapor condensed = m a {(*>\ ~ «z)
Solution
We have
^2=^2 = 0.8721 kPa
P a2 = 100 - 0.8721 = 99.128 kPa
Therefore,
w 2 = 0.622 = 0.0055
Mass of vapor condensed = m a (o>i ~ o) 2 ) = 108.6(0.0255 0.0055)
= 2.172 kg
EXAMPLE 12.4E Calculate the amount of water vapor condensed if the mixture of Example 12.3E is
cooled to 40 F in a constant-pressure process.
Control mass: Mixture.
Initial state: Known (Example 12.3E).
Final state: T known.
Process: Constant pressure.
THE FIRST LAW APPLIED TO GAS- VAPOR MIXTURES B 485
Analysis
At the final temperature, 40 F, the mixture is saturated, since this is below the dew-point
temperature. Therefore,
P V 2=PgZ, P<,2=P-P»2
and
o> 2 - 0.622 P . f~
"al
From the conservation of mass, it follows that the amount of water condensed is equal to
the difference between the initial and final mass of water vapor, or
Mass of vapor condensed = m a ((*>i ~ <o 2 )
Solution
We have
Pvi =P g i = 0.1217 lbf/in. 2
P a2 = 14.7 — 0.12 = 14.58 lbf/in. 2
Therefore,
(o 2 = 0.622 X ^1 = 0.00520
Mass of vapor condensed = m a {<o x - w 2 ) = 139.6(0.02135 - 0.0052)
- 2.25 tbm
123 the First Law Applied to
Gas-Vapor Mixtures
In applying the first law of thermodynamics to gas-vapor mixtures, it is helpful to realize
that because of our assumption that ideal gases are involved, the various components can
be treated separately when calculating changes of internal energy and enthalpy. There-
fore, in dealing with air-water vapor mixtures, the changes in enthalpy of the water vapor
can be found from the steam tables and the ideal-gas relations can be applied to the air.
This is illustrated by the examples that follow.
EXAMPLE 12.5 An air-conditioning unit is shown in Fig. 12.5, with pressure, temperature, and relative
humidity data. Calculate the heat transfer per kilogram of dry air, assuming that changes
in kinetic energy are negligible.
Control volume: Duct, excluding cooling coils.
Met state: Known (Fig. 12.5).
486 ffl Chapter. Twelve gas Mixtures
Air-water vapor
p=105kPa .
r=30*C
(> = 80%
Air-water vapor
p= 100 kPa
T= 15°C
f» = 95%
Cooling coils
©
Liquid water 15°C
FIGURE 12.5 Sketch
for Example 12.5.
t
Exit state: Known (Fig. 12.5).
Process: Steady state with no kinetic or potential energy changes.
Model: Air— ideal gas, constant specific heat, value at 300 K. Water —
steam tables. (Since the water vapor at these low pressures is
being considered an ideal gas, the enthalpy of the water vapor
is a function of the temperature only. Therefore the enthalpy of
slightly superheated water vapor is equal to the enthalpy of sat-
urated vapor at the same temperature.)
Analysis
From the continuity equations for air and water, we have
If we divide this equation by m a , introduce the continuity equation for the water,
and note that m v = com a , we can write the first law in the form
m 0l = m at
The first law gives
^ + h a) + a>Xi = Ki + 02K2 + (wi - OiWrz
Solution
We have
P»i = <l>i p 8 i = 0.80(4.246) = 3.397 kPa
Pv2 = ^ 2 p s2 = 0.95(1.7051) = 1.620 kPa
The First Law Applied to Gas- Vapor Mixtures B 487
Substituting, we obtain
QcJ'K + K\ + 0>Jl vX = }l a2 + <U 2 A«2 + (fi>l ~ <»2)h[2
Q c Jm a = 1.004(15 - 30) + 0.0102(2528.9)
-0.0208(2556.3) + (0.0208 - 0.0102)(62.99)
= -41.76 kj/kg dry air
EXAMPLE 12.6 A tank has a volume of 0.5 m 3 and contains nitrogen and water vapor. The temperature of
the mixture is 50°C 3 and the total pressure is 2 MPa, The partial pressure of the water vapor
is 5 kPa. Calculate the heat transfer when the contents of the tank are cooled to 10°C
Control mass; Nitrogen and water.
Initial state: P b 7*! known; state fixed.
Final state: T 2 known.
Process: Constant volume.
Model: Ideal-gas mixture; constant specific heat for nitrogen; steam ta-
bles for water.
Analysis
This is a constant- volume process. Since the work is zero, the first law reduces to
Q=Ui-U l - m Ni C vm {T 2 - 7*!) + (m 2 u 2 % + (nhu^j ~ (m^X
This equation assumes that some of the vapor condensed. This assumption must
be checked, however, as shown in the solution:
Solution
The mass of nitrogen and water vapor can be calculated using the ideal-gas equation of
state:
», - P ^ V _ 1995 X 0.5 1ft - Q1
~R^T~ 0.2968 X 323.2 ~ 10 " 39 kg
m - P * V - 5X0.5
WbI " SJT ~ 0.4615 X 323.2 " °* 016 76 kg
If condensation takes place, the final state of the vapor will be saturated vapor at
10°C. Therefore,
,» ~ P ^ V ~ 1-2276 X 0.5 ftm7nl
m * "R~f~ 0.4615 X 283.2 = °-°° 4 70 kg
Since this amount is less than the original mass of vapor, there must have been
condensation.
The mass of liquid that is condensed, m a , is
m a = m ol - m vl = 0.016 76 - 0.004 70 = 0.012 06 kg
488 M CHAPTER TWELVE GAS MIXTURES
The internal energy of the water vapor is equal to the internal energy of saturated
water vapor at the same temperature. Therefore,
u 0l — 2443.5 kJ/kg
u 0t = 2389.2 kJ/kg
u n 42.0 kJ/kg
g cv - 10.39 X 0.745(10 - 50) + 0.0047(2389.2)
+ 0.012 06(42.0) - 0.016 76(2443.5)
- -338.8 kJ
12.4 THE ADIABATIC SATURATION PROCESS
An important process for an air-water vapor mixture is theadiabatic saturation process. In
this process, a air-vapor mixture comes in contact with a body of water in a well-
insulated duct (Fig. 12.6). If the initial humidity is less than 100%, some of the water will
evaporate and the temperature of the air-vapor mixture wilt decrease. If the mixture leav-
ing the duct is saturated and if the process is adiabatic, the temperature of the mixture on
leaving is known as the adiabatic saturation temperature. For this to take place as a
steady-state process, makeup water at the adiabatic saturation temperature is added at the
same rate at which water is evaporated. The pressure is assumed to be constant.
Considering the adiabatic saturation process to be a steady-state process, and ne-
glecting changes in kinetic and potential energy, the first law reduces to
h aX + wj/iyj + (ft> 2 ~~ wife = K% + MiKt
o>,(/M " h a ) = c^(r 2 - r,) + w 2 (h v2 - h n )
«i(A„i - hi) = C pa {T 2 - T x ) + o> 2 h m (12.30)
The most significant point to be made about the adiabatic saturation process is that
the adiabatic saturation temperature, the temperature of the mixture when it leaves the
duct, is a function of the pressure, temperature, and relative humidity of the entering
air-vapor mixture and of the exit pressure. Thus, the relative humidity and the humidity
ratio of the entering air-vapor mixture can be determined from the measurements of the
pressure and temperature of the air-vapor mixture entering and leaving the adiabatic satu-
rator. Since these measurements are relatively easy to make, this is one means of deter-
mining the humidity of an air-vapor mixture.
''/////////////////////////////XL
Air+ vapor
FIGURE 12.6 The
adiabatic saturation
process.
Saturated
^ air-vapor
mixture
Water
THE ADIABATJC SATURATION PROCESS B 489
EXAMPLE 12,7 The pressure of the mixture entering and leaving the adiabatic saturator is 0.1 MPa, the
entering temperature is 30°C, and the temperature leaving is 20°C 3 which is the adiabatic
saturation temperature. Calculate the humidity ratio and relative humidity of the
air-water vapor mixture entering.
Control volume; Adiabatic saturator.
Inletstate: P b T] known.
Exitstate: P 2 , T 2 known; tf> 2 = 100%; state fixed.
'Process: Steady state, adiabatic saturation (Fig. 12.6).
Model: Ideal-gas mixture; constant specific heat for air; steam tables
for water.
Analysis
Use continuity and the first law, Eq. 12.30.
Solution
Since the water vapor leaving is saturated, P v2 = P g2 and a> 2 can be calculated.
°' 622 * (lorar) - 00149
a> l can be calculated using Eq. 12.30.
_ C pa {T 2 ~ 7',) + o) 2 h fg2
(hi ~ A/z)
1.004(20 - 30) + 0.0149 X 2454.1
1 2556.3 - 83.96
(o x = 0.0107 = 0.622 X
- 0.0107
.100 ~P vl
P vl = 1.691 kPa
^-^-4^46 = - 398
Example 12.7E
The pressure of the mixture entering and leaving the adiabatic saturator is 14.7 Ibf/in. 2 ,
the entering temperature is 84 F, and the temperature leaving is 70 F, which is the adia-
batic saturation temperature. Calculate the humidity ratio and relative humidity of the
air-water vapor mixture entering.
Control volume: Adiabatic saturator.
Inletstate:: P 1 ,7 , I known.
Exit state: P 2 , T 2 known; <f> 2 = 1 00%; state fixed.
Process: Steady state, adiabatic saturation (Fig. 12.6).
Model: Ideal-gas mixture; constant specific heat for air; steam tables
for water.
490 H CHAPTER TWELVE GAS MIXTURES
Analysis
Use continuity and the first law, Eq. 12.30.
Solution
Since the water vapor leaving is saturated, P v2 = P $2 and <o 2 can be calculated.
0.622 X =0.01573
o) l can be calculated using Eq. 12.30.
_ 0.24(70 - 84) + 0.01573 X 1054.0 = -3.36 + 1660 _ Q Q125
w i ~ 109S.1 - 38.1 1060.0
".-"• 622X (l4^)^ 0125
P vl = 0.289
, P«i_ 0.289 „ 0195
12.5 WET-BULB AND BRY-BULB
TEMPERATURES
The humidity of air-water vapor mixtures has traditionally been measured with a device
called a psychrometer, which uses the flow of air past wet-bulb and dry-bulb thermome-
ters. The bulb of the wet-bulb thermometer is covered with a cotton wick that is saturated
with water. The dry-bulb thermometer is used simply to measure the temperature of the
air. The airflow can be maintained by a fan, as shown in the continuous-flow psychrome-
ter depicted in Fig. 12.7.
The processes that take place at the wet-bulb thermometer are somewhat compli-
cated. First, if the air-water vapor mixture is not saturated, some of the water in the wick
evaporates and diffuses into the surrounding air, which cools the water in the wick. As
soon as the temperature of the water drops, however, heat is transferred to the water from
both the air and the thermometer, with corresponding cooling. A steady state, determined
by heat and mass transfer rates, will be reached, in which the wet-bulb thermometer tem-
perature is lower than the dry-bulb temperature.
It can be argued that this evaporative cooling process is very similar, but not identi-
cal, to the adiabatic saturation process described and analyzed in Section 12.4. In fact, the
adiabatic saturation temperature is often termed the thermodynamic wet-bulb tempera-
ture. It is clear, however, that the wet-bulb temperature as measured by a psychrometer is
influenced by heat and mass transfer rates, which depend, for example, on the airflow ve-
locity and not simply on thermodynamic equilibrium properties. It does happen that the
The Psychrometric Chart m 491
FIGURE 12.7 Steady-
flow apparatus for
measuring wet- and dry-
bulb temperatures.
Wafer reservoir
two temperatures are very close for air-water vapor mixtures at atmospheric temperature
and pressure, and they will be assumed to be equivalent in this text.
In recent years, humidity measurements have been made using other phenomena
and other devices, primarily electronic devices for convenience and simplicity. For exam-
pie, some substances tend to change in length, in shape, or in electrical capacitance, or in
a number of other ways, when they absorb moisture. They are therefore sensitive to the
amount of moisture in the atmosphere. An instrument making use of such a substance can
be calibrated to measure the humidity of air-water vapor mixtures. The instrument output
can be programmed to furnish any of the desired parameters, such as relative humidity,
humidity ratio, or wet-bulb temperature.
12.6 The psychrometric Chart
Properties of air-water vapor mixtures are given in graphical form on psychrometric
charts. These are available in a number of different forms, and only the main features are
considered here. It should be recalled that three independent properties— such as pressure,
temperature, and mixture composition—will describe the state of this binary mixture.
A simplified version of the chart included in Appendix E, Fig. E.4, is shown in Fig.
12.8. This basic psychrometric chart is a plot of humidity ratio (ordinate) as a function of
dry-bulb temperature (abscissa), with relative humidity, wet-bulb temperature, and mix-
ture enthalpy per mass of dry air as parameters. If we fix the total pressure for' which the
chart is to be constructed (which in our chart is 1 bar, or 100 kPa), lines of constant rela-
tive humidity and wet-bulb temperature can be drawn on the chart, because for a given
dry-bulb temperature, total pressure, and humidity ratio, the relative humidity and wet-
bulb temperature are fixed. The partial pressure of the water vapor is fixed by the humid-
ity ratio and the total pressure, and therefore a second ordinate scale that indicates the
partial pressure of the water vapor could be constructed. It would also be possible to in-
clude the mixture-specific volume and entropy on the chart.
Most psychrometric charts give the enthalpy of an air-vapor mixture per kilogram
of dry air. The values given assume that the enthalpy of the dry air is zero at -20°C, and
the enthalpy of the vapor is taken from the steam tables (which are based on the assump-
CHAPTER TWELVE GAS MIXTURES
FIGURE 12.8
Psychrometrie chart.
10 15 20 25 30
Dry bulb temperature <
tion that the enthalpy of saturated liquid is zero at 0*C). The value used in the psychro-
metric chart is then
A - *. - A a (-20°C) + (oh„
This procedure is satisfactory because we are usually concerned onl y will ^««oc«ta
enthatoy. That the lines of constant enthalpy are essentially parallel to hues of constant
wTbulb temperature is evident from the fact that the wet-bulb temperature * essentially
Til to the adiabatic saturation temperature. Thus, in Fig. 12 6, if we neglect the en ha py
oi th liquid entering the adiabatic saturator, the enthalpy of the au-vapor rmxture leaving
a, a given adiabatic saturation temperature fixes the enthalpy of the mixture
The chart plotted in Fig. 12.8 also indicates the human comfort
conditions most agreeable for human well-being. An air cond.tioner should then be able
n 21 an Environment within the comfort zone regardless of the outs.de atmospheric
cond tio is to be considered adequate. Some charts are available that give corrections for
0= 100%
FIGURE 12,9
Processes on a
psychrometrie chart.
Heating
summary i:j 493
s
UMMARY
variation from standard atmospheric pressures. Before using a given chart, one should
fully understand the assumptions made in constructing it and should recognize that it is
applicable to the particular problem at hand.
The direction in which various processes proceed for an air-water vapor mixture
is shown on the psychrometric chart of Fig. 12.9. For example, a constant-pressure
cooling process beginning at state 1 proceeds at constant humidity ratio to the dew
point at state 2, with continued cooling below that temperature moving along the satura-
tion line (100% relative humidity) to point 3. Other processes could be traced out in a
similar manner.
A mixture of gases is treated from the definition of the mixture composition of the various
components based on mass or based on moles. This leads to the mass fractions and mole
fractions, both of which can be called concentrations. The mixture has an overall average
molecular weight and other mixture properties on a mass or mole basis. Another simple
model includes Dalton's model of ideal mixtures of ideal gases, which leads to partial
pressures as the contribution from each component to the total pressure given by the mole
fraction. As entropy is sensitive to pressure, the mole fraction enters into the entropy gen-
eration by mixing. However, for processes other than mixing of different components, we
can treat the mixture as we treat a pure substance by vising the mixture properties.
Special treatment and nomenclature is used for moist air as a mixture of air and
water vapor. The water content is quantified by the relative humidity (how close the water
vapor is to a saturated state) or by the humidity ratio (also called absolute humidity). As
moist air is cooled down, it eventually reaches the dew point (relative humidity is 100%),
where we have saturated moist air. Vaporizing liquid water without external heat transfer
gives an adiabatic saturation process also used in a process called evaporative cooling. In
an actual apparatus, we can obtain wet-bulb and dry-bulb temperatures, indirectly measur-
ing the humidity of the incoming air. These property relations are shown in a psychromet-
ric chart.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to
* Handle the composition of a multicomponent mixture on a mass or mole basis.
* Convert concentrations from a mass to a mole basis and vice versa.
* Compute average properties for the mixture on a mass or mole basis.
* Know partial pressures and how to evaluate them. ' "
* Know how to treat mixture properties such as v, n, h s s, C pmix , M m]x .
* Find entropy generation by a mixing process.
* Formulate the general conservation equations for mass, energy, and entropy for the
case of a mixture instead of a pure substance.
* Know how to use the simplified formulation of the energy equation using the frozen
heat capacities for the mixture.
* Deal with a polytropic process when the substance is a mixture of ideal gases.
* Know the special properties (0, <*>) describing humidity in moist air.
* Have a sense of what changes relative humidity and humidity ratio and recognize
that you can change one and not the other in a given process.
1
494 CHAPTER TWELVE GAS MIXTURES
KEY CONCEPTS Mass concentration
and Formulas
COMPOSITION
Mole concentration
Molecular weight
01 2 yMi
yi ~ " tDl 2 c M
AU t = 2:ftM
Properties
Internal energy
Enthalpy
Gas constant
Heat capacity frozen
Ratio of specific heats
Dalton model
Entropy
Component entropy
Air-Water Mixtures
Relative humidity
Humidity ratio
Enthalpy per kg dry air
"mix = 2 c i u »
^ = 2^",
Rnto = MM*** = 2 c t R i
Cymix = 2 c iCvi>
Cpjmx = 2 C lCpi>
^rak "~ C g mhJCt) mix
J p mix
&
*W = 2 C t ,S i )
5f = if, In [y ; P/P ]
«mix = 2^i = W ^ M ^
/lmix = 2M- = ^ix M mk
Cy m Lx = 2 J^i i
C u mk " mix ~~ ^
^mix = 2 M
12.2 For a mixture how many component concentra-
tions are needed?
12.3 Are any of the properties (P, T, w) for oxygen
and nitrogen in air the same?
12 4 If oxygen is 21% by mole of air, what is the
oxygen state (P, T, v) in a room at 300 K, 100
kPa, of total volume 60 m 3 ?
12,5 A flow of oxygen and one of nitrogen, both 300
K, are mixed to produce 1 kg/s air at 300 K, 1 00
kPa. What are the mass and volume flow rates of
each line?
P <£P E
h = h a + a>h
A flow of gas A and a flow of gas B are mixed in
a 1 : 1 mole ratio with same T. What is the en-
tropy generation per kmole flow out?
12 7 A rigid container has 1 kg argon at 300 K and
1 kg argon at 400 K both at 150 kPa. Now they
are allowed to mix without any external heat
transfer. What is final T, P? Is any s generated?
12 8 A rigid container has 1 kg C0 2 at 300 K and
1 kg argon at 400 K, both at 150 kPa. Now they
are allowed to mix without any heat transfer.
What is final 7, P?
12 9 A flow of 1 kg/s argon at 300 K and another flow
of 1 kg/s C0 2 at 1600 Kboth at 150 kPa are mixed
without any heat transfer. What is the exit T, P?
HOMEWORK PROBLEMS M 495
12.10 What is the rate of entropy increase in Problem
12.9?
12.11 If I want to heat a flow of a 4-component mix-
ture from 300 to 3 10 K at constant P, how many
properties and which ones do I need to know to
find the heat transfer?
12.12 For a gas mixture in a tank, are the partial pres-
sures important?
12.13 What happens to relative and absolute humidity
when moist air is heated?
12.14 I cool moist air; do I reach the dew first in a con-
stant P or constant J 7 process?
12.15 What happens to relative and absolute humidity
when moist air is cooled?
12.16 If I have air at 100 kPa and (a) - 10°C, (b) 45°C,
. and (c) 110°C, what is the maximum absolute
humidity I can have?
12.17 Can moist air below the freezing point, say
— 5°C ( have a dew point?
12.18 Explain in words what the absolute and relative
humidity expresses?
12.19 An adiabatic saturation process changes co,
and T. In which direction?
12.20 I want to bring air at 35°C, $ = 40% to a state
of 25°C, (o - 0.01. Do I need to add or subtract
water?
Homework Problems
Mixture Composition and Properties
12.21 A gas mixture at 120°C and 125 kPa is 50% N 2 ,
30% H 2 0, and 20% 2 on a mole basis. Find the
mass fractions, the mixture gas constant, and the
volume for 5 kg of mixture.
12.22 A mixture of 60% N 2f 30% argon, and 10% 2
on a mass basis is in a cylinder at 250 kPa and
310 K with a volume of 0.5 m 3 . Find the mole
fractions and the mass of argon.
12.23 A mixture of 60% N 2 , 30% Ar, and 1 0% 2 on a
mole basis is in a cylinder at 250 kPa and 310 K
with a volume of 0.5 m 3 . Find the mass fractions
and the mass of argon.
12.24 A new refrigerant R-407 is a mixture of 23%
R-32, 25% R-125, and 52% R-134a on a mass
basis. Find the mole fractions, the mixture gas
constant, and the mixture heat capacities for this
new refrigerant.
12.25 A carbureted internal combustion engine is con-
verted to run on methane gas (natural gas). The
air-fuel ratio in the cylinder is to be 20 to 1 on a
mass basis. How many moles of oxygen per
mole of methane are there in the cylinder?
12.26 Weighing of masses gives a mixture at 60°C,
225 kPa with 0.5 kg 2 , 1.5 kg N 2 , and 0.5 kg
CH 4 . Find the partial pressures of each com-
ponent, the mixture specific volume (mass
basis), mixture molecular weight, and the total
volume.
12.27 A 2-kg mixture of 25% N 2 , 50% 2 , and 25%
C0 2 by mass is at 150 kPa and 300 K. Find the
mixture gas constant and the total volume.
12.28 A 100 m 3 storage tank with fuel gases is at 20°C
and 100 kPa containing a mixture of acetylene
C 2 H 2 , propane C 3 H 8 , and butane C 4 H 10 . A test
shows the partial pressure of the C 2 H 2 is 15 kPa
and that of C 3 H 8 is 65 kPa. How much mass is
there of each component?
12.29 A pipe, of cross-sectional area 0.1 m 2 , carries a
flow of 75% 2 and 25% N 2 by mole with a ve-
locity of 25 m/s at 200 kPa and 290 K. To install
and operate a mass flowmeter, it is necessary to
know the mixture density and the gas constant.
What are they? What mass flow rate should the
meter then show?
12.30 A new refrigerant R-410a is a mixture of R-32
and R-125 in a 1:1 mass ratio. What are the
overall molecular weight, the gas constant, and
the ratio of specific heats for such a mixture?
SimpJe Processes
12.31 At a certain point in a coal gasification process,
a sample of the gas is taken and stored in a 1 L
cylinder. An analysis of the mixture yields the
following results;
Component
H 2
CO
CO,
Percent by mass 2
45
28
25
Determine the mole fractions and total mass in
the cylinder at 100 kPa and 20°C. How much
heat must be transferred to heat the sample at
constant volume from the initial state to 100°C?
12 32 The mixture in Problem 12.27 is heated to 500 K
with constant volume. Find the final pressure and
the total heat transfer needed using Table A.5.
12.33 The mixture in problem 12.27 is heated up to
500 K in a constant-pressure process. Find the
final volume and the total heat transfer using
Table A.5.
12 34 A pipe flows 1.5 kg/s of a mixture with mass
fractions of 40% C0 2 and 60% N 2 at 400 kPa
and 300 K, shown in Fig. P12.34. Heating tape
is wrapped around a section of pipe with in-
sulation added, and 2 kW of electrical power is
heating the pipe flow. Find the mixture exit
temperature.
12.35
12.36
12.37
12.38
FIGURE P12.34
A pipe flows 0.05 kmol/s of a mixture with mole
fractions of 40% C0 2 and 60% N 2 at 400 kPa,
300 K. Heating tape is wrapped around a section
of pipe with insulation added, and 2 kW electri-
cal power is heating the pipe flow. Find the mix-
ture exit temperature.
A rigid insulated vessel contains 12 kg of oxy-
gen at 200 kPa and 280 K separated by a mem-
brane from 26 kg of carbon dioxide at 400 kPa
and 360 K. The membrane is removed, and the
mixture comes to a uniform state. Find the final
temperature and pressure of the mixture.
A mixture of 40% water and 60% carbon diox-
ide by mass is heated from 400 K to 1000 K at a
constant pressure of 120 kPa. Find the total
change in enthalpy and entropy using Table A.5
values.
Do Problem 12.37 but with variable heat capac-
ity using values from Table A. 8.
12.39 An insulated gas turbine receives a mixture of
10% C0 2 , 10% H 2 0, and 80% N 2 on a mass
basis at 1000 K and 500 kPa. The inlet volume
flow rate is 2 mVs, and the exhaust is at 700 K
and 100 kPa. Find the power output in kW using
constant specific heat from Table A.5 at 300 K.
12.40 Solve Problem 12.39 using values of enthalpy
from Table A.8.
12.41 An insulated gas turbine receives a mixture of
10% C0 2 , 10% H 2 0, and 80% N 2 on a mole
basis at 1000 K, 500 kPa. The inlet volume flow
rate is 2 mVs, and the exhaust is at 700 K, 100
kPa. Find the power output in kW using constant
specific heat from A.5 at 300 K.
12.42 Solve Problem 12.41 using values of enthalpy
from Table A.9.
12.43 A piston/cylinder device contains 0.1 kg of a
mixture of 40% methane and 60% propane by
mass at 300 K and 100 kPa. The gas is now
slowly compressed in an isothermal (T = con-
stant) process to a final pressure of 250 kPa.
Show the process in a P-V diagram and find
both the work and heat transfer in the process.
12.44 Consider Problem 12.39 and find the value for
the mixture heat capacity, mass basis, and the
mixture ratio of specific heats, fc mLx , both esti-
mated at 850 K from values (differences) of ft in
Table A.8. With these values make an estimate
for the reversible adiabatic exit temperature of
the turbine at 100 kPa.
12.45 Consider Problem 12.41 and find the value for
the mixture heat capacity, mole basis, and the
mixture ratio of specific heats,
^mix> both est i
mated at 850 K from values (differences) of h in
Table A.9. With these values make an estimate
for the reversible adiabatic exit temperature of
the turbine at 100 kPa.
12 46 A mixture of 0.5 kg of nitrogen and 0.5 kg of
oxygen is at 100 kPa and 300 K in a piston
cylinder keeping constant pressure. Now 800 kJ
is added by heating. Find the final temperature
and the increase in entropy of the mixture using
Table A.5 values.
12.47 Repeat Problem 12.46, but solve using values
from Table A.8.
12.48 New refrigerant R-4 10a is a mixture of R-32 and
R-125 in a 1 : 1 mass ratio. A process brings 0.5
Homework Problems H 497
kg R-410a from 270 K to 320 K at a constant
pressure of 250 kPa in a piston cylinder. Find
the work and heat transfer.
12.49 A piston cylinder contains 0.5 kg of argon and
0.5 kg of hydrogen at 300 K and 100 kPa. The
mixture is compressed in an adiabatic process to
400 kPa by an external force on the piston. Find
the final temperature, the work, and the heat
transfer in the process.
12.50 Natural gas as a mixture of 75% methane and
25% ethane by mass is flowing to a compressor
at 17°C and 100 kPa. The reversible adiabatic
compressor brings the flow to 250 kPa. Find the
exit temperature and the needed work per kg
flow.
12.51 A mixture of 2 kg of oxygen and 2 kg of argon
is in an insulated piston-cylinder arrangement at
100 kPa and 300 K. The piston now compresses
the mixture to half its initial volume. Find the
final pressure, final temperature, and the piston
work.
12.52 The substance R-410a (see Problem 12.48) is at
100 kPa and 290 K. It is now brought to 250 kPa
and 400 K in a reversible polytropic process.
Find the change in specific volume, specific en-
thalpy, and specific entropy for the process,
12.53 Two insulated tanks A and B are connected by a
valve, shown in Fig. P 12.53. Tank A has a vol-
ume of 1 m 3 and initially contains argon at 300
kPa and 10°C. Tank B has a volume of 2 m 3 and
initially contains ethane at 200 kPa and 50°C.
The valve is opened and remains open until the
resulting gas mixture comes to a uniform state.
Determine the final pressure and temperature.
FIGURE P12.53
12.54 A compressor brings R-410a (see Problem
12.48) from -10°C and 125 kPa up to 500 kPa
in an adiabatic reversible compression. Assume
ideal-gas behavior and find the exit temperature
and the specific work.
12.55 A mixture of 50% carbon dioxide and 50%
. water by mass is brought from 1500 K. and 1
MPa to 500 K and 200 kPa in a polytropic
process through a steady-state device. Find the
necessary heat transfer and work involved using
values from Table A.5.
12.56 Solve problem 12.55 using specific heats C p =
A/j/Arfrom Table A.8 at 1000 K.
12.57 A 50/50 (by mass) gas mixture of methane CH 4
and ethylene C 2 H 4 is contained in a cylinder pis-
ton at the initial state of 480 kPa, 330 K, and
1 .05 m 3 . The piston is now moved, compressing
the mixture in a reversible, polytropic process
to the final state of 260 K and 0.03 m 3 . Calculate
the final pressure, the polytropic exponent, the
work and heat transfer, and entropy change for
the mixture.
12.58 The gas mixture from Problem 12.31 is com-
pressed in a reversible adiabatic process from
the initial state in the sample cylinder to a vol-
ume of 0,2 L. Determine the final temperature of
the mixture and the work done during the
process.
Entropy Generation
12.59 A flow of 2 kg/s mixture of 50% C0 2 and 50%
2 by mass is heated in a constant-pressure heat
exchanger from 400 K to 1000 K by a radiation
source at 1400 K. Find the rate of heat transfer
and the entropy generation in the process shown
in Fig.P12.59.
\ 1400 K /
FIGURE P12.59
12.60 Carbon dioxide gas at 320 K is mixed with ni-
trogen at 280 K in an insulated mixing chamber.
Both flows are at 100 kPa, and the mass ratio of
carbon dioxide to nitrogen is 2: 1. Find the exit
temperature and the total entropy generation per
kg of the exit mixture.
498
CHAPTER TWELVE GAS MIXTURES
12.61
12.62
12.63
12.64
12.65
Repeat Problem 12.60 with inlet temperatures of
1400 K for the carbon dioxide and 300 K for the
nitrogen. First estimate the exit temperature with
the specific heats from Table A.5 and use this to
start iterations with values from Table A.8.
Carbon dioxide gas at 320 K is mixed with ni-
trogen at 280 K in an insulated mixing chamber.
Both flows are coming in at 100 kPa and the
mole ratio of carbon dioxide to nitrogen is 2: 1.
Find the exit temperature and the total entropy
generation per kmole of the exit mixture.
Repeat Problem 12.62 with inlet temperature of
1400 K for the carbon dioxide and 300 K for the
nitrogen. First estimate the exit temperature with
the specific heats from Table A.5 and use this to
start iterations with values from A.9.
The only known sources of helium are the
atmosphere (mole fraction approximately
5 X 10~ 6 ) and natural gas. A large unit is being
constructed to separate 100 m 3 /s of natural gas,
assumed to be 0.001 He mole fraction and 0.999
CH 4 . The gas enters the unit at 150 kPa, 10°C.
Pure helium exits at 100 kPa, 20°C, and pure
methane exits at 150 kPa, 30°C. Any heat trans-
fer is with the surroundings at 20°C. Is an elec-
trical power input of 3000 kW sufficient to drive
this unit?
A flow of 1 kg/s carbon dioxide at 1600 K, 100
kPa is mixed with a flow of 2 kg/s water at 800
K, 100 kPa, and after the mixing it goes through
a heat exchanger where it is cooled to 500 K by
a 400 K ambient. How much heat transfer is
taken out in the heat exchanger? What is the en-
tropy generation rate for the whole process?
12.67 Repeat Problem 12.50 for an isentropic com-
pressor efficiency of 82%.
12.68 A large air separation plant takes in ambient air
(79% N 2) 21% 2 by mole) at 100 kPa and 20°C
at a rate of 25 kg/s. It discharges a stream of
pure 2 gas at 200 kPa and 100°C and a stream
of pure N 2 gas at 100 kPa and 20°C. The plant
operates on an electrical power input of 2000
kW, shown in Fig. P12.68. Calculate the net rate
of entropy change for the process.
2000 kW
12.66
H 2 Cr i
FIGURE P12.65
A mixture of 60% helium and 40% nitrogen by
mass enters a turbine at 1 MPa and 800 K at a
rate of 2 kg/s. The adiabatic turbine has an exit
pressure of 100 kPa and an isentropic efficiency
of 85%. Find the turbine work.
12.69
12.70
12.71
12.72
Air-
V
f
n
1
o n o □ □
n
— I —
o nun □
■ o 2
FIGURE P12.68
A steady flow of 0.3 kg/s of 50% carbon dioxide
and 50% water by mass at 1200 K and 200 kPa
is used in a heat exchanger where 300 kW is ex-
tracted from the flow. Find the flow exit temper-
ature and the rate of change of entropy using
Table A.8.
A steady flow of 0.01 kmol/s of 50% carbon
dioxide and 50% water on a mole basis at 1200
K and 200 kPa is used in a heat exchanger where
300 kW is extracted from the flow. Find the flow
exit temperature and the rate of change of en-
tropy using Table A.9.
A flow of 1.8 kg/s steam at 400 kPa, 400°C, is
mixed with 3.2 kg/s oxygen at 400 kPa, 400 K,
in a steady flow mixing-chamber without any
heat transfer. Find the exit temperature and the
rate of entropy generation.
A tank has two sides initially separated by a di-
aphragm, shown in Fig. PI 2.72. Side A con-
tains 1 kg of water and side B contains 1.2 kg
of air, both at 20°C and 10.0 kPa. The di-
aphragm is now broken, and the whole tank is
heated to 600°C by a 700°C reservoir. Find the
final total pressure, heat transfer, and total en-
tropy generation.
homework problems
b 499
FIGURE P12.72
12.73 Three steady flows are mixed in an adiabatic
chamber at 150 kPa. Flow one is 2 kg/s of 2 at
340 K, flow two is 4 kg/s of N 2 at 280 K, and
flow three is 3 kg/s of C0 2 at 310 K. All flows
are at 150 kPa, the same as the total exit pres-
sure. Find the exit temperature and the rate of
entropy generation in the process.
12.74 Reconsider Problem 12.53, but let the tanks
have a small amount of heat transfer so the final
mixture is at 400 K. Find the final pressure, the
heat transfer, and the entropy change for the
process.
Air-Water Vapor Mixtures
12.75 Atmospheric air is at 100 kPa and 25°C with a
relative humidity of 75%. Find the absolute hu-
midity and the dew point of the mixture. If the
mixture is heated to 30°C, what is the new rela-
tive humidity?
12.76 Consider 100 m 3 of atmospheric air, which is an
air-water vapor mixture at 100 kPa, 15°C, and
40% relative humidity. Find the mass of water
and the humidity ratio. What is the dew point of
the mixture?
12.77 The products of combustion are flowing through
a heat exchanger with 12% C0 2l 13% H 2 0, and
75% N 2 on a volume basis at the rate 0.1 kg/s
and 100 kPa. What is the dew-point tempera-
ture? If the mixture is cooled 10°C below the
dew-point temperature, how long will it take to '
collect 10 kg of liquid water?
12.78 A 1 kg/s flow of saturated moist air (relative hu-
midity 100%) at 100 kPa and 10°C goes through
a heat exchanger and comes out at 25°C. What is
the exit relative humidity and how much power
is needed?
12.79 A new high-efficiency home heating system in-
cludes an air-to-air heat exchanger, which uses
energy from outgoing stale air to heat the fresh
incoming air. If the outside ambient temperature
is -10°C and the relative humidity is 30%, how
much water will have to be added to the incom-
ing air, if it flows in at the rate of 1 m 3 /s and
must eventually be conditioned to 20°C and 40%
relative humidity?
12.80 Consider a 1 m 3 /s flow of atmospheric air at 100
kPa, 25°C, and 80% relative humidity. Assume
this flows into a basement room where it cools
to 15°C at 100 kPa. How much liquid water will
condense out?
12.81 A 2 kg/s flow of completely dry air at T x and 100
kPa is cooled down to 10°C by spraying liquid
water at 10°C and 100 kPa into it so it becomes
saturated moist air at 10°C. The process is
steady state with no external heat transfer or
work. Find the exit moist air humidity ratio and
the flow rate of liquid water. Find also the dry
air inlet temperature T h
12.82 A piston cylinder has 100 kg of saturated moist
air at 100 kPa and 5°C. If it is heated to 45°C in
an isobaric process, find x g 2 and the final relative
humidity. If it is compressed from the initial
state to 200 kPa in an isothermal process, find
the mass of water condensing.
12.83 A saturated air-water vapor mixture at 20°C,
100 kPa, is contained in a 5-m 3 closed tank in
equilibrium with 1 kg of liquid water. The tank
is heated to 80°C. Is there any liquid water at the
final state? Find the heat transfer for the process.
12.84 Ambient moist air enters a steady-flow air-
conditioning unit at 102 kPa and 30°C with a
60% relative humidity. The volume flow rate en-
tering the unit is 100 L/s. The moist air leaves
the unit at 95 kPa and 15°C with a relative hu-
midity of 100%. Liquid condensate also leaves
the unit at 15°C. Determine the rate of heat
• transfer for this process.
12.85 Consider at 500 L rigid tank containing an
air- water vapor mixture at 100 kPa and 35°C
with a 70% relative humidity. The system is
cooled until the water just begins to condense.
Determine the final temperature in the tank and
the heat transfer for the process.
12.86 Air in a piston cylinder is at 35°C, 100 kPa, and
a relative humidity of 80%. It is now com-
pressed to a pressure of 500 kPa in a constant-
temperature process. Find the final relative and
specific humidity and the volume ratio V 2 IV X ,
L
500 S Chapter Twelve Gas Mixtures
12.87 A 300 L rigid vessel initially contains moist air
at 150 kPa and 40°C with a relative humidity of
10% A supply line connected to this vessel by a
valve carries steam at 600 kPa and 200°C. The
valve is opened, and steam flows into the vessel
until the relative humidity of the resultant moist
air mixture is 90%. Then the valve is closed.
Sufficient heat is transferred from the vessel so
that the temperature remains at 40°C during the
process. Determine the heat transfer for the
process, the mass of steam entering the vessel,
and the final pressure inside the vessel.
12 88 A rigid container, 10 m 3 in volume, contains
moist air at 45°C and 100 kPa with 3> = 40%.
The container is now cooled to 5°C. Neglect the
volume of any liquid that might be present and
find the final mass of water vapor, final total
pressure, and the heat transfer.
12.89 A water-filled reactor of 1 m 3 is at 20 MPa,
360°C and is located inside an insulated contain-
ment room of 100 m 3 that contains air at 100
kPa and 25°C. Due to a failure, the reactor rup-
tures and the water fills the containment room.
Find the final quality and pressure by iterations.
Tables and Formulas or Psychrometric Chart
12.90 A flow of moist air at 100 kPa, 40°C, and 40%
relative humidity is cooled to 15°C in a con-
stant-pressure device. Find the humidity ratio of
the inlet and the exit flow and the heat transfer in
the device per kg dry air.
12.91 A flow, 0.2 kg/s dry air, of moist air at 40°C and
50% relative humidity flows from the outside
state 1 down into a basement where it cools to
16°C, state 2. Then it flows up to the living room
where it is heated to 25°C, state 3. Find the dew
point for state 1, any amount of liquid that may
appear, the heat transfer that takes place in the
basement, and the relative humidity in the living
room at state 3.
12.92 Two moist air streams with 85% relative humid-
ity, both flowing at a rate of 0. 1 kg/s of dry air,
are mixed in a steady-flow setup. One inlet
stream is at 32.5°C and the other at 16°C, Find
the exit relative humidity.
12.93 The discharge moist air from a clothes dryer is
at 35°C, 80% relative humidity. The flow is
guided through a pipe up through the roof and a
vent to the atmosphere shown in Fig. PI 2.93.
Due to heat transfer in the pipe, the flow is
cooled to 24°C by the time it reaches the vent.
Find the humidity ratio in the flow out of the
clothes dryer and at the vent. Find the heat trans-
fer and any amount of liquid that may be form-
ing per kg dry air for the flow.
FIGURE P12.93
12.94 A steady supply of 1.0 m 3 /s air at 25°C, 100 kPa,
and 50% relative humidity is needed to heat a
building in the winter. The ambient outdoors is
at 10°C, 100 kPa, and 50% relative humidity.
What are the required liquid water input and
heat transfer rates for this purpose?
12.95 A combination air cooler and dehumidification
unit receives outside ambient air at 35°C, 100
kPa, and 90% relative humidity. The moist air is
first cooled to a low temperature T 2 to condense
the proper amount of water; assume all the liq-
uid leaves at T 2 . The moist air is then heated and
leaves the unit at 20°C, 100 kPa, and 30% rela-
tive humidity with a volume flow rate of 0.01
m 3 /s. Find the temperature T 2i the mass of liquid
per kilogram of dry air, and the overall heat
transfer rate.
12.96 Use the formulas and the steam tables to find the
missing property of: *, to, and T A[y> for a total
pressure of 100 kPa; repeat the answers using
the psychrometric chart.
a. * = 50%, to = 0.010
b. r <t[y = 25°c J r^ = 2i°c
12 97 An insulated tank has an air inlet, a>i = 0.0084
and an outlet, T 2 - 22°C, <& 2 - 90%, both at 100
kPa. A third line sprays 0.25 kg/s of water at
Homework problems M 501
12.98
12.99
80°C and 100 kPa, as shown in Fig. P12.97. For
steady operation, find the outlet specific humid-
ity, the mass flow rate of air needed, and the re-
quired air inlet temperature, T\.
FIGURE P12.97
A flow of moist air from a domestic furnace,
state I, is at 45°C, 10% relative humidity with a
flow rate of 0.05 kg/s dry air. A small electric
heater adds steam at 100°C, 100 kPa, generated
from tap water at 15°C shown in Fig. P12.98. Up
tn the living room, the flow comes out at state 4:
30°C, 60% relative humidity. Find the power
needed for the electric heater and the heat trans-
fer to the flow from state 1 to state 4.
3
FIGURE P12.98
A water-cooling tower for a power plant cools
45°C liquid water by evaporation. The tower re-
ceives air at 19.5°C, $ = 30%, and 100 kPa that
is blown through/over the water such that it
leaves the tower at 25°C and <t> = 70%. The re-
maining liquid water flows back to the con-
denser at 30°C having given off 1 MW. Find the
mass flow rate of air, and determine the amount
of water that evaporates.
12.100 A flow of air at 5°C, <£> - 90%, is brought into a
' house, where it is conditioned to 25°C, 60% rel-
ative humidity. This is done with a combined
heater-evaporator where any liquid water is at
10°C. Find any flow of liquid and the necessary
heat transfer, both per kilogram dry air flowing.
Find the dew point for the final mixture.
12.101 In a car's defrost/defog system atmospheric air
at 21°C and 80% relative humidity is taken in
and cooled such that liquid water drips out. The
now dryer air is heated to 41°C and then blown
onto the windshield, where it should have a
maximum of 10% relative humidity to remove
water from the windshield. Find the dew point
of the atmospheric air, specific humidity of air
onto the windshield, the lowest temperature, and
the specific heat transfer in the cooler.
12.102 Atmospheric air at 35°C with a relative humidity
of 10%, is too warm and also too dry. An air
conditioner should deliver air at 21°C and 50%
relative humidity in the amount of 3600 m 3 /h.
Sketch a setup to accomplish this. Find any
amount of liquid (at 20°C) that is needed or dis-
carded and any heat transfer.
12.103 One means of air-conditioning hot summer air is
by evaporative cooling, which is a process simi-
lar to the adiabatic saturation process. Consider
outdoor ambient air at 35°C, 100 kPa, 30% rela-
tive humidity. What is the maximum amount of
cooling that can be achieved by such a tech-
nique? What disadvantage is there to this ap-
proach? Solve the problem using a first law
analysis and repeat it using the psychrometric
chart, Fig. E.4.
12.104 A flow of moist air at 45°C, 10% relative humid-
ity with a flow rate of 0.2 kg/s dry air is mixed
with a flow of moist air at 25°C, and absolute
humidity of <w = 0.018 with a rate of 0.3 kg/s
dry air. The mixing takes place in an air duct at
100 kPa, and there is no significant heat transfer.
After the mixing, there is heat transfer to a final
temperature of 40°C. Find the temperature and
relative humidity after mixing. Find the heat
transfer and the final exit relative humidify.
12.105 An indoor pool evaporates 1.512 kg/h of water,
which is removed by a dehumidifier to maintain
21°C, <J> = 70% in the room. The dehumidifier,
shown in Fig, P12.105, is a refrigeration cycle in
502 H CHAPTER TWELVE GAS MIXTURES
which air flowing over the evaporator cools such
that liquid water drops out, and the air continues
flowing over the condenser. For an airflow rate
of 0.1 kg/s the unit requires 1.4 kW input to a
motor driving a fan and the compressor, and it
has a coefficient of performance, J3 = QJW C = 12
2.0. Find the state of the air as it returns to the
room and the compressor work input.
Evaporator Vaive Condenser
FIGURE P12.105
Psychrometric Chart Only
12.106 Use the psychrometric chart to find the missing
property of: <!>, a>, T^ £li r d[y .
a. = 25°C, <f> = 80%
b. 7^ = 15°C, $ = 100%
c. = 20°C, to = 0.008
d. = 25°C, T va = 23°C
12.107 Use the psychrometric chart to find the missing
property of: 4>, 7^, T d[y .
a. <t> = 50%, (o = 0.012
b. r wet = 15°C, 4> = 60%
c. a> = 0.008, 7^ - 17°C
d. = 10°C, w = 0.006
12.108 Use the formulas and the steam tables to find the
missing property of: o>, and T dr/) total pres^
sure is 100 kPa. Repeat the answers using the
psychrometric chart.
a. # = 50%, to = 0.010
b. T w , t = 15°C, S> = 50%
c. r diy = 25°c, r« t = 2rc
12.109 For each of the states in Problem 12.107 find the
dew-point temperature.
12.110 Compare the weather in two places where it is
cloudy and breezy. At beach A the temperature
is 20°C J the pressure is 103.5 kPa, and the rela-
tive humidity is 90%; beach B has 25°C, 99 kPa,
and 20% relative humidity. Suppose you just
took a swim and came out of the water. Where
would you feel more comfortable, and why?
.111 Ambient air at 100 kPa, 30°C, and 40% relative
humidity goes through a constant-pressure heat
exchanger as a steady flow. In one case it is
heated to 45°C, and in another case it is cooled
until it reaches saturation. For both cases find
the exit relative humidity and the amount of heat
transfer per kilogram of dry air.
12,112 A flow of moist air at 21°C with 60% relative hu-
midity should be produced from mixing two dif-
ferent moist airflows. Flow 1 is at 10°C and 80%
relative humidity; flow 2 is at 32°C and has
j" wet = 27°C. The mixing chamber can be fol-
lowed by a heater or a cooler, as shown in Fig.
PI 2.1 12. No liquid water is added, and P = 100
kPa. Find the two controls— one is the ratio of the
two mass flow rates m a i/m a2 and the other is the
heat transfer in the heater/cooler per kg dry air.
FIGURE P12.112
12.113 In a hot and dry climate, air enters an air-condi-
tioner unit at 100 kPa, 40°C, and 5% relative hu-
midity, at the steady rate of 1.0 m 3 /s. Liquid
water at 20°C is sprayed into the air in the AC
unit at the rate of 20 kg/h, and heat is rejected
from the unit at the rate 20 kW. The exit pres-
sure is 100 kPa. What are the exit temperature
and relative humidity?
12.114 Consider two states of atmospheric air. (1) 35°C,
r wt = 18°C and (2) 26.5°C, <I> - 60%. Suggest a
system of devices that will allow air in a steady
flow to change from (1) to (2) and from (2) to
(1). Heaters, coolers, (de)humidifiers, liquid
traps, and the like are available, and any
liquid/solid flowing is assumed to be at the low-
est temperature seen in the process. Find the
homework Problems M 503
specific and relative humidity for state 1, dew
point for state 2, and the heat transfer per kilo-
gram dry air in each component in the systems.
12.115 To refresh air in a room, a counterflow heat ex-
changer, see Fig. P12.115, is mounted in the
wall, drawing in outside air at 0.5°C, 80% rela-
tive humidity, and pushing out room air, 40°C,
50% relative humidity. Assume an exchange of
3 kg/min dry air in a steady flow, and also as-
sume that the room air exits the heat exchanger
to the atmosphere at 23°C. Find the net amount
of water removed from the room, any liquid
flow in the heat exchanger, and (J 1 , <£) for the
fresh air entering the room.
Outside
air
© air
Review Problems
12.119 A piston/cylinder contains helium at 1 10 kPa at
ambient temperature 20°C, and initial volume of
20 L as shown in Fig. P12.119. The stops are
mounted to give a maximum volume of 25 L,
and the nitrogen line conditions are 300 kPa,
30°C. The valve is now opened, which allows
nitrogen to flow in and mix with the helium. The
valve is closed when the pressure inside reaches
200 kPa, at which point the temperature inside is
40°C. Is this process consistent with the second
law of thermodynamics?
FIGURE P12.115
Availability (Exergy) in Mixtures
12.116 Consider the mixing of a steam flow with an
oxygen flow in Problem 12.71. Find the rate of
total inflowing availability and the rate of exergy
destruction in the process.
12.117 A mixture of 75% carbon dioxide and 25%
water by mass is flowing at 1600 K, 100 kPa
into a heat exchanger where it is used to deliver
energy to a heat engine. The mixture leaves the
heat exchanger at 500 K with a mass flow rate of
2 kg/min. Find the rate of energy and the rate of
exergy delivered to the heat engine.
12.118 Find the second-law efficiency of the heat ex-
changer in Problem 12.59.
He
N 2
line
FIGURE P12.119
12.120 A spherical balloon has an initial diameter of
1 m and contains argon gas at 200 kPa, 40°C.
The balloon is connected by a valve to a 500-L
rigid tank containing carbon dioxide at 100 kPa,
100°C. The valve is opened, and eventually the
balloon and tank reach a uniform state m which
the pressure is 185 kPa. The balloon pressure is
directly proportional to its diameter. Take the
balloon and tank as a control volume, and calcu-
late the final temperature and the heat transfer
for the process.
12.121 An insulated vertical cylinder is fitted with a
frictionless constant loaded piston of cross-
sectional area 0.1 m 2 and the initial cylinder
height of 1 .0 m. The cylinder contains methane
gas at 300 K, 150 kPa, and also inside is a 5-L
capsule containing neon gas at 300 K, 500 kPa
shown in Fig. P 12. 121. The capsule now breaks,
and the two gases mix together in a constant-
pressure process. What is the final temperature,
final cylinder height, and net entropy change for
the process?
504 H CHAPTER TWELVE GAS MIXTURES
rrr
mp
i I
FIGURE P12.121
.122 An insulated rigid 2 m 3 tank^ contains C0 2 gas at
200°C ) 1 MPa. An uninsulated rigid 1 m 3 tank B
contains ethane, QF^, gas at 200 kPa, Toom tem-
perature 20°C. The two are connected by a one-
way check valve that will allow gas from A to B i
but not from B to A shown in Fig. PI2.122. The
valve is opened and gas flows from A to B until
the pressure in B reaches 500 kPa when the valve
is closed. The mixture in B is kept at room temper-
ature due to heat transfer. Find the total number of
moles and the ethane mole fraction at the final
state in B, Find the final temperature and pressure
in tank ,4 and the heat transfer, to/from tank B.
12,124 An air-water vapor mixture enters a steady-flow
heater humidifier unit at state 1: 10°C, 10% rela-
tive humidity, at the rate of 1 mVs. A second
air-vapor stream enters the unit at state 2: 20°C,
20% relative humidity, at the rate of
2 mVs. Liquid water enters at state 3: 10°C, at
the rate of 400 kg per hour. A single air-vapor
flow exits the unit at state 4: 40°C shown in Fig.
P 12. 124. Calculate the relative humidity of the
exit flow and the rate of heat transfer to the unit.
FIGURE P12.124
FIGURE P12.122
.123 A 0.2 m 3 insulated, rigid vessel is divided into
two equal parts A and B by an insulated partition,
as shown in Fig. P. 12.123. The partition will sup-
port a pressure difference of 400 kPa before
breaking. Side A contains methane and side B
contains carbon dioxide. Both sides are initially at
1 MPa, 30°C. A valve on side B is opened, and
carbon dioxide flows out. The carbon dioxide that
remains in B is assumed to undergo a reversible
adiabatic expansion while there is flow out. Even-
tually the partition breaks, and the valve is closed.
Calculate the net entropy change for the process
that begins when the valve is closed.
< ///////////////////////////////{
A
CH 4
ccv
? /////¥///;//////;///;/;/////;// /
FIGURE P12.123
1 2.125 You have just washed your hair and now blow dry
it in a room with 23°C, $ - 60%, (1). The dryer,
500 W, heats the air to 49°C, (2), blows it through
your hair where the air becomes saturated (3), and
then flows on to hit a window where it cools to
15°C (4). Find the relative humidity at state 2, the
heat transfer per kilogram of dry air in the dryer,
the airflow rate, and the amount of water con-
densed on the window, if any.
12.126 Steam power plants often utilize large cooling
towers to cool the condenser cooling water so it
can be recirculated; see Fig. P12.126. The
process is essentially evaporative adiabatic
cooling, in which part of the water is lost and
must therefore be replenished. Consider the
setup shown in Fig. P12.126, in which 1000
kg/s of warm water at 32°C from the condenser
enters the top of the cooling tower and the
cooled water leaves the bottom at 20°C. The
moist ambient air enters the bottom at 100 kPa,
dry-bulb temperature of 18°C and a wet-bulb
temperature of 10°C. The moist air leaves the
tower at 95 kPa, 30°C, and relative humidity of
85%. Determine the required mass flow rate of
dry air, and the fraction of the incoming water
that evaporates and is lost.
Homework Problems I! 505
Air in
FIGURE P12.126
12.127 Ambient air is at a condition of 100 kPa, 35°C,
50% relative humidity. A steady stream of air at
100 kPa, 23°C, 70% relative humidity is to be
produced by first cooling one stream to an appro-
priate temperature to condense out the proper
amount of water and then mix this stream adia-
batically with the second one at ambient condi-
tions. What is the ratio of the two flow rates? To
what temperature must the first stream be cooled?
12.128 A semipermeable membrane is used for the par-
tial removal of oxygen from air that is blown
through a grain elevator storage facility. Ambi-
ent air (79% nitrogen, 21% oxygen on a mole
basis) is compressed to an appropriate pressure,
cooled to ambient temperature 25°C, and then
fed through a bundle of hollow polymer fibers
that selectively absorb oxygen, so the mixture
leaving at 120 kPa, 25°C, contains only 5% oxy-
gen, shown in Fig. P12.128. The absorbed oxy-
gen is bled off through the fiber walls at 40 kPa,
25°C, to a vacuum pump. Assume the process to
0.79 N 2 ] ■ — 2 0.79 N 2
+0.21 2 ' - - 1 *" + ? 2
Will
FIGURE P12.128
he reversible and adiabatic and determine the
minimum inlet air pressure to the fiber bundle.
12.129 A dehumidifier receives a flow of 0.25 kg/s
moist air at 2S°C, 80% relative humidity as
shown in Figure P12.105. It is cooled down to
20°C as it flows over the evaporator and then
heated up again as it flows over the condenser.
The standard refrigeration cycle uses R-22 with
an evaporator temperature of 5°C and a conden-
sation pressure of 1600 kPa. Find the amount of
liquid water removed and the heat transfer in the
cooling process. How much compressor work is
needed? What is the final air exit temperature
and relative humidity?
12.130 A 100-L insulated tank contains N 2 gas at 200
kPa and ambient temperature 25°C. The tank is
connected by a valve to a supply line flowing
C0 2 at 1.2 MPa, 90°C. A mixture of 50% N 2 ,
50% C0 2 by mole should be obtained by open-
ing the valve and allowing C0 2 to flow in until
an appropriate pressure is reached, when the
valve is closed. What is the pressure? The tank
eventually cools to ambient temperature. Find
the net entropy change for the overall process.
12.131 A cylinder/piston loaded with a linear spring
contains saturated moist air at 120 kPa, 0.1 m 3
volume and also 0.01 kg of liquid water, all at
ambient temperature 20°C. The piston area is
0.2 m 2 t and the spring constant is 20 kN/m.
This cylinder is attached by a valve to a line
flowing dry air at 800 kPa, 80°C. The valve is
opened, and air flows into the cylinder until
the pressure reaches 200 kPa, at which point
the temperature is 40°C. Determine the relative
humidity at the final state, the mass of air en-
tering the cylinder, and the work done during
the process.
12.132 Consider the previous problem and additionally
determine the heat transfer. Show that the
process does not violate the second law.
12.133 The air-conditioning by evaporative cooling in
Problem 12.103 is modified by adding a dehu-
midification process before the water spray cool-
ing process. This dehumidification is achieved
as shown in Fig. P12.I33 by using a desiccant
material, which absorbs water on one side of a
rotating drum heat exchanger. The desiccant is
regenerated by heating on the other side of the
506 CHAPTER TWELVE GAS MIXTURES
Exhaust <3-
Rotary- Drum
dehumidifier
T, = 35°C
^=0.30
H 2 in
vVW^
Insulated
heat exchanger
1
Evaporative
cooler
Evaporative
cooler
14*
Air
to room
Return
air
H 2 in
T 2 = 60°C
T B = 80 a C
T 3 = 25°C
T 6 = 20°C
T 4 = 20°C
r g = 25°C
©5= ©4
FIGURE P12.133
dram to drive the water out. The pressure is 100
kPa everywhere, and other properties are on the
diagram. Calculate the relative humidity of the
cool air supplied to the room at state 4, and the
heat transfer per unit mass of air that needs to be
supplied to the heater unit.
English unit problems
Concept Problems
12.134E If oxygen is 21% by mole of air, what is the
oxygen state (P, T, v) in a room at 540 R, 15
psia of total volume 2000 ft 3 ?
12.135E A flow of oxygen and one of nitrogen, both
540 R, are mixed to produce 1 Ibm/s air at
540 R, 15 psia. What are the mass and volume
flow rates of each line?
12.136E A flow of gas A and a flow of gas B are mixed
in a 1 : 1 mole ratio with same T. What is the
entropy generation per kmole flow out?
12.137E A rigid container has 1 Ibm argon at 540 R and
1 lbm argon at 720 R, both at 20 psia. Now they
are allowed to mix without any external heat
transfer. What is final T, P? Is any s generated?
12.138E A rigid container has 1 lbm C0 2 at 540 R and
1 lbm argon at 720 R, both at 20 psia. Now
they are allowed to mix without any heat trans-
fer. What is final T, PI
12.139E A flow of I lbm/s argon at 540 R and another
flow of 1 lbm/s C0 2 at 2800 R, both at 20 psia,
are mixed without any heat transfer. What is
the exit T, P?
12.140E What is the rate of entropy increase in Problem
12.139?
12.141E If I have air at 14.7 psia and (a) 15 F, (b) 115
F, and (c) 230 F, what is the maximum ab-
solute humidity I can have?
English Unit Problems
12.142E A gas mixture at 250 F, 18 lbf/in. 2 is 50% N 2 ,
30% H 2 0, and 20% 2 on a mole basis. Find
the mass fractions, the mixture gas constant,
and the volume for 10 lbm of mixture.
12.143E Weighing of masses gives a mixture at 80 F,
35 lbf/in 2 with 1 lbm 2l 3 lbm N 2 , and 1 lbm
CH 4 . Find the partial pressures of each compo-
nent, the mixture specific volume (mass basis),
the mixture molecular weight, and the total
volume.
12.144E A new refrigerant R-410a is a mixture of R-32
and R-125 in a 1:1 mass ratio. What is the
overall molecular weight, the gas constant, and
the ratio of specific heats for such a mixture?
12.145E A pipe flows 1.5 lbm/s of a mixture with mass
fractions of 40% C0 2 and 60% N 2 at 60
lbf/in. 2 , 540 R. Heating tape is wrapped around
a section of pipe with insulation added, and 2
Btu/s electrical power is heating the pipe flow.
Find the mixture exit temperature.
ENGLISH UNIT PROBLEMS B 507
12.146E An insulated gas turbine receives a mixture of
10% C0 2) 10% H 2 0, and 80% N 2 on a mass
basis at 1800 R, 75 lbf/in. 2 . The inlet volume
flow rate is 70 ft 3 /s, and the exhaust is at 1300
R, 15 lbf/in. 2 . Find the power output in Btu/s
using constant specific heat from F4 at 540 R.
12.147E Solve Problem 12.146 using the values of en- 12.155E
thalpy from Table F.6.
12.148E A piston cylinder device contains 0.3 lbm of a
mixture of 40% methane and 60% propane by
mass at 540 R and 15 psia. The gas is now
slowly compressed in an isothermal (T — con-
stant) process to a final pressure of 40 psia.
Show the process in a P-V diagram, and find 12.156E
both the work and heat transfer in the process.
12. 149 E A mixture of 4 lbm oxygen and 4 lbm of argon 12.157E
is in an insulated piston cylinder arrangement at
14,7 lbf/in. 2 , 540 R. The piston now compresses
the mixture to half its initial volume. Find the
final pressure, temperature, and the piston work.
12.150E Two insulated tanks A and B are connected by 12.158E
a valve. Tank A has a volume of 30 ft 3 and ini-
tially contains argon at 50 lbf/in. 2 , 50 F. Tank
B has a volume of 60 ft 3 and initially contains
ethane at 30 lbf/in. 2 , 120 F. The valve is
opened and remains open until the resulting J2.159E
gas mixture comes to a uniform state. Find the
final pressure and temperature.
12.151E A mixture of 50% carbon dioxide and 50%
water by mass is brought from 2800 R, 150
lbf/in. 2 to 900 R, 30 lbf/in. 2 in a polytropic
process through a steady-flow device. Find the 12.160E
necessary heat transfer and work involved
using values from F.4.
12.152E Carbon dioxide gas at 580 R is mixed with ni-
trogen at 500 R in an insulated mixing cham-
ber. Both flows are at 14.7 lbf/in. 2 , and the 12.161E
mole ratio of carbon dioxide to nitrogen is 2:1.
Find the exit temperature and the total entropy
generation per mole of the exit mixture.
12.153 E A mixture of 60% helium and 40% nitrogen by
mole enters a turbine at 150 lbf/in. 2 , 1500 R at 12.162E
a rate of 4 lbm/s. The adiabatic turbine has an
exit pressure of 15 lbf/in. 2 and an insentropic
efficiency of 85%. Find the turbine work.
12.154E A large air separation plant, see Fig. PI 2.68,
takes in ambient air (79% N 2 , 21% 2 by vol-
ume) at 14.7 lbf/in. 2 , 70 F, at a rate of 2 lb
mol/s. It discharges a stream of pure 2 gas at
30 lbf/in. 2 , 200 F, and a stream of pure N 2 gas
at 14.7 lbf/in. 2 , 70 F. The plant operates on an
electrical power input of 2000 kW. Calculate
the net rate of entropy change for the process.
A tank has two sides initially separated by a di-
aphragm. Side A contains 2 lbm of water, and
side B contains 2.4 lbm of air — both at 68 F,
14.7 lbf/in. 2 . The diaphragm is now broken,
and the whole tank is heated to 1 100 F by a
1300 F reservoir. Find the final total pressure,
heat transfer, and total entropy generation.
Find the entropy generation for the process in
Problem 12.I50E.
Consider a volume of 2000 ft 3 that contains an
air-water vapor mixture at 14.7 lbf/in. 2 , 60 F,
and 40% relative humidity. Find the mass of
water and the humidity ratio. What is the dew
point of the mixture?
A 1 lbm/s flow of saturated moist air (relative
humidity 100%) at 14.7 psia and 50 F goes
through a heat exchanger and comes out at 80
F. What is the exit relative humidity, and how
much power is needed?
Consider a 10-ft 3 rigid tank containing an
air-water vapor mixture at 14.7 lbf/in. 2 , 90 F,
with 70% relative humidity. The system is
cooled until the water just begins to condense.
Determine the final temperature in the tank and
the heat transfer for the process.
Consider at 35 ftVs flow of atmospheric air at
14.7 psia, 80 F, and 80% relative humidity. As-
sume this flows into a basement room where it
cools to 60 F at 14.7 psia. How much liquid
will condense out?
Air in a piston/cylinder is at 95 F, 15 lbf/in. 2
and relative humidity of 80%, It is now com-
pressed to a pressure of 75 lbf/in. 2 in a constant-
temperature process. Find the final relative and
specific humidity and the volume ratio V 2 IV V
A 10-ft 3 rigid vessel initially contains moist air
at 20 lbf/in. 2 , 100 F, with a relative humidity of
10%. A supply line connected to this vessel by
a valve carries steam at 100 lbf/in. 2 , 400 F. The
valve is opened, and steam flows into the ves-
sel until the relative humidity of the resultant
508 M CHAPTER TWELVE GAS MIXTURES
moist air mixture is 90%. Then the valve is
closed. Sufficient heat is transferred from the
vessel so the temperature remains at 100 F dur-
ing the process. Determine the heat transfer for
the process, the mass of steam entering the
vessel, and the final pressure inside the vessel.
12.163E A water-filled reactor of 50 ft 3 is at 2000
lbf/in. 2 , 550 F, and located inside an insulated
containment room of 5000 ft 3 that has air at
1 arm. and 77 F. Due to a failure, the reactor rup-
tures and the water fills the containment room.
Find the final quality and pressure by iterations.
12.164E Two moist air streams with 85% relative hu-
midity, both flowing at a rate of 0.2 lbm/s of
dry air are mixed in a steady flow setup. One
inlet fiowstream is at 90 F, and the other at 61
F. Find the exit relative humidity.
12.165E A flow of moist air from a domestic furnace,
state 1 in Fig. PI 2. 98 is at 120 F, 10% relative
humidity with a flow rate of 0.1 lbm/s dry air.
A small electric heater adds steam at 212 F,
14.7 psia, generated from tap water at 60 F. Up
in the living room the flow comes out at state
4: 90 F, 60% relative humidity. Find the power
needed for the electric heater and the heat
transfer to the flow from state 1 to state 4.
12.166E Atmospheric air at 95 F, relative humidity of
10%, is too warm and also too dry. An air con-
ditioner should deliver air at 70 F, and 50%
relative humidity in the amount of 3600 ft 3 per
hour. Sketch a setup to accomplish this; find
any amount of liquid (at 68 F) that is needed or
discarded and any heat transfer.
12.167E An indoor pool evaporates 3 Ibm/h of water,
which is removed by a dehumidifier to maintain
70 F, <E> = 70% in the room. The dehumidifier
is a refrigeration cycle in which air flowing
over the evaporator cools such that liquid water
drops out, and the air continues flowing over
the condenser, as shown in Fig. PI 2. 105. For an
airflow rate of 0.2 lbm/s, the unit requires 1.2
Btu/s input to a motor driving a fan and the
compressor, and it has a coefficient of perfor-
mance, j3 = QJW C = 2.0. Find the state of the
air after the evaporator, T 2 , o> 2 , <£> 2 , and the heat,
rejected. Find the state of the air as it returns to
the room and the compressor work input.
12.168E To refresh air in a room, a counterflow heat ex-
changer is mounted in the wall, as shown in
Fig. P12.115. It draws in outside air at 33 F,
80% relative humidity, and draws room air,
104 F, 50% relative humidity, out. Assume an
exchange of 6 Ibm/min dry air in a steady-flow
device, and also that the room air exits the heat
exchanger to the atmosphere at 72 F. Find the
net amount of water removed from the room,
any liquid flow in the heat exchanger, and
(7*, $) for the fresh air entering the room.
12.169E Ambient air is at a condition of 14.7 lbf/in. 2 ,
95 F, 50% relative humidity. A steady stream of
air at 14.7 lbf/in. 2 , 73 F, 70% relative humidity is
to be produced by first cooling one stream to an
appropriate temperature to condense out the
proper amount of water and then mix this stream
adiabatically with the second one at ambient con-
ditions. What is the ratio of the two flow rates? To
what temperature must the first stream be cooled?
12.170E A 4-ft 3 insulated tank contains nitrogen gas at 30
lbf/in. 2 and ambient temperature 77 F. The tank
is connected by a valve to a supply line flowing
carbon dioxide at 180 lbf/in. 2 , 190 F. A mixture
of 50 mole percent nitrogen and 50 mole percent
carbon dioxide is to be obtained by opening the
valve and allowing flow into the tank until an ap-
propriate pressure is reached, when the valve is
closed. What is the pressure? The tank eventu-
ally cools to ambient temperature. Calculate the
net entropy change for the overall process.
Computer, design, and open-ended problems
12.171 Write a program to solve the general case of
Problems 12.53/74 in which the two volumes
and the initial state properties of the argon and
the ethane are input variables. Use constant spe-
cific heat from Table A.5.
12.172 Mixing of C0 2 and N 2 in a steady-flow setup
was given in Problem 12.62. If the temperatures
are very different an assumption of constant spe-
cific heat is inappropriate. Study the problem as-
suming the C0 2 enters at 300 K, 100 kPa, as a
Computer, Design, and Open-Ended Problems S 509
function of the N 2 inlet temperature using spe-
cific heat from Table A. 7 or the formula in A.6.
Give the nitrogen inlet temperature for which
the constant specific heat assumption starts to be
more than 1%, 5%, and 10% wrong for the exit
mixture temperature.
12.173 The setup in Problem 12.97 is similar to a
process that can be used to produce dry powder
from a slurry of water and dry material as coffee
or milk. The water flow at state 3 is a mixture of
80% liquid water and 20% dry material on a
mass basis with = 0.4 kj/kg K. After the
water is evaporated, the dry material falls to the
bottom and is removed in an additional line,
exit at state 4. Assume a reasonable T 4 and that
state 1 is heated atmospheric air. Investigate the
inlet flow temperature as a function of state 1
humidity ratio.
12.174 A dehumidifier for household applications is
similar to the system shown in Fig. P 12. 105.
Study the requirements to the refrigeration cycle
as a function of the atmospheric conditions and
include a worst case estimation.
12.175 A clothes dryer has a 60°C, 4> = 90% airflow
out at a rate of 3 kg/min. The atmospheric condi-
tions are 20 o C, relative humidity of 50%. How
much water is carried away and how much
power is needed? To increase the efficiency, a
counterflow heat exchanger is installed to pre-
heat the incoming atmospheric air up with the
hot exit flow. Estimate suitable exit tempera-
tures from the heat exchanger and investigate
the design changes to the clothes dryer. (What
happens to the condensed water?) How much
energy can be saved this way?
12.176 Addition of steam to combustors in gas turbines
and to internal-combustion engines reduces the
peak temperatures and lowers emission of NC^.
Consider a modification to a gas turbine, as
shown in Fig. P 12. 176, where the modified
cycle is called the Cheng cycle. In this example,
it is used for a cogenerating power plant. As-
sume 12 kg/s air with state 2 at 1.25 MPa, un-
known temperature, is mixed with 2.5 kg/s water
at 450°C at constant pressure before the inlet to
the turbine. The turbine exit temperature is T 4 =
500°C J and the pressure is 125 kPa. For a rea-
sonable turbine efficiency, estimate the required
air temperature at state 2. Compare the result to
the case where no steam is added to the mixing
chamber and only air runs through the turbine.
— |— 5-
©
2.5 fcg'a 450'C
Dsirlrt
tisaEng load Q,
FIGURE P12.176
12.177 Consider the district water heater acting as the
condenser for part of the water between states 5
and 6 in Fig. P 12. 176. If the temperature of the
mixture (12 kg/s air, 2.5 kg/s steam) at state 5 is
135°C, make a study of the district heating load,
Q u as a function of the exit temperature T 6 .
Study also the sensitivity of the results with re-
spect to the assumption that state 6 is saturated
moist air.
12.178 The cogeneration gas-turbine cycle can be aug-
mented with a heat pump to extract more energy
from the turbine exhaust gas, as shown in Fig.
P12.178. The heat pump upgrades the energy to
70°C
Heat
pump
Sat. air
to chimney
FIGURE P12.178
CHAPTER TWELVE GAS MIXTURES
be delivered at the 70°C line for district heating.
In the modified application, the first heat ex-
changer has exit temperature T 6a = T la - 45 C,
and the second one has T €b = T lb = 36°C. As-
sume the district heating line has the same exit
temperature as before so this arrangement al-
lows for a higher flow rate. Estimate the increase
in the district heating load that can be obtained
and the necessary work input to the heat pump.
12 179 Several applications of dehumidification do not
rely on water condensation by cooling. A desic-
cant with a greater affinity to water can absorb
water directly from the air accompanied by a
heat release. The desiccant is then regenerated
by heating, driving the water out. Make a list of
several such materials as liquids, gels, and solids
and show examples of their use.
THERMODYNAMIC RELATIONS
We have already defined and used several thermodynamic properties. Among these are
pressure, specific volume, density, temperature, mass, internal energy, enthalpy, en-
tropy, constant-pressure and constant-volume specific heats, and the Joule-Thomson
coefficient. Two other properties, the Heimholtz function and the Gtbbs function, will
also be introduced and will be used more extensively in the following chapters. We
have also had occasion to use tables of thermodynamic properties for a number of dif-
ferent substances.
One important question is now raised: Which of the thermodynamic properties
can be experimentally measured? We can answer this question by considering the
measurements we can make in the laboratory. Some of the properties such as internal
energy and entropy cannot be measured directly and must be calculated from other ex-
perimental data. If we carefully consider all these thermodynamic properties, we con-
clude that there are only four that can be directly measured: pressure, temperature,
volume, and mass.
This leads to a second question: How can values of the thermodynamic properties
that cannot be measured be determined from experimental data on those properties that
can be measured? In answering this question, we will develop certain general thermody-
namic relations. In view of the fact that millions of such equations can be written, our
study will be limited to certain basic considerations, with particular reference to the de-
termination of thermodynamic properties from experimental data. We will also consider
such related matters as generalized charts and equations of state.
13.1 THE CLAPEYRON EQUATION
In calculating thermodynamic properties such as enthalpy or entropy in terms of other
properties that can be measured, the calculations fall into two broad categories: differ-
ences in properties between two different phases and changes within a single, homoge-
neous phase. In this section, we focus our attention on the first of these categories, that of
different phases. Let us assume that the two phases are liquid and vapor, but we will see
that the results apply to other differences as well.
Consider a Camot-cycle heat engine operating across a small temperature differ-
ence between reservoirs at T and T — A7*. The corresponding saturation pressures are P
and P - AP. The Carnot cycle operates with four steady-state devices. In the high tem-
perature heat-transfer process, the working fluid changes from saturated liquid at 1 to sat-
urated vapor at 2, as shown in the two diagrams of Fig. 13.1.
From Fig. 13.1a, for reversible heat transfers,
qH=Ts fg ; q L =(T-AT) S/g
511
512 B Chapter Thirteen thermodynamic relations
t -
T-&T
FIGURE 13,1 A
Camot cycle operating
across a small temperature
difference.
P-AP
(a)
so that
From Fig. 13.16, each process is steady-state and reversible, such that the work in each
process is given by Eq. 9.19,
w =-jvdP
Overall, for the four processes in the cycle,
> NEr = - j v dP + Q - j v dP
2 4
„ - ) (P - LP - P) - (^p) " P + ^
~ AP
(13.2)
(The smaller the AP, the better the approximation.)
Now, comparing Eqs. 13.1 and 13.2 and rearranging,
AP S A. —
AT /u 2 +u 3 \ (v x + v A
(13.3)
2 j \ 2
In the limit as AT~^ 0: u 3 -» v 2 = v p u 4 ^ t>i = v fi which results in
AP = ^t = ^fe
1™&T dT v /g
Since the heat addition process 1 - 2 is at constant pressure as well as constant temperature,
<1h = h fg = Ts fs
and the general result of Eq. 13.3 is the expression
dT v /g Tv fg
(13-4)
The Clapeyron Equation H 513
which is called the Clapeyron equation. This is a very simple relation and yet an ex-
tremely powerful one. We can experimentally determine the left-hand side of Eq. 13.4,
which is the slope of the vapor pressure as a function of temperature. We can also mea-
sure the specific volumes of saturated vapor and saturated liquid at the given temperature,
which means that the enthalpy change and entropy change of vaporization can both be
calculated from Eq. 13.4. This establishes the means to cross from one phase to another in
first- or second-law calculations, which was the goal of this development.
We could proceed along the same lines for the change of phase solid to liquid or for
solid to vapor. In each case, the result is the Clapeyron equation, in which the appropriate
saturation pressure, specific volumes, entropy change, and enthalpy change are involved.
For solid i to liquid/; the process occurs along the fusion line, and the result is
dT v if Tv if
(13.5)
We note that v if - ty - v t is typically a very small number, such that the slope of the fu-
sion line is very steep. (In the case of water, tyis a negative number, which is highly un-
usual, and the slope of the fusion line is not only steep, it is also negative.)
For sublimation, the change from solid / directly to vapor g, the Clapeyron equation
has the values
dT v ig Tv ig
A special case of the Clapeyron equation involving the vapor phase occurs at low
temperatures when the saturation pressure becomes very small. The specific volume v g is
then not only much larger than that of the condensed phase, liquid in Eq. 13.4 or solid in
Eq. 13.6, but is also closely represented by the ideal-gas equation of state. The Clapeyron
equation then reduces to the form
dT Tv fg RT * {LX,)
At low temperatures (not near the critical temperature), h fg does not change very much
with temperature. If it is assumed to be constant, then Eq. 13.7 can be rearranged and inte-
grated over a range of temperatures to calculate a saturation pressure at a temperature at
which it is not known. This point is illustrated by the following example.
EXAMPLE 13.1 Determine the sublimation pressure of water vapor at -60°C using data available in the :
steam tables.
Control mass: Water.
Solution
Appendix Table B. 1.5 of the steam tables does not give saturation pressures for tern- :
peratures less than -40°C. However, we do notice that h lg is relatively constant in
514 W CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
this range; therefore, we proceed to Eq. 13.7 and integrate between the limits -40°C
and -60°C.
J, ? J, R T 1 R Ji T 2
111 F t R\ 7\r 2
Let
Then
P, = 0.0129 kPa T 2 = 233.2 K T x
. ^^j, f 233.2 - 213.2 \ _ 2
M P, ~ 0.461 52 \233.2 X 213.2/
P, = 0.001 09 kPa
213.2 K
.4744
EXAMPLE
13.1E Determine the sublimation pressure of water vapor at -70 F using data available in the
steam tables.
Control mass: Water.
Solution
Appendix Table F.7.4 of the steam tables does not give saturation pressures for tem-
peratures less than -40 F. However, we do notice that h lg is relatively constant m
this range; therefore, we proceed to use Eq. 13.7 and integrate between the limits
-40 F and ™70 F.
Let
Then,
J, P J, R T 2 R Ji T 2
ln /\ r\ m
P 7 = 0.0019 lbf/in. 2
f = 419.7 R 7*! = 389.7 R
, P 2 1218.7 X 778 f 419.7 - 389.7 \ _J 2 027p
^ P~ x ~ 85.76 V 419 - 7 x 389 - 7 '
P. = 0.000 25 lbf/in. 2
Mathematical relations for a homogeneous phase H 515
13,2 Mathematical Relations
for a Homogeneous Phase
In the preceding section, we established the means to calculate differences in enthalpy (and
therefore internal energy) and entropy between different phases, in terms of properties that
are readily measured. In the following sections, we will develop expressions for calculating
differences in these properties within a singlej homogeneous phase (gas, liquid, or solid),
assuming a simple compressible substance. In order to develop such expressions, it is first
necessary to present a mathematical relation that will prove useful in this procedure.
Consider a variable (thermodynamic property) that is a continuous function of* mdy.
z =f(x,y)
It is convenient to write this function in the form
dz = Mdx + Ndy (13.8)
where
= partial derivative of z with respect to x (the variable y being held constant)
— partial derivative of z with respect to y (the variable x being held constant)
The physical significance of partial derivatives as they relate to the properties of a
pure substance can be explained by referring to Fig. 13.2, which shows a P~u-T surface
of the superheated vapor region of a pure substance. It shows constant-temperature,
constant-pressure, and constant specific volume planes that intersect at point b on the
surface. Thus, the partial derivative {dPidv) T is the slope of curve abc at point b. Line
de represents the tangent to curve abc at point b. A similar interpretation can be made
of the partial derivatives {dPidT\ and (dv!dT) p .
If we wish to evaluate the partial derivative along a constant-temperature line, the
rules for ordinary derivatives can be applied. Thus, we can write for a constant-temperature
process:
= d ll
\dvj T dv T
and the integration can be performed as usual. This point will be demonstrated later in a
number of examples.
Let us return to the consideration of the relation
dz = Mdx + Ndy
516 H Chapter Thirteen Thermodynamic Relations
FIGURE 13.2
Schematic representation
of partial derivatives.
If x, y, and z are all point functions (that is, quantities that depend only on the state and are
independent of the path), the differentials are exact differentials. If this is the case, the fol-
lowing important relation holds:
The proof of this is
dy
3M
dy
dN
dx
dN
dx
d"z
dxdy
d 2 z
dydx
(13,9)
Since the order of differentiation makes no difference when point functions are involved,
it follows that
d 2 z
d 2 z
dxdy dydx
dM
dy
dN
dx
13.3 the Maxwell Relations
Consider a simple compressible control mass of fixed chemical composition. The
Maxwell relations, which can be written for such a system, are four equations relating the
properties P,v,T f and s. These will be found to be useful in the calculation of entropy in
terms of the other, measurable properties.
THE MAXWELL RELATIONS M 517
The Maxwell relations are most easily derived by considering the different forms of
the thermodynamic property relation, which was the subject of Section 8.5. The two
forms of this expression are rewritten here as
du = Tds-Pdv (13.10)
and
dh = Tds + vdP (13.11)
Note that in the mathematical representation of Eq. 13.8, these expressions are of the form
u = 11(5, v), h = his, P)
in both of which entropy is used as one of the two independent properties. This is an un-
desirable situation in that entropy is one of the properties that cannot be measured. We
can, however, eliminate entropy as an independent property by introducing two new prop-
erties and thereby two new forms of the thermodynamic property relation. The first of
these is the Helmholtz function^,
A = U-TS, a = u~Ts (13.12)
Differentiating and substituting Eq. 13.10 results in
da = du - Tds - s dT
= -sdT~Pdv (13.13)
which we note is a form of the property relation utilizing Tand v as the independent prop-
erties. The second new property is the Gibbs function G,
G = H-TS, g = h-Ts (13.14)
Differentiating and substituting Eq. 13.11,
dg = dh ~ Tds-sdT
= ~sdT+vdP (13.15)
a fourth form of the property relation, this form using Tand P as the independent properties.
Since Eqs. 13.10, 13.11, 13.13, and 13.15 are all relations involving only proper-
ties, we conclude that these are exact differentials and, therefore, are of the general
form of Eq. 13.8,
dz = PI dx + N dy
in which Eq. 13.9 relates the coefficients M and N }
($M\ =
It follows from Eq. 13.10 that
518 ffl CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
Similarly, from Eqs. 13.11, 13.13, and 13.15 we can write
21) = \m (13.17)
3P\ =(&) (13.18)
3TL \dV T
(13.19)
These four equations are known as the Maxwell relations for a simple compressible mass,
and the great utility of these equations will be demonstrated in later sections of this chap-
ter. As was noted earlier, these relations will enable us to calculate entropy changes in
terms of the measurable properties pressure, temperature, and specific volume.
A number of other useful relations can be derived from Eqs. 13.10, 13.1 1, 13.13,
and 13.15. For example, from Eq. 13.10, we can write the relations
iu\ = t I = -p (13.20)
Similarly, from the other three equations, we have the following
f 3h\ _ ^ (dh
(ft),-'- (
If -
As already noted, the Maxwell relations just presented are written for a simple com-
pressible substance. It is readily evident, however, that similar Maxwell relations can be
written for substances involving other effects, such as surface or electrical effects. For ex-
ample, Eq. 8.9 can be written in the form
dU = TdS - P dV + 3" dh + SP dA + % dZ + • • ' (13.22)
Thus, for a substance involving only surface effects, we can write
dU = TdS+tfdA
and it follows that for such a substance
K dAj s \dS l
Other Maxwell relations could also be written for such a substance by writing the prop-
erty relation in terms of different variables, and this approach could also be extended to
systems having multiple effects. This matter also becomes more complex when we con-
sider applying the property relation to a system of variable composition, a topic that will
be taken up in Section 13.11.
EXAMPLE 13.2 From an examination of the properties of compressed liquid water, as given m Table
B 1 4 of the Appendix, we find that the entropy of compressed liquid is greater than the
entropy of saturated liquid for a temperature of 0°C and is less than that of saturated liq-
THERMODYNAMIC RELATIONS INVOLVING ENTHALPY, INTERNAL ENERGY, AND ENTROPY B
519
V
FIGURE 13.3 Sketch
for Example 13.2.
4°C (39 F)
T
uid for all the other temperatures listed. Explain why this follows from other thermody-
namic data.
Control mass; Water.
Suppose we increase the pressure of liquid water that is initially saturated, while keeping
the temperature constant. The change of entropy for the water during this process can be
found by integrating the following Maxwell relation, Eq. 13.19:
Therefore, the sign of the entropy change depends on the sign of the term (dvIdT) P , The
physical significance of this term is that it involves the change in specific volume of water
as the temperature changes while the pressure remains constant. As water at moderate pres-
sures and 0°C is heated in a constant-pressure process, the specific volume decreases until
the point of maximum density is reached at approximately 4°C, after which it increases.
This is shown on a v~T diagram in Fig. 13.3. Thus, the quantity (du/dT) P is the slope of the
curve in Fig. 13.3. Since this slope is negative at 0°C, the quantity {dsldP) T is positive at
0°C. At the point of maximum density the slope is zero and, therefore, the constant-pressure
line shown in Fig. 8.7 crosses the saturated-liquid line at the point of maximum density.
Let us first derive two equations, one involving C p and the other involving C u .
We have denned C p as
Solution
13.4 Thermodynamic Relations Involving
enthalpy, internal energy, and entropy
We have also noted that for a pure substance
Tds = dh-vdP
Therefore,
(13.23)
520 M Chapter thirteen Thermodynamic Relations
Similarly, from the definition of C v>
C ^ ^
and the relation
it follows that
Tds = du + Pdv
c " f Jrl T \dT
(13.24)
We will now derive a general relation for the change of enthalpy of a pure sub-
stance. We first note that for a pure substance
h = h(T, P)
Therefore,
From the relation
it follows that
Tds = dh-vdP
Substituting the Maxwell relation, Eq. 13.19, we have
On substituting this equation and Eq. 13.23, we have
dh = C p dT +
v-T
dP
(13.25)
(13.26)
Along an isobar we have
and along an isotherm,
dh p = C p dT p
dh T —
dP T
(13.27)
The significance of Eq. 13.26 is that this equation can be integrated to give the
change in enthalpy associated with a change of state
h, - hi
dP
(13.28)
THERMODYNAMIC RELATIONS INVOLVING ENTHALPY, INTERNAL ENERGY, AND ENTROPY H 521
The information needed to integrate the first term is a constant-pressure specific
heat along one (and only one) isobar.. The integration of the second integral requires that
an equation of state giving the relation between P, v, and 7* be known. Furthermore, it is
advantageous to have this equation of state explicit in v, for then the derivative (dvidT) P
is readily evaluated.
This matter can be further illustrated by reference to Fig. 13.4. Suppose we wish to
know the change of enthalpy between states 1 and 2. We might determine this change
along path \-x-2, which consists of one isotherm, 1-x, and one isobar, x-2. Thus, we
could integrate Eq. 13.28:
The second term in this equation gives the change in enthalpy along the isotherm
1-T and the first term the change in enthalpy along the isobar x-2. When these are added
together, the result is the net change in enthalpy between 1 and 2. Therefore, the constant-
pressure specific heat must be known along the isobar passing through 2 and x. The
change in enthalpy could also be found by following path \-y-2 } in which case the con-
stant-pressure specific heat must be known along the \-y isobar. If the constant-pressure
specific heat is known at another pressure, say, the isobar passing through m-n, the
change in enthalpy can be found by following path l-m-n-2. This involves calculating
the change of enthalpy along two isotherms — \~m and n-2.
Let us now derive a similar relation for the change of internal energy. All the steps
in this derivation are given but without detailed comment. Note that the starting point is to
write u = u(T t v), whereas in the case of enthalpy the starting point was h = h(T, P).
Since T x = T x and P 2 = P„ this can be written
u
Tds
du+Pdu
P = constant
FIGURE 13.4 Sketch
showing various paths by
which a given change of
state can take place.
T
P = constant
s
522 M Chapter thirteen thermodynamic Relations
Therefore,
du\ = T (ds\ „ p
dv) T \dvJ T
(13.29)
Substituting the Maxwell relation, Eq. 13.18, we have
Therefore,
dv
du= C u dT +
dP
ST
dv
(13.30)
Along an isometric this reduces to
du„ = C„ dT„
and along an isotherm we have
du T =
dv.
(13.31)
In a manner similar to that outlined earlier for changes in enthalpy, the change of in-
ternal energy for a given change of state for a pure substance can be determined from Eq.
13 30 if the constant-volume specific heat is known along one isometric and an equation
of state explicit in P [to obtain the derivative (dP/dT),] is available in the region involved.
A diagram similar to Fig. 13.4 could be drawn, with the isobars replaced with isometrics,
and the same general conclusions would be reached.
To summarize, we have derived Eqs. 13.26 and 13.30:
dh = C p dT +
du = C„ dT +
Hi.
dP
dv
The first of these equations concerns the change of enthalpy, the constant-pressure specific
heat, and is particularly suited to an equation of state explicit in v. The second equation
concerns the change of internal energy and the constant-volume specific heat, and is partic-
ularly suited to an equation of state explicit in P. If the first of these equations is used to de-
termine the change of enthalpy, the internal energy is readily found by noting that
"2
= h 2 -h l -(P 2 v 2 --PiVi)
If the second equation is used to find changes of internal energy, the change of enthalpy is
readily found from this same relation. Which of these two equations is used to determine
changes in internal energy and enthalpy will depend on the information available for spe-
cific heat and an equation of state (or other P-v-T data).
Two parallel expressions can be found for the change of entropy.
s =s(T,P)
THERMODYNAMIC RELATIONS INVOLVING ENTHALPY, INTERNAL ENERGY, AND ENTROPY B 523
Substituting Eqs. 13.19 and 13.23, we have
Along an isobar we have
and along an isotherm
W^f -Hi!)/' 03.33)
Note from Eq. 13.33 that if a constant-pressure specific heat is known along one iso-
bar and an equation of state explicit in v is available, the change of entropy can be evalu-
ated. This is analogous to the expression for the change of enthalpy given in Eq. 13.26.
s =s(T,v)
Substituting Eqs. 13.18 and 13.24 gives
d S ^C v f±(^jj v (13t34)
----- J>f + f(S)L*
^ This expression for change of entropy concerns the change of entropy along an iso-
metric where the constant-volume specific heat is known and along an isotherm where an
equation of state explicit in P is known. Thus, it is analogous to the expression for change
of internal energy given in Eq. 13.30.
EXAMPLE 13.3 Over a certain small range of pressures and temperatures, the equation of state of a cer-
tain substance is given with reasonable accuracy by the relation
RT p
or
RT C
V = P
where C and C are constants.
Derive an expression for the change of enthalpy and entropy of this substance in
an isothermal process.
Conti-ol mass'. Gas.
524 H CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
Solution
Since the equation of state is explicit in v, Eq. 13.27 is particularly relevant to the
change in enthalpy. On integrating this equation, we have
(A 2 ~ Ai)r =
From the equation of state,
f[-'(-
dT) P _
dP 7
dTjp P T*
Therefore,
dP T
p p p T 3J
dP T
QH-h^W-fdP^-f^P^r
For the change in entropy we use Eq. 13.33, which is particularly relevant for an
equation of state explicit in v.
133 VOLUME EXPANSIVITY AND ISOTHERMAL
AND AMABATIC COMPRESSIBILITY
The student has most likely encountered the coefficient of linear expansion in his or he
studies of strength of materials. This coefficient indicates how the length of a solid boc
is influenced by a change in temperature while the pressure remains constant In terms .
the notation of partial derivatives, the coefficient of linear expansion, S r , is defined as
A similar coefficient can be defined for changes in volume. Such a coefficient is a
plicable to liquids and gases as well as to solids. This coefficient of volume expansion, a
also called the volume expansivity, is an indication of the change in volume as temper
ture changes while the pressure remains constant. The definition of volume expansivity
ap ~V\dT} P v\ d T) P T
VOLUME EXPANSIVITY AND ISOTHERMAL AND ADIABATIC COMPRESSIBILITY M 525
and it equals three times the coefficient of linear expansion. You differentiate V = L^L^
with temperature to prove that which is left as a homework exercise. Notice that it is the
volume expansivity whicii enters into the expressions for calculating changes in enthalpy,
Eq. 13.26, and entropy, Eq. 13.32.
The isothermal compressibility, B T , is an indication of the change in volume as pres-
sure changes while the temperature remains constant. The definition of the isothermal
compressibility is
The reciprocal of the isothermal compressibility is called the isothermal bulk modu-
lus, B T .
The adiabatic compressibility, p st is an indication of the change in volume as pres-
sure changes while the entropy remains constant; it is defined as
A-"i(f) s (13.40)
The adiabatic bulk modulus, B„ is the reciprocal of the adiabatic compressibility.
' -"(£), ( i3 - 41 >
The velocity of sound, c, in a medium is defined by the relation
it, < 13 ' 42 >
This can also be expressed as
* = -*$)r vB > (13 - 43)
in terms of the adiabatic bulk modulus B s . For a compressible medium such as a gas the
speed of sound becomes modest, whereas in an incompressible state such as a liquid or a
solid it can be quite large.
The volume expansivity and isothermal and adiabatic compressibility are thermody-
namic properties of a substance, and for a simple compressible substance are functions of
two independent properties. Values of these properties are found in the standard hand-
books of physical properties. The following examples give an indication of the use and
significance of the volume expansivity and isothermal compressibility.
The pressure on a block of copper having a mass of 1 kg is increased in a reversible
process from 0.1 to 100 MPa while the temperature is held constant at 15°C. Determine
the work done on the copper during this process, the change in entropy per kilogram of
copper, the heat transfer, and the change of internal energy per kilogram.
526 B Chapter Thirteen Thermodynamic relations
Over the range of pressure and temperature in this problem, the following data can
be used:
Volume expansivity = cc p = 5.0 X lO^KT 1
Isothermal compressibility = j3 r - 8.6 X 10" 12 m 2 /N
Specific volume = 0.000 1 14 m 3 /kg
Analysis
Control mass; Copperblock.
States: Initial and final states known.
Process: Constant temperature, reversible.
The work done during the isothermal compression is
w = j P dv T
The isothermal compressibility has been defined as
*-4(g) r
v{3 T dP T = -dv T
Therefore, for this isothermal process,
w = -j 2 v(3 T P dP T
Since u and /3 r remain essentially constant, this is readily integrated:
w = — ^~ ("2 ~ ^
The change of entropy can be found by considering the Maxwell relation, Eq.
13.19, and the definition of volume expansivity.
ds T — —va P dP T
This equation can be readily integrated, if we assume that v and a P remain constant:
(J2 - *i) r = -va p (P 2 - Pi) r
The heat transfer for this reversible isothermal process is
q = T(s 2 - j t )
The change in internal energy follows directly from the first law.
Real-Gas Behavior and equations of State H 527
Solution
= _ Q-Q00 114X^.6X10-" (10Q2 _ Q l2) x 10U
= -4.9 J/kg
(s 2 - = ~va p (P 2 - P 3 ) r
- -0.000 114 X 5.0 X 10~ s (100 - 0.1) X 10 6
= —0.5694 J/kg K
q = Tfe - - -288.2 X 0.5694 = - 164.1 J/kg
(«z - «i) = ? - w = -164.1 - (-4.9) = -159.2 J/kg
13,6 Real-Gas Behavior
and equations of state
In Section 3.4, we examined the P-u-r behavior of gases, and we defined the compres-
sibility factor in Eq. 3.6,
We then proceeded to develop the generalized compressibility chart, presented in Ap-
pendix Fig. D.l in terms of the reduced pressure and temperature. The generalized chart
does not apply specifically to any one substance, but is instead an approximate relation
that is reasonably accurate for many substances, especially those that are fairly simple
in molecular structure. In this sense, the generalized compressibility chart can be
viewed as one aspect of generalized behavior of substances, and also as a graphical
form of equation of state representing real behavior of gases and liquids over a broad
range of variables.
To gain additional insight into the behavior of gases at low density, let us examine
the low-pressure portion of the 'generalized compressibility chart in greater detail. This
behavior is as shown in Fig. 13.5. The isotherms are essentially straight lines in this re-
gion, and their slope is of particular importance. Note that the slope increases as T r in-
creases until a maximum value is reached at a T r of about 5, and then the slope decreases
toward the Z = 1 line for higher temperatures. That single temperature, about 2.5 times
the critical temperature, for which
s(§l = (13 - 44 >
is defined as the Boyle temperature of the substance. This is the only temperature at which
a gas behaves exactly as an ideal gas at low, but finite pressures, since all other isotherms
528 a CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
Z
1.0
T ~ 0.7
FIGURE 13.5 Low-
pressure region of
compressibility chart.
P
go to zero pressure on Fig. 13.5 with a nonzero slope. To amplify this point, let us con-
sider the residual volume a,
Multiplying this equation by P we have
aP = RT-Pv
Thus, the quantity aP is the difference between RT- and Pv. Now as P -> 0, Pv -» RT.
However, it does not necessarily follow that a -> as P -» 0. Instead, it is only required
that a remain finite. The derivative in Eq. 13.44 can be written as
from which we find that a tends to zero as P -* only at the Boyle temperature, since that
is the only temperature for which the isothermal slope is zero on Fig. 13.5. It is perhaps a
somewhat surprising result that in the limit as P 0, Pv -» RT. In general, however, the
quantity (RT/P - v) does not go to zero but is instead a small difference between two
large values. This does have an effect on certain other properties of the gas.
The compressibility behavior of low-density gases as noted in Fig. 13.5 is the result
of intermolecular interactions and can be expressed in the form of equation of state called
the virial equation, which is derived from statistical thermodynamics. The result is
a
(13.45)
(13.46)
(13.47)
where 5(7), C(J), D(T) are temperature dependent and are called virial coefficients. B(T)
is termed the second virial coefficient and is due to binary interactions on the molecular
Real-gas behavior and equations of State B 529
0.050
FIGURE 13.6 The
second virial coefficient
for nitrogen.
-0.200
100
300 500
T, K
700
level. The general temperature dependence of the second virial coefficient is as shown for
nitrogen in Fig. 13.6. If we multiply Eq. 13.47 by RTiP, the result can be rearranged to the
form
f ~v = a^-B(T)% (13.48)
In the limit, as P -> 0,
lima = -B(T) (13.49)
P-K)
and we conclude from Eqs. 13.44 and 13.46 that the single temperature at which B{T) = 0,
Fig. 13.6, is the Boyle temperature. The second virial coefficient can be viewed as the
first-order correction for nonideality of the gas, and consequently becomes of considerable
importance and interest. In fact, the low-density behavior of the isotherms shown in Fig.
13.5 is directly attributable to the second virial coefficient.
Another aspect of generalized behavior of gases is the behavior of isotherms in the
vicinity of the critical point. If we plot experimental data on P~v coordinates, it is found
that the critical isotherm is unique in that it goes through a horizontal inflection point at
the critical point as shown in Fig. 13.7. Mathematically, this means that the first two de-
rivatives are zero at the critical point.
dP
dv
d 2 P
= at CP. (13.50)
<rfJ Tt = atCP ' ( 1151 )
a feature that is used to constrain many equations of state.
To this point, we have discussed the generalized compressibility chart, a graphical
form of equation of state, and the virial equation, a theoretically founded equation of state.
We now proceed to discuss other analytical equations of state, which may be either gener-
alized behavior in form, or empirical equations, relying on specific P-u-T data of their
constants. The oldest generalized equation, the van der Waals equation, was presented in
530 m CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
FIGURE 13.7 Plot of
isotherms in the region of
the critical point on
pressure-volume
coordinates for a typical v c
pure substance.
1873 as a semitheoretical improvement over the ideal-gas model. The van der Waals
equation of state has two constants and is written as
i> = -*£« (13.52)
v ™ b v 2
The constant b is intended to correct for the volume occupied by the molecules, and the
term a/v 2 is a correction that accounts for the intermolecular forces of attraction. As might
be expected in the case of a generalized equation, the constants a and b are evaluated from
the general behavior of gases. In particular, these constants are evaluated by noting that
the critical isotherm passes through a point of inflection at the critical point and that the
slope is zero at this point. Thus, for the van der Waals equation of state we have
dP] = _. RT + 2a (B . 53)
dvjr ( V - bf v 2
dv 2 jT (v - bf v A
Since both of these derivatives are equal to zero at the critical point we can write
RT C
la
(v c ~bf
2RT C
6a
(v c -b?
>, we find
(13.55)
RT*
a
(Vc ~ b) vl
= 3&
' (13.56)
64 P c
Real-Gas behavior and equations of state M 531
The compressibility factor at the critical point for the van der Waals equation is
7 - PcVc - 3
c RT C 8
which is considerably higher than the actual value for any substance.
A simple equation of state that is considerably more accurate than the van der
Waals equation is that proposed byRedlich and Kwong in 1949.
P = ~ V (v tb)t a (13 - 57)
with
R i T m
a = 0.427 48—^ (13.58)
RT
6 = 0.086 64—^ (13.59)
The numerical values in the constants have been determined by a procedure similar
to that followed in the van der Waals equation. Because of its simplicity, this equation
could not be expected to be sufficiently accurate to find use in the calculation of precision
tables of thermodynamic properties. It has, however, been used frequently for mixture
calculations and phase equilibrium correlations with reasonably good success. A number
of modified versions of this equation have also been utilized in recent years.
One of the best known empirical equations of state is the Benedict-Webb-Rubin
equation, often termed the BWR equation. The original equation, proposed in 1940, con-
taining eight empirical constants, was given in Chapter 3 as Eq. 3.9. The constants for a
number of substances are given in Appendix Table D.2. This equation, and particularly a
number of modifications to it, have been widely used over the years.
One particularly interesting modification of the BWR equation of state is the
Lee-Kesler equation, which was proposed in 1975. This equation has 12 constants and is
written in terms of generalized properties as
Z =
B ~ bl x n~n
c = c i~y + ^ (13.60)
fl = rf, + £
1 r
in which the variable v' r is not the true reduced specific volume but is instead defined as
' _ v
r RTJP C
(13.61)
Empirical constants for simple fluids for this equation are also given in Appendix
Table D.3.
532 m CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
13,7 THE GENERALIZED CHART FOR CHANGES
OF ENTHALPY AT CONSTANT TEMPERATURE
In Section 13.4, Eq. 13.27 was derived for the change of enthalpy at constant temperature.
This equation is appropriately used when a volume-explicit equation of state is known.
Otherwise, it is more convenient to calculate the isothermal change m internal energy
fromEq. 13.31
and then calculate the change in enthalpy from its definition as
(A 2 -/ll) = (K2-«l) + (^2-^l)
= (»2 - «l) + RT & ~ 2 l)
To determine the change in enthalpy behavior consistent with the generalized chart
Fie D 1 we follow the second of these approaches, since the Lee-Kesler generalized
equation' of state, Eq. 13.60, is a pressure-explicit form in terms of specific volume and
temperature. Equation 13.60 is expressed in terms of the compressibility factor Z, so we
write
D ZRT (dP\ ^ZR + BZ(di\
p = -*r> \dT) u v v W) u
Therefore, substituting into Eq. 13.31, we have
*=*RSf *
But
» v' r T T r
so that, in terms of reduced variables,
This expression is now integrated at constant temperature from any given state (P , v' r ) to
the ideal-gas limit (P* -» 0, v'* -* -)(the superscript * wilt always denote an ideal-gas
state or property), causing an internal energy change or departure from the ideal-gas value
at the given state,
« = CIl {m dv > (13.62)
The Generalized Chart for Changes of Enthalpy at Constant temperature M 533
The integral on the right-hand side of Eq. 13.62 can be evaluated from the Lee-Kesler
equation, Eq. 13.60. The corresponding enthalpy departure at the given state (P r> v' r ) is
then found from integrating Eq. 13.62 to be
Following the same procedure as for the compressibility factor, we can evaluate Eq. 13.63
with the set of Lee-Kesler simple-fluid constants to give a simple-fluid enthalpy depar-
ture. The values for the enthalpy departure are shown graphically in Fig. D.2. Use of the
enthalpy departure function is illustrated in the following example.
EXAMPLE 13,5 Nitrogen is throttled from 20 MPa, -70°C, to 2 MPa in an adiabatic, steady-state,
steady-flow process. Determine the final temperature of the nitrogen.
Control volume: Throttling valve.
Inlet state: P 1} T x known; state fixed.
Exit state: P 2 known.
Process: Steady-state, throttling process.
Diagram: Figure 13.8.
Model: Generalized charts, Fig. D.2.
Analysis
First law:
h x = h 2
Solution
Using values from Table A.2, we have
P, =20 MPa p H =^= 5 ,9
T, ~ 203.2 K T rl = f§|f = 1-61
P 2 — 2 MPa ^ = ^- = 0.59
From the generalized charts, Fig. D.2, for the change in enthalpy at constant tempera-
ture, we have
~ -= 2 1
RT C
A? - K = 2.1 X 0.2968 X 126.2 = 78.7 kJ/kg
1
534 m CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
P = 20 MPa
FIGURE 13.8 Sketch
for Example 13.5.
h = constant
It is now necessary to assume a final temperature and to check whether the net
change in enthalpy for the process is zero. Let us assume that T 2 = 146 K. Then the
change in enthalpy between 1* and 2* can be found from the zero-pressure, specific-
heat data.
hf - ht = C^Tf - Ti) = 1.0416(203.2 - 146) = +59.6 kJ/kg
(The variation in with temperature can be taken into account when necessary.)
We now find the enthalpy change between 2* and 2.
Therefore, from the enthalpy departure chart, Fig. D.2, at this state
^ = 0.5
f$ - A 2 = 0.5 X 0.2968 X 126.2 = 19,5 kJ/kg
We now check to see whether the net change in enthalpy for the process is zero.
kl - hl = = -(ftf - A,) + {hf - hf) + (ht - h 2 )
= -78.7 + 59.6 + 19.5 «
It essentially checks. We conclude that the final temperature is approximately 146 K. It
is interesting that the thermodynamic tables for nitrogen, Table B.6, give essentially this
same value for the final temperature.
The Generalized Chart for Changes of Entropy at Constant Temperature B 535
13.8 The Generalized Chart for Changes
of Entropy at constant Temperature
to this section we wish to develop a generalized chart giving entropy departures from
ideal-gas values at a given temperature and pressure, in a manner similar to that followed
lor enthalpy m the previous section. Once again, we have two alternatives. From Eq
1 5.5l } at constant temperature,
dT ft
which is convenient for use with a volume-explicit equation of state. The Lee-Kesler ex-
pression, Eq. 13.60, is, however, a pressure-explicit equation. It is therefore more appro-
priate to use Eq. 13.34, which is, along an isotherm,
In the Lee-Kesler form, in terms of reduced properties, this equation becomes
ds ( 3P T
R UT' dvl
JVhen this expression is integrated from a given state (P n v' r ) to the ideal-gas limit
V r U,v r -> oo), there is a problem because ideal-gas entropy is a function of pres-
sure and approaches infinity as the pressure approaches zero. We can eliminate this prob-
lem with a two-step procedure. First, the integral is taken only to a certain finite P* v'*
which gives the entropy change " r '
This integration by itself is not entirely acceptable, because it contains the entropy at
some arbrtrary, low-reference pressure. A value for the reference pressure would have to
be specified. Let us now repeat the integration over the same change of state, except this
time for a hypothetical ideal gas. The entropy change for this integration is
Sp* S n
^T~^ +hl p^ (13.65)
If we now subtract Eq. 13.65 from Eq. 13.64, the result is the difference in entropy of a
hypothetical ideal gas at a given state <T„ P f ) and that of the real substance at the same
state, or
s p~ s p_ , P , dP\
— r— - "in ^ + dv r (13.66)
Here the values associated with the arbitrary reference state P* cancel out of the
r i f^Sf S1 ? ° f the eqUatbn - (The first term of the inte g f al includes the term
+ln(P/P ) } which cancels the other term. The three different states associated with the de-
velopment of Eq. 13.66 are shown in Fig. 13.9.
The same procedure that was given in Section 13.7 for enthalpy departure values is
followed for generalized entropy departure values. The Lee-Kesler simple-fluid constants
are y^d in evaluating the integral of Eq. 1 3.66 and yield a simple-fluid entropy departure,
ine values for the entropy departure are shown graphically in Fig. D.3.
536 m CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
FIGURE 13.9 Real
and ideal gas states and
entropies.
EXAMPLE 13.6 Nitrogen at 8 MPa, 150 K, is throttled to 0.5 MPa. After ^ ^ K^DetenSn^the^eat
length of pipe, its temperature is measured and found to be 125 K. Determine me neai
transfer and the change of entropy using the generalized charts. Compare these results
with those obtained by using the nitrogen tables.
Control volume: Throttle and pipe.
Inlet state: P u T x known; state fixed.
Exit state: P 2t T 2 known; state fixed.
Process: Steady state.
Diagram: Figure 13.10.
Mode!: Generalized charts, results to be compared with those obtained
with nitrogen tables.
Analysis
There is no work done, and we neglect changes in kinetic and potential energies. There-
fore, per kilogram,
First law:
q + h x = h 2
q = h 2 -h = -(A? - h 2 ) + 0% - hi) + {hi - hi)
FIGURE 13.10
Sketch for Example 13.6.
P = 8 MPa
= 0.5 MPa
THE GENERALISED CHART FOR CHANGES OF ENTROPY AT CONSTANT TEMPERATURE H 537
Solution
Using values from Table A. 2, we have
From Fig. D.2,
^ = 3^2.36 ^=^=1.189
~ = 25
RT C ^
hf -h x = 2.5 X 0.2968 X 126.2 = 93.6kJ/kg
hf - h 2
= 0.15
RT C
ht~h 2 = 0.15 X 0,2968 X 126.2 - 5.6 kJ/kg
Assuming a constant specific heat for the ideal gas, we have
ht ~ /if = C^iTi - T x ) = 1.0416(125 - 150) = -26.0kJ/kg
q = -5.6 - 26.0 + 93.6 = 62.0kJ/kg
From the nitrogen tables, Table B.6, we can find the change of enthalpy directly.
q = k 2 -h x = 123.77 - 61.92 = 61.85 kJ/kg
To calculate the change of entropy using the generalized charts, we proceed as follows:
s 2 ~Si = -(4 >7i - s 2 ) + (y^ - sfa) + (sfa - Jj)
From Fig. D.3
S Pi.Ti S P„Tt
^ -=1.6
4i,t, ~ Sp t , Tl = 1-6 X 0.2968 = 0.475 kJ/kg K
~ = 0.1 X 0.2968 = 0.0297 kJ/kgK
538 II CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
Assuming a constant specific heat for the ideal gas, we have
= 1.0416 In HI - 0.2968 ln^
- 0.6330 kJ/kg K
s 2 - Sl = -0.0297 + 0.6330 + 0.475
= 1.078 kJ/kgK
From the nitrogen tables, Table B.6,
Sj _ Sl = -5.4282 - 4.3522 = 1.0760 kJ/kg K
13.9 developing Tables
OF THERMODYNAMIC PROPERTIES
FROM EXPERIMENTAL DATA
For a given pure substance, tables of thermodynamic properties can be developed from
experimental data in many ways. This section conveys some general principles and con-
cepts by considering only the liquid and vapor phases.
Let us assume that the following data for a pure substance have been obtained in the
laboratory.
1. Vapor-pressure data. That is, saturation pressures and temperatures have been mea-
sured over a wide range.
2. Pressure, specific volume, and temperature data in the vapor region. These data are
usually obtained by determining the mass of the substance in a closed vessel (which
means a fixed specific volume) and then measuring the pressure as the temperature
is varied. This is done for a large number of specific volumes.
3. Density of the saturated liquid and the critical pressure and temperature.
4. Zero-pressure specific heat for the vapor. This might be obtained either calorimetri-
cally or from spectroscopic data and statistical thermodynamics (see Appendix C).
From these data a complete set of thermodynamic tables for the saturated liquid,
saturated vapor, and superheated vapor can be calculated. The first step is to determine an
equation for the vapor-pressure curve that accurately fits the data. It may be necessary to
use one equation for one portion of the vapor-pressure curve and a different equation for
another portion.
One form of equation that has been used is
\nP^ = A + f+C\nT+DT
Once an equation has been found that accurately represents the data, the saturation pressure
for any given temperature can be found by solving this equation. Thus, the saturation pres-
Developing Tables of Thermodynamic Properties from Experlmental Data
ia 539
sures in Table B.l.l of the Steam Tables would be determined for the given temperatures.
The second step is to determine an equation of state for the vapor region that accurately rep-
resents the P-o-T data. There are many possible forms of the equation of state that may be
selected. The important considerations are that the equation of state accurately represents
the data, and that it be of such a form that the differentiations required can be performed.
That is, though it may be desirable to have an equation of state that is explicit in v, as a func-
tion of TandP, in order to use Eq. 13.27, such a representation is inherently not the most ac-
curate. Instead, the most accurate form is one explicit in P, as a function of Tand v.
Once an equation of state has been determined, the specific volume of superheated
vapor at given pressures and temperatures can be determined by solving the equation and
tabulating the results as in the superheat tables for steam, ammonia, and the other sub-
stances listed in the appendix. The specific volume of saturated vapor at a given tempera-
ture may be found by determining the saturation pressure from the vapor-pressure curve
and substituting this saturation pressure and temperature into the equation of state.
The procedure followed in determining enthalpy and entropy is best explained with
the aid of Fig. 13,11. Let us assume that the enthalpy and entropy of saturated liquid in
state 1 are zero. The enthalpy of saturated vapor in state 2 can be found from the Clapeyron
equation.
The left side of this equation is found by differentiating the vapor-pressure curve. The
specific volume of the saturated vapor is found by the procedure outlined in the last para-
graph, and it is assumed that the specific volume of the saturated liquid has been mea-
sured. Thus, the enthalpy of evaporation, hj gi can be found for this particular temperature,
and the enthalpy at state 2 is equal to the enthalpy of evaporation (since the enthalpy in
state 1 is assumed to be zero). The entropy at state 2 is readily found, since
From state 2 we proceed along this isotherm into the superheated vapor region. The
specific volume at 3 is found from the equation of state at this pressure (by iteration, since
sat
T{v s -v f )
T
FIGURE 13.11
Sketch showing procedure
for developing a table of
thermodynamic properties
from experimental data.
s
540 M Chapter thirteen thermodynamic relations
the equation is explicit in P, not u). The internal energy and entropy are calculated by in-
tegrating Eqs. 13.31 and 13.35, and the enthalpy is then calculated from its definition:
The properties at point 4 are found in exactly the same manner. Pressure P 4 is sufficiently
low that the real superheated vapor behaves essentially as an ideal gas (perhaps 1 kPa).
Thus, we use this constant-pressure line to make all temperature changes for our calcula-
tions, as, for example, to point 5. Since the specific heat is known as a function of
temperature, the enthalpy and entropy at 5 are found by integrating the ideal-gas relations
The properties at points 6 and 7 are found from those at 5 in the same manner as those at
points 3 and 4 were found from 2. (The saturation pressure P 7 is calculated from the
vapor-pressure equation.) Finally, the enthalpy and entropy for saturated liquid at point 8
are found from the properties at point 7 by applying the Ciapeyron equation.
Thus, values for the pressure, temperature, specific volume, enthalpy, entropy, and
internal energy of saturated liquid, saturated vapor, and superheated vapor can be tabulated
for the entire region for which experimental data were obtained. The accuracy of such a
table depends both on the accuracy of the experimental data and the degree to which the
equation for the vapor pressure and the equation of state represent the experimental data.
Finally, it should be noted that most present-day development of thermodynamic
property tables follows a somewhat different line, beginning with representation of the
Helmholtz function a (Eqs. 13.12 and 13.13) as an empirical function of T and u (or p).
This representation implicitly includes the ideal-gas specific heat. The function typically
includes 40 or 50 terms altogether, and the empirical constants are determined from spe-
cific heat terms and P-v-T data. Differentiating this expression with respect to v yields
the equation of state explicit in P (see Eq. 13.21), while differentiating with respect to T
yields s (also Eq. 13.21). Finally, u can be calculated from Eq. 13.12, and h from its defin-
ition. This approach to property calculation requires no mathematical integrations, such as
are required in Eqs. 13.31 and 13.35.
13.10 the property relation
for Mixtures
In Chapter 12 our consideration of mixtures was limited to ideal gases. There was no need
at that point for further expansion of the subject. We now continue this subject with a
view toward developing the property relations for mixtures. This subject will be particu-
larly relevant to our consideration of chemical equilibrium in Chapter 15.
h 2 = Ui-u 2 + P 3 u 3 - P z v 2
The Property Relation for Mixtures m 541
For a mixture, any extensive property X is a function of the temperature and pres-
sure of the mixture and the number of moles of each component. Thus, for a mixture of
two components,
X=f(T,P,n Al n B )
Therefore,
Since at constant temperature and pressure an extensive property is directly propor-
tional to the mass, Eq. 13.67 can be integrated to give
X TP ^ X A n A + X B n B (13.68)
where
Here X is defined as the partial molal property for a component in a mixture. It is
particularly important to note that the partial molal property is defined under conditions of
constant temperature and pressure.
The partial molal property is particularly significant when a mixture undergoes a
chemical reaction. Suppose a mixture consists of components^ and B, and a chemical re-
action takes place so that the number of moles of A is changed by dn A and the number of
moles of B by dn B . The temperature and the pressure remain constant. What is the change
in internal energy of the mixture during this process? From Eq. 13.67 we conclude that
dU ZP = U A dn A + U B dn B (13.69)
where U A and U B are the partial molal internal energy of A and B, respectively. Equation
13.69 suggests that the partial molal internal energy of each component can also be de-
fined as the internal energy of the component as it exists in the mixture.
In Section 13.3 we considered a number of property relations for systems of fixed
mass such as
dU = TdS~PdV
In this equation, temperature is the intensive property or potential function associated
with entropy, and pressure is the intensive property associated with volume. Suppose we
have a chemical reaction such as described in the last paragraph. How would we modify
this property relation for this situation? Intuitively, we might write the equation
dU =TdS~PdV+ n A dn A + ijl b dn B (13.70)
where fi A is the intensive property or potential function associated with n A , and similarly
fi B for n B . This potential function is called the chemical potential.
To derive an expression for this chemical potential, we examine Eq. 13.70 and con-
clude that it might be reasonable to write an expression for Um the form
U=f(S t V,n A , n B )
542 B CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
Therefore,
dU Jm ds+m) dv+lf] dn A + [m <*«
Since the expressions
h) "* (¥)
imply constant composition, it follows from Eq. 13.20 that
m = T and m) =-P
Thus
dU = TdS~PdV + (jg) dn A + (§) dn B (13.71)
On comparing this equation with Eq. 13.70, we find that the chemical potential can
be defined by the relation
We can also relate the chemical potential to the partial molat Gibbs function. We
proceed as follows.
G = U+PV- TS
dG = dU + PdV+VdP-TdS~SdT
Substituting Eq. 13.70 into this relation, we have
dG = S dT + V dP + fx A dn A + {x B dn B (13.73)
This equation suggests that we write an expression for G in the following form.
G=f(T,P, n A ,n B )
Proceeding as we did for a similar expression for internal energy, we have
dG = i d A dT-v( d A dP + (f) dn A + l^) dn B
\*T) PM ai +\9P)T M Wr.P.* Witt,,
When this equation is compared with Eq. 13.73, it follows that
Because partial molal properties are defined at constant temperature and pres-
sure, the quantities (3G/dn A ) T ^ g and {dGidn B ) T ^ A are the partial molal Gibbs functions
Pseudopure Substance Models for Real-Gas Mixtures n 543
for the two components. That is, the chemical potential is equal to the partial molal
Gibbs function.
^-feL; *-^teL (13 ' 74)
Although ft can also be defined in terms of other properties, such as inEq. 13.72, this ex-
pression is not the partial molal internal energy, since the pressure and temperature are
not constant .in this partial derivative. The partial molal Gibbs function is an extremely
important property in the thermodynamic analysis of chemical reactions, for at constant
temperature and pressure (the conditions under which many chemical reactions occur) it
is a measure of the chemical potential or the driving force that tends to make a chemical
reaction take place.
13.11 Pseudopure substance models
for Real-Gas Mixtures
A basic prerequisite to the treatment of real-gas mixtures in terms of pseudopure sub-
stance models is the concept and use of appropriate reference states. As an introduction to
this topic, let us consider several preliminary reference state questions for a pure sub-
stance undergoing a change of state, for which it is desired to calculate the entropy
change. We can express the entropy at the initial state 1 and also at the final state 2 in
terms of a reference state 0, in a manner similar to that followed when dealing with the
generalized-chart corrections. It follows that
*i = j + (sf tTo - s ) + (j* ri - sfc) + (s v - st (T ) (13.75)
*2 = *o + (s%t, ~ %> + (st 2 r, ~ sl T ) + {s 2 - sl T ) (13.76)
These are entirely general expressions for the entropy at each state in terms of an arbitrary
reference state value and a set of consistent calculations from that state to the actual de-
sired state. One simplification of these equations would result from choosing the reference
state to be a hypothetical ideal-gas state atP and T Q , thereby making the term
C*Vi " %> = (13.77)
in each equation, which results in
*o = J* (13.78)
It should be apparent that this choice is a reasonable one, since whatever value is chosen
for the correction term, Eq. 13.77, it will cancel out of the two equations when the change
S2 - Si is calculated, and the simplest value to choose is zero. In a similar manner, the
simplest value to choose for the ideal-gas reference value, Eq. 13.78, is zero, and we
would commonly do that if there are no restrictions on choice, such as occur in the case of
a chemical reaction.
Another point to be noted concerning reference states is related to the choice of P
and T . For this purpose, let us substitute Eqs. 13.77 and 13.78 into Eqs. 13.75 and 13.76,
544 B Chapter Thirteen thermodynamic relations
and also assume constant specific heat, such that those equations can be written in the
form
4 + C p0 In - R In (jg + - s% T ) (13-79)
W + q- m g) - R m (§) + - -fc) (13-80)
Since the choice for P and T a is arbitrary if there are no restrictions, such as would be the
case with chemical reactions, it should be apparent from examining Eqs. 13.79 and 13.80
that the simplest choice would be for
P = Pi or P 2 To^T, or T 2
It should be emphasized that inasmuch as the reference state was chosen as a hypothetical
ideal gas at P , r , Eq. 13.77, it is immaterial how the real substance behaves at that pres-
sure and temperature. As a result, there is no need to select a low value for the reference
state pressure P .
Let us now extend these reference state developments to include real-gas mixtures.
Consider the mixing process shown in Fig. 13.12, with the states and amounts of each
substance as given on the diagram. Proceeding with entropy expressions as was done ear-
lier, we have
- H + *U i« (£) ^ * ft) + < 5 > - ^ (1381)
H = St. + <U In (I) - 1 ta ft) + ft - -4,r,)s (13.82)
i + C^.„ft)-^nft) + ft-^. 03.83)
$3 ^mi
in which
^ = yA^+yA- *<y* inxi + ^ ^) ( 13 - 84)
c^=y^y s c p% ( 13 - 85 >
When Eqs. 13.81-13.83 are substituted into the equation for the entropy change,
the arbitrary reference values tf , sf , P , and 7 all cancel out of the result, which is, of
course necessary in view of their arbitrary nature. An ideal-gas entropy of mixing expres-
sion, the final term in Eq. 13.84, remains in the result, establishing, in effect, the mixture
reference value relative to its components. The remarks made earlier concerning the
choices for reference state and the reference state entropies apply in this situation as well.
FIGURE 13.12
Example of mixing
process.
Pure A
71 1
Pure B
"2
atP 2 ,T 2
.. Mixing ■
chamber
Real mix .
'at P 3 ,
PSEUDOPURE SUBSTANCE MODELS FOR REAL-GAS MIXTURES M 545
To summarize the development to this point, we find that a calculation of real mix-
ture properties, as, for example, using Eq. 13.83, requires the establishment of a hypothet-
ical ideal gas reference state, a consistent ideal-gas calculation to the conditions of the
real mixture, and finally a correction that accounts for the real behavior of the mixture at
that state. This last term is the only place that the real behavior is introduced, and this is
therefore the term that must be calculated by the pseudopure substance model to be used.
In treating a real-gas mixture as a pseudopure substance, we will follow two ap-
proaches to represent the P~v~T behavior: rise of the generalized charts and use of an ana-
lytical equation of state. With the generalized charts, we need to have a model that
provides a set of pseudocritical pressure and temperature in terms of the mixture compo-
nent values. Many such models have been proposed and utilized over the years, but the
simplest is that suggested by W. B. Kay in 1936, in which
CcXnl* =2,y l P e „ (n)mk = (13.86)
i i
This is the only pseudocritical model that we will consider in this chapter. Other models
are somewhat more complicated to evaluate and use, but considerably more accurate.
The other approach to be considered is that of using an analytical equation of state,
in which the equation for the mixture must be developed from that for the components. In
other words, for an equation in which the constants are known for each of the compo-
nents, we must develop a set of empirical combining rules that will then give a set of con-
stants for the mixture as though it were a pseudopure substance. This problem has been
studied for many equations of state, using experimental data for the real-gas mixtures, and
various empirical rules have been proposed. For example, for both the van der Waals
equation, Eq. 13.52, and the Redlich-Kwong equation, Eq. 13.57, the two pure substance
constants a and b are commonly combined according to the relations
= (l) cfi)^j b m = 2 ctb t (13.87)
The following example illustrates the use of these two approaches to treating real-gas
mixtures as pseudopure substances.
EXAMPLE 13,7 A mixture of 80% C0 2 and 20% CH 4 (mass basis) is maintained at 310.94 K, 86.19 bar,
at which condition the specific volume has been measured as 0.006757 m 3 /kg. Calculate
■, the percent deviation if the specific volume had been calculated by (a) Kay's rule and (b)
van der Waals' equation of state.
Control mass: Gas mixture.
State: P, u, T known.
Model: (a) Kay's rule, (b) van der Waals' equation.
Solution
Let subscript A denote C0 2 and B denote CH 4 ; then from Tables A.2 and A. 5
74 = 304. 1 K P Ci = 7.38 MPa R A = 0. 1 889 kJ/kg K
T Cb = 190.4 K P Cs = 4.60 MPa R B = 0.5 1 83 kJ/kg K
546 B CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
The gas constant from Eq. 12.15 becomes
R m = 2) cfc = 0-8 X 0.1889 + 0.2 X 0.5183 = 0.2548 kJ/kg K
and the mole fractions are
>>a = W2 (cM) = C o.8/44.0^) 8 + 4 (0.2/16.043) = °' 5932
^=1-^ = 0.4068
a. For Kay's rule, Eq. 13.86,
^ = 2^, = ^ + ^
= 0.5932(304.1) + 0.4068(190.4)
= 257.9 K
= 0.5932(7.38) + 0.4068(4.60)
= 6.249 MPa
Therefore, the pseudoreduced properties of the mixture are
= X= 31flLM =1.206
r - T £n 257.9
P = A =8^=1379
^ P Cn 6.249
From the generalized chart, Fig. D. 1
Z» = 0-7
and
= W = 0.7 X 0.2548 X 310.94 = 00fi435 3/kg
i> 8619
The percent deviation from the experimental value is
Percent deviation = (^06^206435) x 100 = 4.8%
The major factor contributing to this 5% error is the use of the linear Kay's rule
pseudocritical model, Eq. 13.86. Use of an accurate pseudocritical model and the
generalized chart would reduce the error to approximately 1%.
Pseudopure Substance Models for Real-Gas Mixtures B 547
b. For van der Waals' equation, the pure substance constants are
27R\2% kPa m 6
^la^ 18864 ^
b A = - 0.000973 m 3 /kg
and
= -h^ = 0-002682 m 3 /kg
Therefore, for the mixture, from Eq. 13.87,
= <0.8V0.18864 + 0.2VO8931) 2 = 0.2878 ^J?i!
kg
b m = c A b A + Cgbg
= 0.8 X 0.000973 + 0.2 X 0.002682 - 0.001315 rnVkg
The equation of state for the mixture of this composition is
P -
8619 - 0.2548 X 310.94 0.2878
v ~ 0.001315 w 2
Solving for v by trial and error,
v = 0.006326 m 3 /kg
Percent deviation = (^g^ 0026 ) X 100 = 6.4%
As a point of interest from the ideal-gas law, v = 0.00919 m 3 /kg, which is a devia-
tion of 36% from the measured value. Also, if we use the Redlich-Kwong equation of
state, and follow the same procedure as for the van der Waals equation, the calculated
specific volume of the mixture is 0.00652 mVkg, which is in error by 3 .5%.
We must be careful not to draw too general a conclusion from the results of this
example. We have calculated percent deviation in v at only a single point for only one
mixture. We do note, however, that the various methods used give quite different re-
sults. From a more general study of these models for a number of mixtures, we find that
the results found here are fairly typical, at least qualitatively. Kay's rule is very useful
548 B CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
because it is fairly accurate and yet relatively simple. The van der Waals equation is too
simplified an expression to accurately represent P-v-T behavior, but it is useful to
demonstrate the procedures followed in utilizing more complex analytical equations or
state. The Redlich-Kwong equation is considerably better and is still relatively simple
t0 US< As noted in the example, the more sophisticated generalized behavior models and
empirical equations of state will represent mixture P-u-T behavior to within abou I A
over a wide range of density, but they are of course more difficult to use than the methods
considered in Example 13.7. The generalized models have the advantage of being easier
to use and they are suitable for hand computations. Calculations with the complex empir-
ical equations of state become very involved, but have the advantage of expressing the
P-v-T composition relations in analytical form, which is of great value when usmg a dig-
ital computer for such calculations.
SUMMARY As an intro duction to the development of property information that can be obtained exper-
imentally, we derive the Clapeyron equation. This equation relates the slope of the two-
phase boundaries in the P-T diagram to the enthalpy and specific volume change going
from one phase to the other. If we measure pressure, temperature, and the specific vol-
umes for liquid and vapor in equilibrium, we can calculate the enthalpy of evaporation.
Because thermodynamic properties are functions of two variables, a number of relations
can be derived from the mixed second derivatives and the Gibbs relations, which are
known as Maxwell relations. Many other relations can be derived, and those that are use-
ful let us relate thermodynamic properties to those that can be measured directly like P, v,
r, and indirectly like the heat capacities.
Changes of enthalpy, internal energy, and entropy between two states are presented
as integrals over properties that can be measured and thus obtained from experimental
data Some of the partial derivatives are expressed as coefficients like expansivity and
compressibility, with the process as a qualifier like isothermal or isentropic (adiabatic).
These coefficients, as single numbers, are useful when they are nearly constant over some
range of interest, which happens for liquids and solids and thus are found in various hand-
books. The speed of sound is also a property that can be measured, and it relates to a par-
tial derivative in a nonlinear fashion.
The experimental information about a substance behavior is normally correlated in
an equation of state relating P-*-T to represent part of the thermodynamic surface. Start-
ing with the general compressibility and its extension to the virial equation of state, we
lead up to other more complex equations of state (EOS). We show the most versatile
equations such as the van der Waals EOS, the Redlich-Kwong EOS, and the -Lee-Keslei
EOS which is shown as an extension of Benedict-Webb-Rubin (BWR), with others that
are presented in Appendix D. The most accurate equations are too complex for hand cal-
culations and are used on computers to generate tables of properties. Therefore, we do not
cover those details. , ,
As an application of Lee-Kesler EOS for a simple fluid, we present the develop-
ment of the generalized charts that can be used for substances for which we do not
have a table. The charts express the deviation of the properties from an ideal gas m
terms of a compressibility factor (Z) and the enthalpy and entropy departure terms
These charts are in dimensionless properties based on the properties at the critical
point.
key concepts and formulas H 549
Properties for mixtures are introduced in general, and the concept of a partial molal
property leads to the chemical potential derived from the Gibbs function. Real mixtures
are treated on a mole basis, and we realize that a model is required to do so. We present a
pseudocritical model of Kay that predicts the critical properties for the mixture and then
uses the generalized charts. Other models predict EOS parameters for the mixture and
then use the EOS as for a pure substance. Typical examples here are the van der Waals
and Redlich-Kwong equations of state.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to:
• Apply and understand the assumptions for the Clapeyron equation.
• Use the Clapeyron equation for all three two-phase regions.
• Have a sense of what a partial derivative means.
• Understand why Maxwell relations and other relations are relevant.
• Know that the relations are used to develop expression for changes in h, u, and s.
• Know that coefficients of linear expansion and compressibility are common data
"useful for describing certain processes.
• Know that speed of sound is also a property.
• Be familiar with various equations of state and their use.
• Know the background for and how to use the generalized charts.
• Know that a model is needed to deal with a mixture.
• Know the pseudocritical model of Kay and the equation of state models for a mixture.
Key Concepts ™
AND FORMULAS ^P 6 * 10 * ec * uatl0 *
Maxwell relations
Change in enthalpy
Change in energy
Change in entropy
Virial equation
Van der Waals
dP^ t _^ }f _ ^'
~dT~ = T(v" — v'Y ^~^ anc * regions
dz = Mdx + Ndy (^\ =\ML
dP
dv
*-*--fc' ff+ r[ B - , M
*7«.-/, , w/ l 1 [r(g) > -p
= h 2 ~h l - (P 2 v 2 - P lVt )
— rw;(§)-
Z = A- 1+ C(T) D{T)
RT " v~ — ~ + — T~ ' ' ' ( mass basis)
P =
RT
^Zrjj-^ (mass basis)
550 ■ Chapter thirteen thermodynamic relations
Redlich Kwong
RT
(mass basis)
~ v - b v (v + b)7 i/1
Other equations of state See Appendix D.
Generalized charts for h h 2 - h = (** - A*W. " RT c(&h ' A/l i)
AA = (A* - h)IRT c \ h* value for ideal gas
j 2 - = (4 ~ Si)id.g. ~ R(^2 - AJ,)
As = (s* - s)fR; s* value for ideal gas
Pseudocritical pressure P c mi* = 2 ^ «
Pseudocritical temperature T c
mix X-> si* ci
i
Pseudopure substance « m = (x = 2 (mass basis)
Enthalpy departure
Generalized charts for j
Entropy departure
Concept-Study guide problems
13.1 Mention two uses of the Clapeyron equation.
13.2 The slope dPIdT of the vaporization line is finite
as you approach the critical point, yet h fg and u fg
both approach zero. How can that be?
13.3 In view of Clapeyron' s equation and Fig. 3.7, is
there something special about ice / versus the
other forms of ice?
13.4 If we take a derivative as (dPidT) v in the two-
phase region (see Figs. 3.18 and 3.19), does it
matter what v is? How about T?
13.5 Sketch on a P-T diagram how a constant v line
behaves in the compressed liquid region, the
two-phase L-V region, and the superheated
vapor region?
13.6 If I raise the pressure in an isentropic process,
does h go up or down? Is that independent upon
the phase?
13.7 If I raise the pressure in an isothermal process,'
does h go up or down for a liquid or solid? What
do you need to know if it is a gas phase?
13.8 The equation of state in Example 13.3 was used
as explicit in v. Is it explicit in P?
13.9 Over what range of states are the various coeffi-
cients in Section 13.5 most useful?
13.10 For a liquid or a solid, is v more sensitive to T or
PI How about an ideal gas?
13.11 If I raise the pressure in a solid at constant T t does
5 go up or down?
13.12 Most equations of state are developed to cover
which range of states?
13.13 Is an equation of state valid in the two-phase
regions?
13.14 As P -> 0, the specific volume v -» «. For P -»
does v —> 0?
13.15 Must an equation of state satisfy the two condi-
tions in Eqs. 13.50 and 13.51?
13.16 At which states are the departure terms for h and s
small? What is Z there?
13.17 What is the benefit of the generalized charts?
Which properties must be known besides the
charts themselves?
13.18 What does it imply if the compressibility factor is
larger than 1?
13.19 The departure functions for h and s as defined are
always positive. What does that imply for the
real substance h and s values relative to ideal-gas
values?
13.20 What is the benefit of Kay's rule versus a mixture
equation of state?
Homework Problems M 551
Homework Problems
Clapeyron Equation
13.21
13.22
13.23
13.24
13.25
13.26
13.27
13.28
13.29
A special application requires R-12 at -140°C. It
is known that the triple-point temperature is
— 157°C. Find the pressure and specific volume of
the saturated vapor at the required condition.
Ice (solid water) at -3°C, 100 kPa, is compressed
isothermally until it becomes liquid. Find the re-
quired pressure.
An approximation for the saturation pressure can be
In P^ t = A — BIT, where A and B are constants.
Which phase transition is that suitable for, and what
kind of property variations are assumed?
In a Carnot heat engine, the heat addition changes
the working fluid from saturated liquid to satu-
rated vapor at T t P. The heat rejection process oc-
curs at lower temperature and pressure {T ~ AT),
(P — AP). The cycle takes place in a piston cylin-
der arrangement where the work is boundary
work. Apply both the first and second law with
simple approximations for the integral equal to
work. Then show that the relation between AP
and AT results in the Clapeyron equation in the
limit Ar-> dT.
Calculate the values h fg and Sf g for nitrogen at
70 K and at 110 K from the Clapeyron equation,
using the necessary pressure and specific volume
values from Table B.6.1.
Ammonia at — 70°C is used in a special applica-
tion at a quality of 50%. Assume the only table
available is B.2 that goes down to -50°C. To size
a tank to hold 0.5 kg with x = 0.5, give your best
estimate for the saturated pressure and the tank
volume.
The saturation pressure can be approximated as -
In P^ t = A — B/T, where A and B are constants.
Use the steam tables and determine A and B from
properties at 25°C only. Use the equation to pre-
dict the saturation pressure at 30°C and compare
to table value.
Using the properties of water at the triple point,
develop an equation for the saturation pressure
along the fusion line as a function of temperature.
Helium boils at 4.22 K at atmospheric pressure,
101.3 kPa, with h fg =83.3 kJ/kmol. By pumping
a vacuum over liquid helium, the pressure can be
lowered, and it may then boil at a lower tempera-
ture. Estimate the necessary pressure to produce a
boiling temperature of 1 K and one of 0.5 K.
13.30 A certain refrigerant vapor enters a steady-flow
constant-pressure condenser at 150 kPa, 70°C, at
a rate of 1.5 kg/s, and it exits as saturated liquid.
Calculate the rate of heat transfer from the con-
denser. It may be assumed that the vapor is an
ideal gas and also that at saturation, v f < v g . The
following is known:
tnP g = 8.15 - 1000/r = 0.7 kj/kg K
with pressure in kPa and temperature in K. The
molecular weight is 100.
13.31 Using thermodynamic data for water from Tables
B.I.I and B.1.5, estimate the freezing tempera-
ture of liquid water at a pressure of 30 MPa.
13.32 Small solid particles formed in combustion
should be investigated. We would like to know
the sublimation pressure as a function of tempera-
ture. The only information available is 7*, h fs for
boiling at 101.3 kPa and T, h if for melting at 101.3
kPa. Develop a procedure that will allow a deter-
mination of the sublimation pressure, P^ h (T).
13.33 A container has a double wall where the wall cav-
ity is filled with carbon dioxide at room tempera-
ture and pressure. When the container is filled
with a cryogenic liquid at 100 K, the carbon diox-
ide will freeze so that the wall cavity has a mix-
ture of solid and vapor carbon dioxide at the
sublimation pressure. Assume that we do not have
data for C0 2 at 100 K, but it is known that at
-90°C: P mb = 38.1 kPa, h ig = 574.5 kj/kg. Esti-
mate the pressure in the wall cavity at 100 K.
Property Relations, Maxwell, and those for
Enthalpy, Internal Energy, and Entropy
13.34 Use Gibbs relation du-Tds-P dv and one of
Maxwell's relations to find an expression for
{du!dP) T that only has properties P, v, and T in-
volved. What is the value of that partial derivative
if you have an ideal gas?
13.35 Start from Gibbs relation dh = T ds + v dP and
use one of Maxwell's equations to get {dhldv) T m
552 H Chapter thirteen thermodynamic relations
terms of properties P, v, and T. Then use Eq.
13.24 to also find an expression for (dh!ST) v .
13.36 From Eqs. 13.23 and 13.24 and the knowledge
that C p > C„, what can you conclude about the
slopes of constant v and constant P curves in a
T-s diagram? Notice that we are looking at func-
tions T{s) P or T(s) v .
13.37 Derive expressions for {dT/dv) a and for {dhids) u
that do not contain the properties h, u, or s. Use
Eq. 13.30 with du = 0.
13.38 Develop an expression for the variation in tem-
perature with pressure in a constant entropy
process, {0TidP) v that only includes the proper-
ties P-v-T and the specific heat, C p . Follow the
development of Eq. 13.32.
13.39 Use Eq. 13.34 to get an expression for the deriva-
tive (dTidv) s . what is the general shape of a con-
stant s process curve in a T-v diagram? For an
ideal gas can you say a little more about the
shape?
13.40 Evaluate the isothermal changes in the internal
energy, the enthalpy, and the entropy for an ideal
gas. Confirm the results in Chapters 5 and 8.
Volume Expansivity and Compressibility
13.41 Determine the volume expansivity, a Pi and the
isothermal compressibility, p T , for water at
20°C, 5 MPa and at 300°C, 15 MPa using the
steam tables.
13.42 What are the volume expansivity a p , the isother-
mal compressibility j3 r , and the adiabatic com-
pressibility ft for an ideal gas?
13.43 Find the speed of sound for air at 20°C, 100 kPa,
using the definition in Eq. 13.43 and relations for
polytropic processes in ideal gases.
13.44 Assume a substance has uniform properties in all
directions with V = and show that volume
expansivity a p = 3S T . Hint: Differentiate with re-
spect to T and divide by V,
13.45 A cylinder fitted with a piston contains liquid
methanol at 20°C, 100 kPa, and volume 10 L. The
piston is moved, compressing the methanol to 20
MPa at constant temperature. Calculate the work
required for this process. The isothermal com-
pressibility of liquid methanol at 20°C is 1.22 X
10^m 2 /N.
13.46 Use Eq. 13.32 to solve for {dTidP) s in terms of T,
v, C and a p . How large a temperature change
does i? 25°C water {a p - 2.1 X 10 H KT 1 ) have,
when compressed from 100 kPa to 1000 kPa in an
isentropic process?
13.47 Sound waves propagate through media as pres-
sure waves that cause the media to go through
isentropic compression and expansion processes.
The speed of sound c is defined by c 2 = {dPldp) $
and it can be related to the adiabatic compressi-
bility, which for liquid ethanol at 20°C is 9.4 X
10^ 10 m 2 /N. Find the speed of sound at this
temperature.
13.48 For commercial copper at 25°C (see Table A.3),
the speed of sound is about 4800 m/s. What is the
adiabatic compressibility ft?
13.49 Consider the speed of sound as defined in Eq.
13.43. Calculate the speed of sound for liquid
water at 20°C, 2.5 MPa, and for water vapor at
200°C, 300 kPa, using the steam tables.
13.50 Soft rubber is used as a part of a motor mount-
ing. Its adiabatic bulk modulus is B s - 2.82 X
10 6 kPa, and the volume expansivity is a p =
4.86 X 10" 4 K" 1 . What is the speed of sound vi-
brations through the rubber, and what is the rel-
ative volume change for a pressure change of
1 MPa?
13.51 Liquid methanol at 25°C has an adiabatic com-
pressibility of 1.05 X 10~ 9 m 2 /N. What is the
speed of sound? If it is compressed from 100 kPa
to 10 MPa in an insulated piston/cylinder, what is
the specific work?
13.52 Use Eq. 13.32 to solve for {dTldP) s in terms of T,
v, C p and a p . How much higher does the temper-
ature become for the compression of the methanol
in Problem 13.51? Use a p = 2.4 X 10" 4 KT 1 for
methanol at 25°C.
Equations of State
13.53 Use the equation of state as shown in Example
13.3 where changes in enthalpy and entropy were
found. Find the isothermal change in internal en-
ergy in a similar fashion; do not .compute it from
enthalpy.
13.54 Evaluate changes in an isothermal process for u,
h, and s for a gas with an equation of state as
P(v ~b) = RT.
HOMEWORK PROBLEMS M 553
13.55 Two uninsulated tanks of equal volume are con-
nected by a valve. One tank contains a gas at a
moderate pressure P lt and the other tank is
evacuated. The valve is opened and remains
open for a long time. Is the final pressure P 2
greater than, equal to, or less than iV2? Hint:
Recall Fig. 13.5.
13.56 Determine the reduced Boyle temperature as pre-
dicted by an equation of state (the experimentally
observed value is about 2.5), using the van der
Waals equation and the Redlich-Kwong equation.
Note: It is helpful to use Eqs. 13.45 and 13.46 in
addition to Eq. 13.44.
13.57 Develop expressions for isothermal changes in in-
ternal energy, enthalpy, and entropy for a gas
obeying the van der Waals equation of state.
13.58 Develop expressions for isothermal changes in in-
ternal energy, enthalpy, and entropy for a gas
obeying the Redlich-Kwong equation of state.
13.59 Consider the following equation of state, ex-
pressed in terms of reduced pressure and tempera-
ture: 2=1 + {P r l\AT r )[\ - 6T; 2 ]. What does
this predict for the reduced Boyle temperature?
13.60 What is the Boyle temperature for the following
equation of state: P = RTIv - b - a/v 2 T where a
and b are constants.
13.61 Show that the van der Waals equation can be
written as a cubic equation in the compressibility
factor involving the reduced pressure and reduced
temperature as
13.62 Determine the second virial coefficient B{T) using
the van der Waals equation of state. Also find its
value at the critical temperature where the experi-
mentally observed value is about -0.34 RTJP C .
13.63 Determine the second virial coefficient B(T) using
the Redlich-Kwong equation of state. Also find
its value at the critical temperature where the ex-
perimentally observed value is about -0.34
RTJP C .
13.64 One early attempt to improve on the van der Waals
equation of state was an expression of the form
P = RT - _A_
v - b V 2 T
Solve for the constants a, b, and v c using the same
procedure as for the van der Waals equation.
13.65 Calculate the difference in internal energy of the
ideal-gas value and the real-gas value for carbon
dioxide at state 20°C, 1 MPa, as determined using
the virial equation of state, including second virial
coefficient terms. For carbon dioxide we have:
B = -0.128 mVkmol, T{dBidT) = 0.266
nrVkmol, both at 20°C.
13.66 Calculate the difference in entropy of the ideal-
gas value and the real-gas value for carbon diox-
ide at the state 20°C, 1 MPa, as determined using
the virial equation of state. Use numerical values
given in Problem 13.65.
13.67 A rigid tank contains 1 kg oxygen at 160 K, 4 MPa,
Determine the volume of the tank assuming we can
use the Redlich-Kwong equation of state for oxy-
gen. Compare the result with the ideal-gas law.
13.68 A flow of oxygen at 230 K, 5 MPa, is throttled to
100 KPa in a steady-flow process. Find the exit
temperature and the specific entropy generation
using the Redlich-Kwong equation of state and
ideal-gas heat capacity. Notice that this becomes
iterative due to the nonlinearity coupling h, P, v,
and 7*.
Generalized Charts
13.69 A 200-L rigid tank contains propane at 9 MPa,
280°C. The propane is then allowed to cool to
50°C as heat is transferred with the surroundings.
Determine the quality at the final state and the
mass of liquid in the tank, using the generalized
compressibility chart, Fig. D.l.
13.70 A rigid tank contains 5 kg of ethylene at 3 MPa,
30°C. It is cooled until the ethylene reaches the sat-
urated vapor curve. What is the final temperature?
13.71 Refrigerant- 123, dichlorotrifiuoroethane, which is
currently under development as a potential re-
placement for environmentally hazardous refrig-
erants, undergoes an isothermal steady-flow
process in which the R-123 enters a heat ex-
changer as saturated liquid at 40°C and exits at
100 kPa. Calculate the heat transfer per kilogram
of R-123, using the generalized charts, Fig. D.2.
13.72 An ordinary lighter is nearly full of liquid
propane with a small amount of vapor, the vol-
ume is 5 cm 3 , and temperature is 23°C. The
propane is now discharged slowly such that heat
554 H Chapter thirteen thermodynamic relations
transfer keeps the propane and valve flow at 23°C.
Find the initial pressure and mass of propane and
the total heat transfer to empty the lighter.
13.73 A piston/cylinder contains 5 kg of butane gas at
500 K, 5 MPa, The butane expands in a reversible
polytropic process to 3 MPa, 460 K. Determine
the polytropic exponent n and the work done dur-
ing the process.
13.74 Calculate the heat transfer during the process de-
scribed in Problem 13.73.
13.75 A cylinder contains ethylene, C 2 H 4 , at 1 .536 MPa,
- 13°C. It is now compressed in a reversible iso-
baric (constant P) process to saturated liquid.
Find the specific work and heat transfer.
13.76 Carbon dioxide collected from a fermentation
process at 5°C, 100 kPa, should be brought to 243
K, 4 MPa in a steady- flow process. Find the mini-
mum amount of work required and the heat trans-
fer. What devices are needed to accomplish this
change of state?
13.77 Consider the following equation of state, ex-
pressed in terms of reduced pressure and tempera-
ture: Z = 1 + (P r /14T r )[l - 6T; 2 ]. What does
this predict for the enthalpy departure at P r = 0.4
and T r = 0.9?
13.78 Consider the following equation of state, ex-
pressed in terms of reduced pressure and tempera-
ture: Z = 1 + (iV147;)[l - 677 2 ]. What does
this predict for the entropy departure at P r = 0.4
and T r = 0.9?
13.79 A flow of oxygen at 230 K, 5 MPa is throttled to
100 kPa in a steady-flow process. Find the exit
temperature and the entropy generation.
13.80 A cylinder contains ethylene, C 2 H 4) at 1.536 MPa,
— 13°C. It is now compressed isothermally in a re-
versible process to 5.12 MPa. Find the specific
work and heat transfer.
13.81 Saturated vapor R-22 at 30°C is throttled to 200
kPa in a steady-flow process. Calculate the exit
temperature assuming no changes in the kinetic
energy, using the generalized charts, Fig. D.2, and
the R-22 tables, Table B.4.
13.82 A 250-L tank contains propane at 30°C, 90%
quality. The tank is heated to 300°C. Calculate the
heat transfer during the process.
13.83 The new refrigerant fluid R-123 (see Table A.2) is
used in a refrigeration system that operates in the
ideal refrigeration cycle, except the compressor is
neither reversible nor adiabatic. Saturated vapor
at — 26.5°C enters the compressor, and super-
heated vapor exits at 65 °C. Heat is rejected from
the compressor as 1 kW, and the R-123 flow rate
is 0.1 kg/s. Saturated liquid exits the condenser
at 37.5°C. Specific heat for R-123 is = 0.6
kJ/kg K. Find the coefficient of performance.
13.84 An uninsulated piston/cylinder contains propene,
C 3 H 6 , at ambient temperature, 19°C, with a qual-
ity of 50% and a volume of 10 L. The propene
now expands slowly until the pressure drops to
460 kPa. Calculate the mass of propene, the work,
and heat transfer for this process.
13.85 A geothermal power plant on the Raft River uses
isobutane as the working fluid. The fluid enters
the reversible adiabatic turbine, as shown in Fig.
P13.85, at 160°C, 5.475 MPa, and the condenser
exit condition is saturated liquid at 33°C. Isobu-
tane has the properties T c = 408.14 K, P c = 3.65
MPa, Cfi = 1.664 kJ/kg K, and ratio of specific
heats k = 1.094 with a molecular weight as
58.124. Find the specific turbine work and the
specific pump work.
Ogeothemia!
A II I
Evaporator
vn
V! / Purnp
Condenser
FIGURE P13.85
13.86 A line with a steady supply of octane, C g H 18 , is at
400°C, 3 MPa. What is your best estimate for the
homework Problems 9 555
availability in a steady-flow setup where changes
in potential and kinetic energies may be neglected?
13.87 An insulated piston/cylinder contains saturated
vapor carbon dioxide at 0°C and a volume of
20 L. The external force on the piston is slowly
decreased, allowing the carbon dioxide to expand
until the temperature reaches -30°C. Calculate
the work done by the C0 2 during this process.
13.88 An evacuated 1O0-L rigid tank is connected to a
line flowing R-142b gas, chlorodifluoroethane, at
2 MPa, 100°C. The valve is opened, allowing the
gas to flow into the tank for a period of time, and
then it is closed. Eventually, the tank cools to am-
bient temperature, 20°C ) at which point it contains
50% liquid, 50% vapor, by volume. Calculate the
quality at the final state and the heat transfer for
the process. The ideal-gas specific heat of R- 142b
is C p = 0.787 kJ/kgK.
13.89 Saturated liquid ethane at 2.44 MPa enters a heat
exchanger and is brought to 611 K at constant
pressure, after which it enters a reversible adia-
batic turbine where it expands to 100 kPa. Find
the specific heat transfer in the heat exchanger,
the turbine exit temperature, and turbine work.
13.90 A control mass of 10 kg butane gas initially at
80°C, 500 kPa, is compressed in a reversible
isothermal process to one-fifth of its initial vol-
ume. What is the heat transfer in the process?
13.91 An uninsulated compressor delivers ethylene,
C 2 H 4 , to a pipe, D = 10 cm, at 10.24 MPa, 94°C,
and velocity 30 m/s. The ethylene enters the com-
pressor at 6.4 MPa, 20.5°C, and the work input re-
quired is 300 kj/kg. Find the mass flow rate, the
total heat transfer, and entropy generation, assum-
ing the surroundings are at 25°C.
13.92 A distributor of bottled propane, C 3 H S , needs to
bring propane from 350 K, 100 kPa, to saturated
liquid at 290 K in a steady-flow process. If this
should be accomplished in a reversible setup
given the surroundings at 300 K, find the ratio of
the volume flow rates V-JV out , the heat specific
transfer, and the work involved in the process.
13.93 The environmentally safe refrigerant R-152a is to
be evaluated as the working fluid for a heat pump
system that will heat a house. It uses an evapora-
tor temperature of -20°C and a condensing
temperature of 30°C. Assume all processes are
ideal and R-152a has a heat capacity of C p =
■ 0.996 kJ/kg K. Determine the cycle coefficient
of performance.
13,94 Rework the previous problem using an evaporator
temperature of 0°C.
Mixtures
13.95 A 2-kg mixture of 50% argon and 50% nitrogen
by mole is in a tank at 2 MPa, 1 80 K. How large
is the volume using a model of (a) ideal gas and
(b) Kay's rule with generalized compressibility
charts.
13.96 A 2-kg mixture of 50% argon and 50% nitrogen
by mass is in a tank at 2 MPa, 180 K. How large
is the volume using a model of (a) ideal gas and
(b) van der Waals equation of state with a, b for
a mixture?
13.97 A 2-kg mixture of 50% argon and 50% nitrogen
by mass is in a tank at 2 MPa, 1 80 K. How large
is the volume using a model of (a) ideal gas and
(b) Redlich-Kwong equation of state with a, b
for a mixture.
13.98 Saturated liquid ethane at 7/ E = 14°C is throttled
into a steady-flow mixing chamber at the rate of
0.25 kmol/s. Argon gas at T 2 = 25°C, 800 kPa,
enters the chamber at the rate 0.75 kmol/s. Heat
is transferred to the chamber from a constant
temperature source at 150°C at a rate such that a
gas mixture exits the chamber at T 3 = 120°C,
800 kPa. Find the rate of heat transfer and the
rate of entropy generation.
13.99 A modem jet engine operates so that the fuel is
sprayed into air at P, T higher than the fuel criti-
cal point. Assume we have a rich mixture of
50% n-octane and 50% air by moles at 500 K
and 3.5 MPa near the nozzle exit. Do I need to
treat this as a real-gas mixture, or is an ideal-gas
assumption reasonable? To answer, find Z and
the enthalpy departure for the mixture assuming
Kay's rule and the generalized charts.
13.100 A mixture of 60% ethylene and 40% acetylene
by moles is at 6 MPa, 300 K. The mixture flows
through a preheater where it is heated to 400 K
at constant P. Using the Redlich-Kwong equa-
tion of state with a, b for a mixture, find the inlet
specific volume. Repeat using Kay's rule and
the generalized charts.
556 H CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
13.101 For the previous problem, find the specific heat
transfer using Kay's rale and the generalized
charts.
13.102 One kmol/s of saturated liquid methane, CH 4 , at
1 MPa and 2 kmol/s of ethane, C 2 H 6 , at 250°C,
1 MPa, are fed to a mixing chamber with the re-
sultant mixture exiting at 50°C, 1 MPa. Assume
that Kay's rule applies to the mixture and deter-
mine the heat transfer in the process.
13.103 A piston/cylinder initially contains propane at
7 = -7"C ) quality 50%, and volume 10 L. A
valve connecting the cylinder to a line flowing
nitrogen gas at T t = 20°C, P t = 1 MPa, is
opened and nitrogen flows in. When the valve is
closed, the cylinder contains a gas mixture of
50% nitrogen, 50% propane, on a mole basis at
T 2 = 20°C, P 2 = 500 kPa. What is the cylinder
volume at the final state, and how much heat
transfer took place?
13.104 Consider the following reference state conditions:
the entropy of real saturated liquid methane at
- 100°C is to be taken as 100 kJ/kmol K, and the
entropy of hypothetical ideal gas ethane at
-100°C is to be taken as 200 kJ/kmol K. Calcu-
late the entropy per kmol of a real-gas mixture of
50% methane, 50% ethane (mole basis) at 20°C,
4 MPa, in terms of the specified reference state
values, and assuming Kay's rule for the real mix-
ture behavior.
13.105 A cylinder/piston contains a gas mixture, 50%
C0 2 and 50% C 2 H 6 (mole basis) at 700 kPa,
35°C, at which point the cylinder volume is 5 L.
The mixture is now compressed to 5.5 MPa in a
reversible isothermal process. Calculate the heat
transfer and work for the process, using the fol-
lowing model for the gas mixture:
a. Ideal-gas mixture.
b. Kay's rule and the generalized charts.
13.106 A cylinder/piston contains a gas mixture, 50%
C0 2 and 50% C 2 H 6 (mole basis) at 700 kPa,
35°C, at which point the cylinder volume is 5 L.
The mixture is now compressed to 5.5 MPa in a
reversible isothermal process. Calculate the heat
transfer and work for the process, using the fol-
lowing model for the gas mixture:
a. Ideal-gas mixture.
b. The van der Waals equation of state.
Review Problems
13.107 Consider a straight line connecting the point
P = o, Z ~ 1 to the critical point P - P Ci Z = Z c
on a Z versus P compressibility diagram. This
straight line will be tangent to one particular
isotherm at low pressure. (The experimentally
determined value is about 0.8 T e .) Determine
what value of reduced temperature is predicted
by an equation of state, using the van der Waals
equation and the Redlich-Kwong equation. See
also note for Problem 13.56.
13.108 A 200-L rigid tank contains propane at 400 K,
3.5 MPa. A valve is opened, and propane flows
out until half the initial mass has escaped, at
which point the valve is closed. During this
process the mass remaining inside the tank ex-
pands according to the relation Pv lA - constant.
Calculate the heat transfer to the tank during the
process.
13.109 A newly developed compound is being consid-
ered for use as the working fluid in a small
Rankine-cycle power plant driven by a supply of
waste heat. Assume the cycle is ideal, with satu-
rated vapor at 200°C entering the turbine and
saturated liquid at 20°C exiting the condenser.
The only properties known for this compound
are molecular weight of 80 kg/kmol, ideal-gas
heat capacity C p0 = 0.80 kJ/kg K and T c = 500
K, P c = 5 MPa. Calculate the work input, per
kilogram, to the pump and the cycle thermal
efficiency.
13.110 A piston/cylinder contains propane initially at
67°C and 50% quality with a volume of 2 L. The
piston cross-sectional area is 0,2 m 2 , The exter-
nal force on the piston is gradually reduced to a
final value of 85 kN during which process the
propane expands to ambient temperature, 4°C.
Any heat transfer to the propane comes from a
constant-temperature reservoir at 67°C, while
any heat transfer from the propane goes to the
ambient. It is claimed that the propane does 30
kJ of work during the process. Does this violate
the second law?
13.111 One kilogram per second water enters a solar
collector at 40°C and exits at 190°C, as shown in
Fig. P. 13. 111. The hot water is sprayed into a
direct-contact heat exchanger (no mixing of the
HOMEWORK PROBLEMS ■ 557
Hot
water
Vapor
butane
Solar
collector
Heat,
exchanger
Water
out
FIGURE P13.111
Liquid
butane
Pump
Condenser -•••£> -Q^^
■W P
two fluids) used to boil the liquid butane. Pure
saturated-vapor butane exits at the top at 80°C
and is fed to the turbine. If the butane condenser
temperature is 30°C and the turbine and pump
isentropic efficiencies are each 80%, determine
the net power output of the cycle.
13.112 A piston/cylinder contains ethane gas initially at
500 kPa, 100 L, and at ambient temperature 0°C.
The piston is moved, compressing the ethane
until it is at 20°C with a quality of 50%. The
work required is 25% more than would have
been needed for a reversible polytropic process
between the same initial and final states. Calcu-
late the heat transfer and the net entropy change
for the process.
13.113 An experiment is conducted at — 100°C inside a
rigid sealed tank containing liquid R-22 with a
small amount of vapor at the top. When the ex-
periment is done, the container and the R-22
warm up to room temperature of 20°C. What is
the pressure inside the tank during the experi-
ment? If the pressure at room temperature
should not exceed 1 MPa, what is the maximum
percent of liquid by volume that can be used
during the experiment?
13.114 The refrigerant R-152a, difluoroethane, is tested
by the following procedure. A 1 0-L evacuated
tank is connected to a line flowing saturated-
vapor R-152a at 40°C. The valve is opened, and
the fluid flows in rapidly, so the process is essen-
tially adiabatic. The valve is to be closed when
the pressure reaches a certain value P 2f and the
tank will then be disconnected from the line.
After a period of time, the temperature inside the
• tank will return to ambient temperature, 25°C,
through heat transfer with the surroundings. At
this time, the pressure inside the tank must be
500 kPa. What is the pressure P 2 at which the
valve should be closed during the filling
process? The ideal-gas specific heat of R-152a is
C p - 0.996 kJ/kg K.
13.115 Carbon dioxide gas enters a turbine at 5 MPa,
100°C, and exits at 1 MPa. If the isentropic
efficiency of the turbine is 75%, determine the
exit temperature and the second-law efficiency.
13.116 A 4-m 3 uninsulated storage tank, initially
evacuated, is connected to a line flowing
ethane gas at 10 MPa, 100°C. The valve is
opened, and ethane flows into the tank for a
period of time, after which the valve is closed.
Eventually, the whole system cools to ambient
temperature, 0°C, at which time it contains
one-fourth liquid and three-fourths vapor, by
volume. For the overall process, calculate the
heat transfer from the tank and the net change
of entropy.
13.117 A 10-m 3 storage tank contains methane at low
temperature. The pressure inside is 700 kPa,
and the tank contains 25% liquid and 75%
vapor, on a volume basis. The tank warms very
slowly because heat is transferred from the
ambient.
a. What is the temperature of the methane when
the pressure reaches 10 MPa?
b. Calculate the heat transferred in the process,
using the generalized charts.
c. Repeat parts (a) and (b), using the methane
tables, Table B.7. Discuss the differences in
the results.
13.118 A gas mixture of a known composition is re-
quired for the calibration of gas analyzers. It is
desired to prepare a gas mixture of 80% ethyl-
ene and 20% carbon dioxide (mole basis) at 1
MPa, 25°C in an uninsulated, rigid 50-L tank.
The tank is initially to contain C0 2 at 25°C and
some pressure P x . The valve to a line flowing
C 2 H 4 at 25°C, 10 MPa, is now opened slightly
and remains open until the tank reaches 10
MPa, at which point the temperature can be as-
sumed to be 25°C. Assume that the gas mixture
558 M CHAPTER THIRTEEN THERMODYNAMIC RELATIONS
so prepared can be represented by Kay's rule
and the generalized charts. Given the desired
final state, what is the initial pressure of the car-
bon dioxide, P{!
ENGLISH UNIT PROBLEMS
13.120E A special application requires R-22 at - 150 F.
It is known that the triple-point temperature is
less than - 1 50 F. Find the pressure and spe-
cific volume of the saturated vapor at the re-
quired condition.
13.121E Ice (solid water) at 27 F, 1 atm, is compressed
isothermally until it becomes liquid. Find the
required pressure.
13.122E The saturation pressure can be approximated
as In P^ = A - BIT, where A and B are con-
stants. Use the steam tables and determine A
and B from properties at 70 F only. Use the
equation to predict the saturation pressure at 80
F and compare to table value.
13.123E Using thermodynamic data for water from Ta-
bles F.7.1 and F.7.4, estimate the freezing tem-
perature of liquid water at a pressure of 5000
lbf/in. 2 .
13.124E Determine the volume expansivity, a p , and the
isothermal compressibility, j3 r , for water at
50 F, 500 lbf/in. 2 and at 500 F, 1500 lbf/in. 2
using the steam tables.
13.125E A cylinder fitted with a piston contains liquid
methanol at 70 F, 15 lbf/in. 2 and volume 1 ft 3 .
The piston is moved, compressing the meth-
anol to 3000 lbf/in. 2 at constant temperature.
Calculate the work required for this process.
The isothermal compressibility of liquid meth-
anol at 70 F is 8.3 X 10^ 6 in. 2 /lbf.
13.126E Sound waves propagate through media as pres-
sure waves that cause the media to go through
isentropic compression and expansion pro-
cesses. The speed of sound c is defined by c 2 =
(dP/dp) s , and it can be related to the adiabatic
compressibility, which for liquid ethanol at
70 F is 6.4 X 10" 6 in 2 /lbf. Find the speed of
sound at this temperature.
13.127E Consider the speed of sound as defined in Eq.
13.43. Calculate the speed of sound for liquid
water at 50 F, 250 lbf/in. 2 , and for water vapor
at 400 F, 80 lbf/in. 2 , using the steam tables.
13.119 Determine the heat transfer and the net entropy
change in the previous problem. Use the initial
pressure of the carbon dioxide to be 4.56 MPa
before the ethylene is flowing into the tank.
13.128E Liquid methanol at 77 F has an adiabatic com-
pressibility of 7.1 X 10" 6 inVlbf. What is the
speed of sound? If it is compressed from 15 psia
to 1500 psia in an insulated piston/cylinder,
what is the specific work?
13.129E Calculate the difference in internal energy of
the ideal-gas value and the real-gas value
for carbon dioxide at the state 70 F, 150
lbf/in. 2 , as determined using the virial equation
of state. At this state B = -2.036 ftVfb mol,
T(dBldT) = 4.236 ftVlb mol.
13.130E A 7-ft 3 rigid tank contains propane at 1300
lbf/in. 2 , 540 F. The propane is then allowed to
cool to 120 F as heat is transferred with the
surroundings. Determine the quality at the final
state and the mass of liquid in the tank, using
the generalized compressibility chart,
13.131E A rigid tank contains 5 Ibm of ethylene at 450
lbf/in. 2 , 90 F. It is cooled until the ethylene
reaches the saturated vapor curve. What is the
final temperature?
13.132E A piston/cylinder contains 10 lbm of butane
gas at 900 R, 750 lbf/in. 2 . The butane expands
in a reversible polytropic process to 450 lbf/in 2
and 820 R. Determine the polytropic exponent
and the work done during the process.
13.133E Calculate the heat transfer during the process
described in Problem 13.132E.
13.134E A cylinder contains ethylene, C 2 H 4 , at 222.6
lbf/in. 2 , 8 F. It is now compressed in a reversible
isobaric (constant P) process to saturated liquid.
Find the specific work and heat transfer.
13.135E Carbon dioxide collected from a fermentation
process at 40 F, 15 lbf/in. 2 should be brought to
438 R, 590 lbf/in. 2 , in a steady-flow process.
Find the minimum amount of work required
and the heat transfer. What devices are needed
to accomplish this change of state?
13.136E Saturated vapor R-22 at 90 F is throttled to 30
lbf/in. 2 in a steady-flow process. Calculate the
exit temperature assuming no changes in the
kinetic energy, using the generalized charts,
Fig. D.2, and repeat using the R-22 tables,
Table F.9.
13.137E A 10-ft 3 tank contains propane at 90 F, 90%
quality. The tank is heated to 600 F. Calculate
the heat transfer during the process.
13.138E A cylinder contains ethylene, C 2 H 4 , at 222.6
lbf/in. 2 , 8 F. It is now compressed isothermally
in a reversible process to 742 lbf/in. 2 . Find the
specific work and heat transfer.
13.139E A geothermal power plant on the Raft River
uses isobutane as the working fluid as shown
in Fig. PI 3.85. The fluid enters the reversible
adiabatic turbine at 320 F, 805 lbf/in. 2 , and
the condenser exit condition is saturated
liquid at 91 F. Isobutane has the properties
T c = 734.65 R, P c = 537 lbf/in. 2 , C p0 =
0.3974 Btu/lbm R and ratio of specific heats
k = 1.094 with a molecular weight as 58,124.
Find the specific turbine work and the spe-
cific pump work.
13.140E A line with a steady supply of octane, C g H 18 , is
at 750 F, 440 lbf/in. 2 . What is your best esti-
mate for the availability in a steady-flow setup
where changes in potential and kinetic energies
may be neglected?
13. 14 IE A control mass of 10 Ibm butane gas initially
at 180 F, 75 lbf/in, 2 , is compressed in a re-
versible isothermal process to one-fifth of its
initial volume. What is the heat transfer in the
process?
Computer, design, and Open-Ended Problems M 559
13.142E A distributor of bottled propane, C 3 H S) needs to
bring propane from 630 R, 14.7 lbf/in. 2 to satu-
rated liquid at 520 R in a steady-flow process.
If this should be accomplished in a reversible
setup given the surroundings at 540 R, find the
ratio of the volume flow rates VJV oui , the heat
transfer, and the work involved in the process.
13.143E A 4-lbm mixture of 50% argon and 50% nitro-
gen by mole is in a tank at 300 psia, 320 R.
How large is the volume using a model of (a)
ideal gas and (b) Kay's rule with generalized
compressibility charts,
13.144E A 7-ft 3 rigid tank contains propane at 730 R,
500 lbf/in. 2 . A valve is opened, and propane
flows out until half the initial mass has es-
caped, at which point the valve is closed. Dur-
ing this process the mass remaining inside the
tank expands according to the relation Pv XA ~
constant. Calculate the heat transfer to the tank
during the process.
13.145E A newly developed compound is being consid-
ered for use as the working fluid in a small
Rankine-cycle power plant driven by a supply
of waste heat. Assume the cycle is ideal, with
saturated vapor at 400 F entering the turbine
and saturated liquid at 70 F exiting the. con-
denser. The only properties known for this
compound are molecular weight of 80
lbm/lbmo], ideal-gas heat capacity C p0 = 0.20
Btu/lbm R and T c = 900 R, P c = 750 lbf/in. 2 .
Calculate the work input, per Ibm, to the pump
and the cycle thermal efficiency.
Computer, design, and Open-Ended Problems
13.146 Write a program to obtain a plot of pressure ver-
sus specific volume at various temperatures (all
on a generalized reduced basis) as predicted by
the van der Waals equation of state. Tempera-
tures less than the critical temperature should be
included in the results.
13.147 We wish to determine the isothermal compressi-
bility, jS r , for a range of states of liquid water. Use
the menu-driven software or write a program to
determine this at a pressure of 1 MPa and at 25
MPa for temperatures of 0°C, 100°C, and 300°C.
13.148 Consider the small Rankine-cycle power plant
in Problem 13.109. What single change would
you suggest to make the power plant more
realistic?
13.149 Supercritical fluid chromatography is an experi-
mental technique for analyzing compositions of
mixtures. It utilizes a carrier fluid, often C0 2 , in
the dense fluid region just above the critical tem-
perature. Write a program to express the fluid
density as a function of reduced temperature and
pressure in the region of 1.0 < T r < 1.2 in re-
duced temperature and 2 < P r < 8 in reduced
pressure. The relation should be an expression
curve-fitted to values consistent with the gener-
alized compressibility charts.
560 O Chapter Thirteen thermodynamic Relations
13.150 It is desired to design a portable breathing sys-
tem for an average-sized adult. The breather will
store liquid oxygen sufficient for a 24-hour sup-
ply, and include a heater for delivering oxygen
gas at ambient temperature. Determine the size
of the system container and the heat exchanger.
13.151 Liquid nitrogen is used in cryogenic experi-
ments and applications where a nonoxidizing
gas is desired. Size a tank to hold 500 kg to be
placed next to a building and estimate the size of
an environmental (to atmospheric air) heat ex-
changer that can deliver nitrogen gas at a rate of
10 kg/h at roughly ambient temperature.
13.152 List a number of requirements for a substance
that should be used as the working fluid in a re-
frigerator. Discuss the choices and explain the
requirements.
13.153 The speed of sound is used in many applica-
tions. Make a list of the speed of sound at P Q , T
for gases, liquids, and solids. Find at least three
different substances for each phase. List a num-
ber of applications where knowledge about the
speed of sound can be used to estimate other
quantities of interest.
13.154 Propane is used as a fuel distributed to the end
consumer in a steel bottle. Make a list of design
specifications for these bottles and give charac-
teristic sizes and the amount of propane they can
hold.
13.155 Carbon dioxide is used in soft drinks and comes
in a separate bottle for large-volume users such
as restaurants. Find typical sizes of these, the
pressure they should withstand, and the amount
of carbon dioxide they can hold.
Chemical reactions
Many thermodynamic problems involve chemical reactions. Among the most familiar
of these is the combustion of hydrocarbon fuels, for this process is utilized in most of
our power-generating devices. However, we can all think of a host of other processes
involving chemical reactions, including those that occur in the human body.
This chapter considers a first- and second-law analysis of systems undergoing a
chemical reaction. In many respects, this chapter is simply an extension of our previous
consideration of the first and second laws. However, a number of new terms are intro-
duced, and it will also be necessary to introduce the third law of thermodynamics.
In this chapter the combustion process is considered in detail. There are two rea-
sons for this emphasis. First, the combustion process is of great significance in many
problems and devices with which the engineer is concerned. Second, the combustion
process provides an excellent vehicle for teaching the basic principles of the thermody-
namics of chemical reactions. The student should keep both of these objectives in mind
as the study of this chapter progresses.
Chemical equilibrium will be considered in Chapter 15 and, therefore, the matter of
dissociation will be deferred until then.
14.1 Fuels
A thermodynamics textbook is not the place for a detailed treatment of fuels. How-
ever, some knowledge of them is a prerequisite to a consideration of combustion, and
this section is therefore devoted to a brief discussion of some of the hydrocarbon
fuels. Most fuels fall into one of three categories — coal, liquid hydrocarbons, or gaseous
hydrocarbons.
Coal consists of the remains of vegetation deposits of past geologic ages, after sub-
jection of biochemical actions, high pressure, temperature, and submersion. The charac-
teristics of coal vary considerably with location, and even within a given mine there is
some variation in composition.
The analysis of a sample of coal is given on one of two bases: the proximate analy-
sis specifies, on a mass basis, the relative amounts of moisture, volatile matter, fixed car-
bon, and ash; the ultimate analysis specifies, on a mass basis, the relative amounts of
carbon, sulfur, hydrogen, nitrogen, oxygen, and ash. The ultimate analysts may be given
on an "as-received" basis or on a dry basis. In the latter case, the ultimate analysis does
not include the moisture as determined by the proximate analysis.
A number of other properties of coal are important in evaluating a coal for a given
use. Some of these are the fusibility of the ash, the grindabiiity or ease of pulverization,
the weathering characteristics, and size.
561
562 Ll CHAPTER FOURTEEN CHEMICAL REACTIONS
Table 14.1
Characteristics of Some of the Hydrocarbon Families
Family
Formula
Structure
Saturated
Paraffin
Chain
Yes
Olefin
Chain
No
Diolefin
Chain
No
Naphthene
Ring
Yes
Aromatic
Benzene
Ring
No
Naphthene
Ring
No
Most liquid and gaseous hydrocarbon fuels are a mixture of many different hydro-
carbons. For example, gasoline consists primarily of a mixture of about 40 hydrocarbons,
with many others present in very small quantities. In discussing hydrocarbon fuels, there-
fore, brief consideration should be given to the most important families of hydrocarbons,
which are summarized in Table 14.1.
Three concepts should be defined. The first pertains to the structure of the mole-
cule. The important types are the ring and chain structures; the difference between
the two is illustrated in Fig. 14.1. The same figure illustrates the definition of saturated
and unsaturated hydrocarbons. An unsaturated hydrocarbon has two or more adjacent
carbon atoms joined by a double or triple bond, whereas in a saturated hydrocarbon
all the carbon atoms are joined by a single bond. The third term to be defined is an
isomer. Two hydrocarbons with the same number of carbon and hydrogen atoms
and different structures are called isomers. Thus, there are several different octanes
(C 8 H 18 ), each having 8 carbon atoms and 18 hydrogen atoms, but each with a different
structure.
The various hydrocarbon families are identified by a common suffix. The com-
pounds comprising the paraffin family all end in "-ane" (as propane and octane). Simi-
larly, the compounds comprising the olefin family end in "-ylene" or "-ene" (as propene
and octene), and the diolefin family ends in "-diene" (as butadiene). The naphthene family
has the same chemical formula as the olefin family but has a ring rather than chain struc-
ture. The hydrocarbons in the naphthene family are named by adding the prefix "cyclo-"
(as cyclopentane).
The aromatic family includes the benzene series (QH^) and the naphthalene se-
ries (CA- 12 ). The benzene series has a ring structure and is unsaturated.
FIGURE 14.1
some hydrocarbon fuels.
HU V V VV
H-C-C-C-C-H H~C-C=C-C-H <\ JX
A H H H H H H H ^ )< H
H H
Molecular structure of Chain structure Chain structure Ring structure
saturated unsaturated saturated
Fuels W 563
Alcohols are sometimes used as a fuel in internal combustion engines. Characteris-
tically, in the alcohol family, one of the hydrogen atoms is replaced by an OH radical.
Thus, methyl alcohol, also called methanol, is CH 3 OH.
Most liquid hydrocarbon fuels are mixtures of hydrocarbons that are derived
from crude oil through distillation and cracking processes. Thus, from a given crude
oil, a variety of different fuels can be produced, some of the common ones being gaso-
line, kerosene, diesel fuel, and fuel oil. Within each of these classifications there is a
wide variety of grades, and each is made up of a large number of different hydrocar-
bons. The important distinction between these fuels is the distillation curve, Fig. 14.2.
The distillation curve is obtained by slowly heating a sample of fuel so that it vapor-
izes. The vapor is then condensed and the amount measured. The more volatile hydro-
carbons are vaporized first, and thus the temperature of the nonvaporized fraction
increases during the process. The distillation curve, which is a plot of the temperaUire
of the nonvaporized fraction versus the amount of vapor condensed, is an indication of
the volatility of the fuel.
For the combustion of liquid fuels, it. is convenient to express the composition in
terms of a single hydrocarbon, even though it is a mixture of many hydrocarbons. Thus,
gasoline is usually considered to be octane, C 8 H IS , and diesel fuel is considered to be do-
decane, C I2 H 26 . The composition of a hydrocarbon fuel may also be given in terms of per-
centage of carbon and hydrogen.
The two primary sources of gaseous hydrocarbon fuels are natural gas wells and
certain chemical manufacturing processes. Table 14.2 gives the composition of a number
of gaseous fuels. The major constituent of natural gas is methane, which distinguishes it
from manufactured gas.
At the present time, considerable effort is being devoted to develop more economi-
cal processes for producing gaseous and also liquid hydrocarbon fuels from coal, oil
shale, and tar sands deposits. Several alternative techniques have been demonstrated to be
feasible, and these resources promise to provide an increasing proportion of our fuel sup-
ply in future years.
564 B CHAPTER FOURTEEN CHEMICAL REACTIONS
Table 14.2
Volumetric Analys es of Some Typical Gaseous Fuels
Methane
Ethane
Propane
Butanes plus"
Ethene
Benzene
Hydrogen
Nitrogen
Oxygen
Carbon monoxide
Carbon dioxide
Various Natural Gases
A
B
C
D
93.9
60.1
67.4
54.3
3.6
14.8
16.8
16.3
1.2
13.4
15.8
16.2
1.3
4.2
7.4
7.5
5.8
Producer
Gas from
Bituminous
Coal
14.0
50.9
0.6
27.0
4.5
Carbureted
Water
Gas
10.2
6.1
2.8
40.5
2.9
0.5
34.0
3.0
Coke-
Oven
Gas
32.1
3.5
0.5
46.5
8.1
0.8
6.3
2.2
'This includes butane and all heavier hydrocarbons
14,2 THE COMBUSTION PROCESS
The combustion process consists of the oxidation of constituents in the fuel that are capa-
ble of bebg oxidized and can therefore be represented by a chemical equation. During
combustion process, the mass of each element remains the same. Thus, writing chemical
Ztns and solving problems concerning quantities of the various constituents basically
involve the conservation of mass of each element. This chapter presents a brief review of
this subject, particularly as it applies to the combustion process.
Consider first the reaction of carbon with oxygen.
Reactants Products
c + o 2 -> co 2
This equation states that 1 kmol of carbon reacts with 1 kmol of oxygen to form 1 kmol of
carbon d oxide. This also means that 12 kg of carbon react with 32 kg of oxygen to form
44 kg of carbon dioxide. All the initial substances that undergo the combustion proce
are called the reactants, and the substances that result from the combustion process are
^"^meratyLcarbon fuel is burned, both the carbon and the hydrogen are oxidized.
Consider the combustion of methane as an example.
CH 4 + 20 2 ->C0 2 + 2H 2 (14-1)
Here the products of combustion include both carbon dioxide and water The water may
be in the vapor, liquid, or solid phases, depending on the temperature and.pressure of the
^tT^L process, many intermediate products are formed during the
chemical reaction. In this book we are concerned with the initial and final products and
the Combustion process M 565
not with the intermediate products, but this aspect is very important in a detailed consider-
ation of combustion.
In most combustion processes, the oxygen is supplied as air rather than as pure oxy-
gen. The composition of air on a molal basis is approximately 21% oxygen, 78% nitro-
gen, and 1% argon. We assume that the nitrogen and the argon do not undergo chemical
reaction (except for dissociation, which will be considered in Chapter 15). They do leave
at the same temperature as the other products, however, and therefore undergo a change
of state if the products are at a temperature other than the original air temperature. At the
high temperatures achieved in internal-combustion engines, there is actually some reac-
tion between the nitrogen and oxygen, and this gives rise to the air pollution problem as-
sociated with the oxides of nitrogen in the engine exhaust.
In combustion calculations concerning air, the argon is usually neglected, and the
air is considered to be composed of 21% oxygen and 79% nitrogen by volume. When this
assumption is made, the nitrogen is sometimes referred to as atmospheric nitrogen. At-
mospheric nitrogen has a molecular weight of 28.16 (which takes the argon into account)
as compared to 28.013 for pure nitrogen. This distinction will not be made in this text, and
we will consider the 79% nitrogen to be pure nitrogen.
The assumption that air is 21.0% oxygen and 79.0% nitrogen by volume leads to the
conclusion that for each mole of oxygen, 79.0/21.0 = 3.76 moles of nitrogen are in-
volved. Therefore, when the oxygen for the combustion of methane is supplied as air, the
reaction can be written
CH 4 + 20 2 + 2(3.76)N 2 C0 2 + 2H 2 + 7.52N 2 (14.2)
The minimum amount of air that supplies sufficient oxygen for the complete com-
bustion of all the carbon, hydrogen, and any other elements in the fuel that may oxidize is
called the theoretical air. When complete combustion is achieved with theoretical air, the
products contain no oxygen. A general combustion reaction with a hydrocarbon fuel and
air is thus written
CJty + v 0i (0 2 + 3.76N 2 ) v COi C0 2 + v Ui0 H 2 + v N N 2 (14.3)
with the coefficients to the substances called stoichiometric coefficients. The balance of
atoms yields the theoretical amount of air as
C: v COi = x
H: 2v Hfi =y
N 2 : v Ni = 3.76 X v 0i
2 : v 0i = v C o 2 + v Hj0 /2 - x + y/4
and the total number of moles of air for 1 mole of fuel becomes
"air = v 0i X 4.76 = 4.76(x + y4)
This amount of air is equal to 100% theoretical air. In practice, complete combustion is
not likely to be achieved unless the amount of air supplied is somewhat greater than the
theoretical amount. Two important parameters often used to express the ratio of fuel
and air are the air-fuel ratio (designed AF) and its reciprocal, the fuel-air ratio (desig-
566 9 CHAPTER FOURTEEN CHEMICAL REACTIONS
nated FA). These ratios are usually expressed on a mass basis, but a mole basis is used
at times.
(M - 5)
They are related through the molecular weights as
,„ _ '"air _ thijMak _ , F
and a subscript s is used to indicate the ratio for 100% theoretical air, also called a stoi-
chiometric mixture. In an actual combustion process, an amount of air is expressed as a
fraction of the theoretical amount, called percent theoretical air. A similar ratio named
the equivalence ratio equals the actual fuel-air ratio divided by the theoretical fuel-air
ratio as
$ - FA/FA, = AFJAF (14.6)
the reciprocal of percent theoretical air. Since the percent theoretical air and the equiva-
lence ratio are both ratios of the stoichiometric air-fuel ratio and the actual air-fuel ratio,
the molecular weights cancel out and they are the same whether a mass basis or a mole
basis is used.
Thus, 150% theoretical air means that the air actually supplied is 1.5 times the theo-
retical air and the equivalence ratio is % The complete combustion of methane with
150% theoretical air is written
CH 4 + 1.5 X 2(0 2 + 3.76N 2 ) C0 2 + 2H 2 + 2 + 1 1.28N 2 (14.?)
having balanced all the stoichiometric coefficients from conservation of all the atoms.
The amount of air actually supplied may also be expressed in terms of percent ex-
cess an. The excess air is the amount of air supplied over and above the theoretical air.
Thus, 150% theoretical air is equivalent to 50% excess air. The terms "theoretical air"
"excess air," and "equivalence ratio" are all in current usage and give an equivalent infor-
mation about the reactant mixture of fuel and air.
When the amount of air supplied is less than the theoretical air required, the com-
bustion is incomplete. If there is only a slight deficiency of air, the usual result is that
some of the carbon unites with the oxygen to form carbon monoxide (CO) instead of car-
bon dioxide (C0 2 ). If the air supplied is considerably less than the theoretical air, there
may also be some hydrocarbons in the products of combustion.
Even when some excess air is supplied, small amounts of carbon monoxide may be
present, the exact amount depending on a number of factors including the mixing and tur-
bulence during combustion. Thus, the combustion of methane with 1 10% theoretical air
might be as follows:
CH 4 + 2(U)0 2 + 2(l.l)3.76N 2 ->
+0.95 C0 2 + 0.05 CO + 2H 2 + 0.225 2 +■ 8.27 N 2 (14.8)
The material covered so far in this section is illustrated by the following examples.
The Combustion Process M 567
EXAMPLE 14.1 Calculate the theoretical air-fuel ratio for the combustion of octane, C g H Is .
Solution
The combustion equation is
C 8 H 18 + 12.5 2 + 1 2.5(3.76) N 2 -> 8C0 2 + 9H 2 + 47.0N 2
The air-fuel ratio on a mole basis is
AF = 12.5 + 47.0 = 595 kmol air/kmoI foe!
The theoretical air-fuel ratio on a mass basis is found by introducing the molecular
weight of the air and fuel,
AF = 59, ^ 8 2 97> = 1 5.0 kg air/kg fuel
EXAMPLE 14.2 Detenriine the molal analysis of the products of combustion when octane, C s Hi 8 , is
burned with 200% theoretical air, and determine the dew point of the products if the
pressure is 0.1 MPa.
Solution
The equation for the combustion of octane with 200% theoretical air is
C 8 H 18 + 1 2.5(2) 2 + 12.5(2)(3.76)N 2 -»■ 8C0 2 + 9H 2 + 12.5 2 + 94.0N 2
Total kmols of product = 8 + 9 + 12.5 + 94.0 = 123.5
Molal analysis of products:
C0 2 = 8/123.5 = 6.47%
H 2 = 9/123.5 = 7.29
2 = 12.5/123.5 = 10.12
N 2 = 94/123.5 = 76.12
100.00%
The partial pressure of the water is 100(0.0729) = 7.29 kPa.
The saturation temperature corresponding to this pressure is 39.7°C, which is also
the dew-point temperature.
The water condensed from the products of combustion usually contains some dis-
solved gases and therefore may be quite corrosive. For this reason the products of com-
bustion are often kept above the dew point until discharged to the atmosphere.
Example 14.2E
Determine the molal analysis of the products of combustion when octane, C 8 H lgj is
burned with 200% theoretical air, and determine the dew point of the products if the
pressure is 14.7 lbf/in. 2 .
568 H chapter fourteen chemical reactions
Solution
Hie equation for the combustion of octane with 200% theoretical air is
C g H Ig + 1 2.5(2) Oj + 12.5(2)(3.76)N 2 -» 8C0 2 + 9H 2 + 12.50 2 + 94.0N 2
Total moles of product = 8 + 9 + 12,5 + 94.0 = 123.5
Molal analysis of products:
C0 2 = 8/123.5 = 6.47%
H 2 = 9/123.5 = 7.29
2 - 12.5/123.5 - 10.12
N 2 = 94/123.5 - 76.12
100.00%
The partial pressure of the H 2 is 14.7(0.0729) - 1.072 lbf/in. 2
The saturation temperature corresponding to this pressure is 104 F, which is also
the dew-point temperature.
The water condensed from the products of combustion usually contains some dis-
solved gases and therefore may be quite corrosive. For this reason the products of com-
bustion are often kept above the dew point until discharged to the atmosphere.
EXAMPLE 14.3 Producer gas from bituminous coal (see Table 14.2) is burned with 20% excess air. Cal-
culate the air-fuel ratio on a volumetric basis and on a mass basis.
Solution
To calculate the theoretical air requirement, let us write the combustion equation for the
combustible substances in 1 kmol of fuel.
0.14H 2 + 0.070O 2 -> 0.14H 2 O
0.27CO + 0. 1 350 2 -» 0.27CO 2
0.03CH 4 + 0.06O 2 -> 0.03CO 2 + 0.06H 2 O
0.265 = kmol oxygen required/kmol fuel
— 0.006 = oxygen in fuel/kmol fuel
0.259 = kmol oxygen required from air/kmol fuel
Therefore, the complete combustion equation for 1 kmol of fuel is
f uel
O.I4H 2 + 0.27CO + 0.03CH 4 + 0.006O 2 + 0.509N 2 + 0.045CO 2
air
+0.259O 2 + 0.259(3.76)N 3 -^0.20H 2 O + 0.345 C0 2 + 1.482 N 2
i^TT^ = 0.259 X^=1.233
\kmol fuel J^ 0.2 1
The Combustion Process H 569
If the atr and fuel are at the same pressure and temperature, this also represents the
ratio of the volume of air to the volume of fuel.
For 20% excess air, , km ° l J f ir , = 1.233 X 1.200 = 1.48
kmol fuel
The air— fuel ratio on a mass basis is
1.48(28.97)
AF =
0.14(2) + 0.27(28) + 0.03(16) + 0.006(32) + 0.509(28) + 0.045(44)
1.48(28.97)
24.74
= 1.73 kg air/kg fuel
An analysis of the products of combustion affords a very simple method for calcu-
lating the actual amount of air supplied in a combustion process. There are various experi-
mental methods by which such an analysis can be made. Some yield results on a "dry"
basis, that is, the fractional analysis of all the components, except for water vapor. Other
experimental procedures give results that include the water vapor. In this presentation we
are not concerned with the experimental devices and procedures, but rather with the use
of such information in a thermodynamic analysis of the chemical reaction. The following
examples illustrate how an analysis of the products can be used to determine the chemical
reaction and the composition of the fuel.
The basic principle in using the analysis of the products of combustion to obtain the
actual fuel-air ratio is conservation of the mass of each of the elements. Thus, in changing
from reactants to products, we can make a carbon balance, hydrogen balance, oxygen bal-
ance, and nitrogen balance (plus any other elements that may be involved). Furthermore,
we recognize that there is a definite ratio between the amounts of some of these elements.
Thus, the ratio between the nitrogen and oxygen supplied in the air is fixed, as well as the
ratio between carbon and hydrogen if the composition of a hydrocarbon fuel is known.
Often the combustion of a fuel uses atmospheric air as the oxidizer, in which case
the reactants also hold some water vapor. Assuming we know the humidity ratio for the
moist air, w, then we would like to know the composition of air per mole of oxygen as
1 2 + 3.76N 2 + xH 2
Since the humidity ratio is, io — mjm ai the number of moles of water is
_ m„ _ a>m a M a
and the number of moles of dry air per mole of oxygen is (1 + 3.76)/l, so we get
-w4.76~- 7.655w (14.9)
^oxygen M,,
This amount of water is found in the products together with the water produced by the ox-
idation of the hydrogen in the fuel.
570 M CHAPTER FOURTEEN CHEMICAL REACTIONS
EXAMPLE 14.4 Methane (CHO is burned with atmospheric air. The analysis of the products on a dry
basis is as follows:
co 2 10.00%
2 2.37
CO 0.53
N 2 87.10
100.00%
Calculate the air-fuel ratio and the percent theoretical air, and determine the com-
bustion equation.
Solution
The solution consists of writing the combustion equation for 100 kmol of dry products,
introducing letter coefficients for the unknown quantities, and then solving for them.
From the analysis of the products, the following equation can be written, keeping
in mind that this analysis is on a dry basis.
aCH 4 + b0 2 + cN 2 ^ 10.0CO 2 + 0.53CO + 2.370 2 + dK 2 + 87.1N 2
A balance for each of the elements will enable us to solve for all the unknown co-
efficients:
Nirogen balance: c = 87.1
Since all the nitrogen comes from the air,
T = 3.76 b = f^U 23.16
o J./o
Carbon balance: a = 10.00 + 0.53 = 10.53
Hydrogen balance: d — 2a = 21.06
Oxygen balance: All the unknown coefficients have been solved for, and therefore
the oxygen balance provides a check on the accuracy. Thus, b can also be determined by
an oxygen balance
b = 10.00 + + 237 + = 23 - 16
Substituting these values for a, b, c, and d, we have
10.53 CH 4 + 23.160 2 + 87.1N 2 ->
10.0CO 2 + 0.53CO + 2.370 2 + 21.06H 2 O + 87.1N 2
Dividing through by 10.53 yields the combustion equation per kmol of fuel.
CH 4 + 2.2 2 + 8.27N 2 -> 0.95 C0 2 + 0.05 CO + 2H 2 + 0.225 2 + 8.27N 2
The air-fuel ratio on a mole basis is
2.2 + 8.27 = 10.47 kmol air/kmol fuel
The combustion process H 571
The air-fuel ratio on a mass basis is found by introducing the molecular weights.
at? 10-47 X 28.97 lfi o-7^^ /i^-r.i
AF = - -^ = 18.97 kg air/kg fuel
The theoretical air-fuel ratio is found by writing the combustion equation for the-
oretical air.
CH 4 + 20 2 + 2(3.76)N 2 -> C0 2 + 2H 2 + 7.52N 2
(2 + 7.52)28.97
18 97
The percent theoretical air is ' • = 1 10%
EXAMPLE 14.5 Coal from Jenkin, Kentucky, has the following ultimate analysis on a dry basis, percent
by mass:
Component
Percent by Mass
Sulfur
0.6
Hydrogen
5,7
Carbon
79.2
Oxygen
10.0
Nitrogen
1.5
Ash
3.0
; This coal is to be burned with 30% excess air. Calculate the air-fuel ratio on a mass
! basis.
j Solution
One approach to this problem is to write the combustion equation for each of the com-
bustible elements per 100 kg of fuel. The molal composition per 100 kg of fuel is found
first.
I
! kmolS/lOOkgfuel
i
!
! kmol H 2 /100 kg fuel
j
! kmol C/100 kg fuel
kmol O 2 /100 kg fuel
kmol N 2 /100 kg fuel
06
32
2
79.2
12
= 0.02
= 2.85
= 6.60
^ = 0.31
32
f =0.05
572 M CHAPTER FOURTEEN CHEMICAL REACTIONS
The combustion equations for the combustible elements are now written, which
enables us to find the theoretical oxygen required,
0.02 S + 0.02 ? -> 0.02 SO;
2.85 H 2 + 1.420 2 ••-> 2.85 H ?
6.60 C 4- 6.60 2 -> 6.60 C0 2
8.04 kmol 2 required/ 1 00 kg fuel
-0.31 kmol 2 in fuel/100 kg fuel
7.73 kmol 2 from air/100 kg fuel
[7.73 + 7.73(3.76)]28.97 . _ „ ■
^theo = 10Q = 10.63 kg air/kg fuel
For 30% excess air the air-fuel ratio is
AF = 1.3 X 10.63 = 13.82 kg air/kg fuel
14.3 ENTHALPY OF FORMATION
In the first thirteen chapters of this book, the problems always concerned a fixed chemical
composition and never a change of composition through a chemical reaction. Therefore,
in dealing with a thermodynamic property, we used tables of thermodynamic properties
for the given substance, and in each of these tables the thermodynamic properties were
given relative to some arbitrary base. In the steam tables, for example, the internal energy
of saturated liquid at 0.0 1°C is assumed to be zero. This procedure is quite adequate when
there is no change in composition because we are concerned with the changes in the prop-
erties of a given substance. The properties at the condition of the reference state cancel
out in the calculation. When dealing with the matter of reference states in Section 13.1 1,
we noted that for a given substance (perhaps a component of a mixture), we are free to
choose a reference state condition, for example, a hypothetical ideal gas, as long as we
then carry out a consistent calculation from that state and condition to the real desired
state. We also noted that we are free to choose a reference state value, as long as there is
no subsequent inconsistency in the calculation of the change in a property because of a
chemical reaction with a resulting change in the amount of a given substance. Now that
we are to include the possibility of a chemical reaction, it will become necessary to
choose these reference state values on a common and consistent basis. We will use as our
reference state a temperature of 25°C, a pressure of 0. 1 MPa, and a hypothetical ideal-gas
condition for those substances that are gases.
Consider the simple steady-state combustion process shown in Fig. 14.3. This ideal-
ized reaction involves the combustion of solid carbon with gaseous (ideal gas) oxygen,
each of which enters the control volume at the reference state, 25°C and 0.1 MPa. The
carbon dioxide (ideal gas) formed by the reaction leaves the chamber at the reference
state, 25°C and 0.1 MPa. If the heat transfer could be accurately measured, it would be
Enthalpy of formation H 573
cv = -393 522 kJ
f '
1 krnoi C0 2
25°C, 0.1 MPa
found to be -393 522 kJ/kmol of carbon dioxide formed. The chemical reaction can be
written
c + o 2 -> co 2
Applying the first law to this process we have
Q CmY . + H R =H P (14.10)
where the subscripts R and P refer to the reactants and products, respectively. W will find
it convenient to also write the first law for such a process in the form
&v +2«A- = 2«A 04.11)
R P
where the summations refer, respectively, to all the reactants or all the products.
Thus, a measurement of the heat transfer would give us the difference between the
enthalpy of the products and the reactants, where each is at the reference state condition.
Suppose, however, that we assign the value of zero to the enthalpy of all the elements at
the reference state. In this case, the enthalpy of the reactants is zero, and
= H p = -393 522 kJ/kmol
The enthalpy of (hypothetical) ideal-gas carbon dioxide at 25°C, 0.1 MPa pressure (with
reference to this arbitrary base in which the enthalpy of the elements is chosen to be zero),
is called the enthalpy of formation. We designate this with the symbol h f . Thus, for car-
bon dioxide
h} = -393 522 kJ/kmol
The enthalpy of carbon dioxide in any other sate, relative to this base in which the
enthalpy of the elements is zero, would be found by adding the change of enthalpy be-
tween ideal gas at 25°C, 0.1 MPa, and the given state to the enthalpy of formation. That is,
the enthalpy at any temperature and pressure, h TPs is
(14.12)
where the term (AA) 2 9s, o.iMPa-»r,/> represents the difference in enthalpy between any given
state and the enthalpy of ideal gas at 298. 15 K, 0. 1 MPa. For convenience we usually drop
the subscripts in the examples that follow.
The procedure that we have demonstrated for carbon dioxide can be applied to any
compound.
Table A. 10 gives values of the enthalpy of formation for a number of substances in
the units kJ/kmol (or Btu/lb mol in F. 1 1).
1 kmo! C
— >
25°C, 0.1 MPa
FIGURE 14.3 kmQ| Q
Example of combustion L__
process. 25°C, 0.1 MPa
574 H CHAPTER FOURTEEN CHEMICAL REACTIONS
Three further observations should be made in regard to enthalpy of formation.
1. We have demonstrated the concept of enthalpy of formation in terms of the mea-
surement of the heat transfer in an idealized chemical reaction in which a compound
is formed from the elements. Actually, the enthalpy of formation is usually found
by the application of statistical thermodynamics, using observed spectroscopic data.
2. The justification of this procedure of arbitrarily assigning the value of zero to the en-
thalpy of the elements at 25°C, 0.1 MPa, rests on the fact that in the absence of nu-
clear reactions the mass of each element is conserved in a chemical reaction. No
conflicts or ambiguities arise with this choice of reference state, and it proves to be
very convenient in studying chemical reactions from a thermodynamic point of view.
3. In certain cases an element or compound can exist in more than one state at 25°C,
0.1 MPa. Carbon, for example, can be in the form of graphite or diamond. It is es-
sential that the state to which a given value is related be clearly identified. Thus, in
Table A. 1 0, the enthalpy of formation of graphite is given the value of zero, and the
enthalpy of each substance that contains carbon is given relative to this base. An-
other example is that oxygen may exist in the monatomic or diatomic form, and also
as ozone, 3 . The value chosen as zero is for the form that is chemically stable at
the reference state, which in the case of oxygen is the diatomic form. Then each of
the other forms must have an enthalpy of formation consistent with the chemical re-
action and heat transfer for the reaction that produces that form of oxygen.
It will be noted from Table A. 10 that two values are given for the enthalpy of forma-
tion for water; one is for liquid water and the other for gaseous (hypothetical ideal gas)
water, both at the reference state of 25°C, 0.1 MPa. It is convenient to use the hypothetical
ideal-gas reference in connection with the ideal-gas table property changes given in Table
A.9, and to use the real liquid reference in connection with real water property changes as
given in the steam tables, Table B.l. The real-liquid reference state properties are obtained
from those at the hypothetical ideal-gas reference by following the procedure of calculation
described in Section 13.11. The same procedure can be followed for other substances that
have a saturation pressure less than 0.1 MPa at the reference temperature of 25°C.
Frequently, students are bothered by the minus sign when the enthalpy of formation
is negative. For example, the enthalpy of formation of carbon dioxide is negative. This is
quite evident because the heat transfer is negative during the steady-flow chemical reac-
tion, and the enthalpy of the carbon dioxide must be less than the sum of enthalpy of the
carbon and oxygen initially, both of which are assigned the value of zero. This is quite
analogous to the situation we would have in the steam tables if we let the enthalpy of satu-
rated vapor be zero at 0. 1 MPa pressure. In this case the enthalpy of the liquid would be
negative, and we would simply use the negative value for the enthalpy of the liquid when
solving problems.
14.4 first-Law analysis
of Reacting Systems
The significance of the enthalpy of formation is that it is most convenient in performing a
first-law analysis of a reacting system, for the enthalpies of different substances can be
added or subtracted, since they are all given relative to the same base.
first-Law analysis of reacting systems H 575
In such problems, we will write the first law for a steady-state, steady- flow process
in the form
0,y. + H R = W ey , + H P
or
£?c.v. + E «A = f ^.v. +2 «A
where and P refer to the reactants and products, respectively. In each problem it is nec-
essary to choose one parameter as the basis of the solution. Usually this is taken as 1 kmol
of fuel.
EXAMPLE 14.6 Consider the following reaction, which occurs in a steady-state, steady-flow process.
CH 4 + 2 2 ~> C0 2 + 2 H 2 0(/)
The reactants and products are each at a total pressure of 0.1 MPa and 25°C. Determine
the heat transfer per kilomole of fuel entering the combustion chamber.
Control volume: Combustion chamber.
Inlet state: P and T known; state fixed.
Exit state: P and T known, state fixed.
Process: Steady state.
Model: Three gases ideal gases; real liquid water.
Analysis
First law:
Qcv. + 2«A* = 2«A
R P
Solution
Using values from Table A. 10, we have
2 »A "74 873 W
R
2^A = #)co 2 + 2C^) Hl0 (/)
p
= -393 522 + 2(-285 830) = -965 182 kj
a.v. = "965 182 - (-74 873) = -890 309 kJ
In most instances, however, the substances that comprise the reactants and products
in a chemical reaction are not at a temperature of 25°C and a pressure of 0. 1 MPa (the
state at which the enthalpy of formation is given). Therefore, the change of enthalpy be-
tween 25°C and 0.1 MPa and the given state must be known. For a solid or liquid, this
576 S Chapter Fourteen CirEMicAL Reactions
change of enthalpy can usually be found from a table of thermodynamic properties or
from specific heat data. For gases, the change of enthalpy can usually be found by one of
the following procedures.
1. Assume ideal-gas behavior between 25°C, 0.1 MPa, and the given state. In this
case, the enthalpy is a function of the temperature only and can be found by an
equation of C p0 or from tabulated values of enthalpy as a function of temperature
(which assumes ideal-gas behavior). Table A.6 gives an equation for C pQ for a num-
ber of substances and Table A.9 gives values of h° — /*2 9S (that is, the A/z of Eq.
14.1 1) in kJ/kmol, (h% 3 refers to 25°C or 298.15 K, For simplicity this is designated
A°9S') The superscript is used to designate that this is the enthalpy at 0.1 MPa pres-
sure, based on ideal-gas behavior, that is, the standard-state enthalpy.
2. If a table of thermodynamic properties is available, A/i can be found directly from
these tables if a real substance behavior reference state is being used, such as that
described above for liquid water. If a hypothetical ideal-gas reference state is being
used, then it is necessary to account for the real substance correction to properties at
that sate to gain entry to the tables.
3. If the deviation from ideal-gas behavior is significant, but no tables of thermody-
namic properties are available, the_value of Lh can be found from the generalized
tables or charts and the values for or Lh at 0. 1 MPa pressure as indicated above.
Thus, in general, for applying the first law to a steady-state process involving a
chemical reaction and negligible changes in kinetic and potential energy, we can write
&.V. + 2 n l Qt}+ AA); = JF C . V . +■ 2 njh} + Ah) e (14.13)
R P
EXAMPLE 14.7 Calculate the enthalpy of water (on a kilomole basis) at 3.5 MPa, 300°C, relative to- the
25°C and 0.1 MPa base, using the following procedures.
1. Assume the steam to be an ideal gas with the value of C pQ given in the Appendix,
Table A.6.
2. Assume the steam to be an ideal gas with the value for Ah as given in the Appendix,
Table A.9.
3. The steam tables.
4. The specific heat behavior given in 2 above and the generalized charts.
Solution
For each of these procedures, we can write
h TJ > = (hf + Ah)
The only difference is in the procedure by which we calculate Ah. From Table A. 10 we
note that
(hfh&g) = -241 826kJ/lcmol
first-Law analysis of reacting Systems II 577
1. Using the specific heat equation for H 2 0(g) from Table A.6,
C p0 - 1.79 + 0.1079 + O.5860 2 - O.2O0 3 , 9 = 771000
The specific heat at the average temperature
T _ 298.15 + 57 3.15 „
iovg — — = 435.65 K
= 1.79 + 0.107(0.43565) + 0.5 86(0.43 565) 2 - 0.2(0.43565) 3
kJ
= 1.9313
Therefore,
kgK
Ah = MC p0 AT
= 18.015 X 1.9313(573.15 - 298.15) - 9568 W
kmol
h T!P = -241 826 + 9568 = ~232 258 kJ/kmol
2. Using Table A.9 for H 2 0(g),
AA - 9494 kJ/kmol
^ = -241 826 + 9494 = -232 332 kJ/kmol
3. Using the steam tables, either the liquid reference or the gaseous reference state may
be used.
For the liquid, ~
Ah = 18.015(2977.5 - 104.9) = 51 750 kJ/kmol
h TtP - -285 830 + 51 750 = -234 080 kJ/kmol
For the gas,
AA = 18.015(2977.5 - 2547.2) = 7752 kJ/kmol
Ajy = -241 826 + 7752 = -234 074 kJ/kmol
The very small difference results from using the enthalpy of saturated vapor at 25°C
(which is almost but not exactly an ideal gas) in calculating the Ah.
4. When using the generalized charts, we use the notation introduced in Chapter 13.
hrs = A? - (A? ~ hi) + (A? - Af) + (Af - A,)
where subscript 2 refers to the state at 3.5 MPa, 300°C, and state 1 refers to the state
at 0.1 MPa,25°C. _
From part 2, h 2 - Af = 9494 kJ/kmol.
Af — h l =0 (ideal-gas reference)
578 H Chapter fourteen chemical Reactions
From the generalized enthalpy chart, Fig. D.2.
= 0,21, hf - /t 2 = 0.21 X 8.3145 X 647.3 = 1130kJ/kmol
h TJ > = -241 826 - 11 30 + 9494 = -233 462 kJ/kmol
The particular approach that is used in a given problem will depend on the data
available for the given substance.
EXAMPLE 14.8 A small gas-turbine uses C 8 H IS (/) for fuel and 400% theoretical air. The air and fuel
enter at 25°C, and the products of combustion leave at 900 K. The output of the engine
and the fuel consumption are measured, and it is found that the specific fuel consump-
tion is 0.25 kg/s of fuel per megawatt output. Determine the heat transfer from the en-
gine per kilomole of fuel. Assume complete combustion.
Control volume: Gas-turbine engine.
Met states: T known for fuel and air.
Exit state: 7* known for combustion products.
Process: Steady state.
Model: All gases ideal gases, Table A.9; liquid octane, Table A.10.
Analysis
The combustion equation is
C 8 Hi S C0 + 4(12.5)0 2 + 4(12.5)(3.76)N 2 8C0 2 + 9H 2 + 37.50 2 + 188.0N 2
First law:
R P
Solution
Since the air is composed of elements and enters at 25°C, the enthalpy of the reactants is
equal to that of the fuel,
2«,-$r + AA), = (h/)c t H,M = -250 105 kJ/kmol fuel
R
Considering the products, we have
2 nj$ + AA), = n COi (h} + AA) C0] + n u Jh} + AA^o
p „ —
+ k Oi (A/0 Oj + »n 2 CA^)n,
= 8(-393 522 + 28 030) + 9(-241 826 + 21 892)
+ 37.5(19 249) + 188(18 222)
= -755 769 kJ/kmol fuel
= 1000 kJ/s x H4.23 kg = 92Q
c v - 0.25 kg/s kmol
First-Law Analysis of Reacting Systems H 579
Therefore, from the first law,
2c.v. = ~755 769 + 456 920 - (-250 105)
= -48 744kJ/kmolfuel
EXAMPLE 14.8E A small gas-turbine uses C 8 H 18 (/) for fuel and 400% theoretical air. The air and fuel
enter at 77 F, and the products of combustion leave at 1100 F. The output of the engine
and the fuel consumption are measured, and it is found that the specific fuel consump-
tion is one pound of fuel per horsepower-hour. Determine the heat transfer from the en-
gine per pound mole of fuel. Assume complete combustion.
Control volume: Gas-turbine engine.
Inlet states: Tknown for fuel and air.
Exit state: T known for combustion products.
Process: Steady state.
Model: All gases ideal gases, Table F.6; liquid octane, Table F. 11.
Analysis
The combustion equation is
C 8 H 18 (/) + 4(12.5)0 2 + 4(12.5)(3.76)N 2 8C0 2 + 9H 2 + 37.50 2 + 188.0N 2
First law;
0e.v. + 2 nf$ + M), = + 2"M + M%
Solution
Since the air is composed of elements and enters at 77 F, the enthalpy of the reactants is
equal to that of the fuel.
2 nlh} + AA] ( - = (^) Q h is( o = " 526 Btu/lb mol
R
Considering the products
2 + AA), = n COi (h} + M) C02 + n Hi0 (h} + Ah) Hi0 + ^(AA)^ + « Nl (Afi) Nj
= 8(-169 184 + 11 291) + 9(-103 966 + 8858)
+ 37.5(7778) 4- 188(7374)
= -441 129 Btu/lb mol fuel.
JF C . V . = 2544 X 114.23 = 290 601 Btu/lb mol fuel
580 m Chapter Fourteen Chemical Reactions
Therefore, from the first law,
ficv. = -441 129 + 290 601 - (-107 526)
= -43 002 Btu/lbmol fuel
EXAMPLE 14.9 A mixture of 1 kmol of gaseous ethene and 3 kmol of oxygen at 25°C reacts in a
constant-volume bomb. Heat is transferred until the products are cooled to 600 K. Deter-
mine the amount of heat transfer from the system.
Control mass: Constant- volume bomb.
Initial state: T known.
Final state: T known.
Process: Constant volume.
Model: Ideal-gas mixtures, Tables A.9 } A. 10.
Analysis
The chemical reaction is
C,H 4 + 3 2 -» 2 C0 2 + 2 H 2 0(g)
First law:
Q+U R =U P
Q + ^n(h} + Mi-RT) = ^n($ + Mi-RT)
R P
Solution
Using values from Tables A.9 and A. 10, gives
= 52 467 - 4 X 8.3145 X 298.2 = 42 550 kJ
+ Aft - RT) = 2{(h}) COl + Aft COi ] + 2[(^ Hi ofe) + AA^,] " 4RT
p
= 2(-393 522 + 12 899) + 2(-241 826 + 10 463)
-4 X 8.3145 X 600
= -1 243 927 kJ
Therefore,
Q =
- 1 243 927 - 42 550 = - 1 286 477 kJ
Enthalpy and Internal energy of Combustion; Heat of Reaction ffl 581
For a real-gas mixture, a pseudocritical method such as Kay's rule, Eq. 13.86 could
be used to evaluate the nonideal-gas contribution to enthalpy at the temperature and pres-
sure of the mixture and this value added to the ideal-gas mixture enthalpy at that tempera-
ture, as in the procedure developed in Section 13.11.
14.5 Enthalpy and Internal Energy
of combustion; heat of reaction
The enthalpy of combustion, hgp, is defined as the difference between the enthalpy of the
products and the enthalpy of the reactants when complete combustion occurs at a given
temperature and pressure. That is,
hgp — Hp — H R
T'rp = + AA) tf - 2 n$f + AA), (14.14)
P R
The usual parameter for expressing the enthalpy of combustion is a unit mass of
fuel, such as a kilogram (hgp) or a kilomole (hgp) of fuel.
As the enthalpy of formation is fixed, we can separate the terms as
H=rf + AH
where
t% = 2 Aflp = 2 «i Aft,
p p
Now the difference in enthalpies is written
H P - H R = Hi - H° R + AH F - AH R
- hgp + AH P - AH R (14.15)
explicitly showing the reference enthalpy of combustion, ft^, and the two departure
terms Ai/p and AH R . The latter two terms for the products and reactants are nonzero if
they exist at a state other than the reference state.
The tabulated values of the enthalpy of combustion of fuels are usually given for a
temperature of 25°C and a pressure of 0. 1 MPa. The enthalpy of combustion for a number
of hydrocarbon fuels at this temperature and pressure, which we designate h Rm> is given
in Table 14.3.
The internal energy of combustion is defined in a similar manner.
Usp=Up~ U R
= 2«.C*/ + Aft - Pv) e - 2^/ + Aft - Fu), (14.16)
P R
582 @ CHAPTER FOURTEEN CHEMICAL REACTIONS
TABLE 14.3
Enthalpy of Combustion of Some Hydrocarbons at 25°C
UNITS:
Hydrocarbon
Paraffins
Methane
Ethane
Propane
n-Butane
n-Pentane
n-Hexane
n-Heptane
n-Octane
n-Decane
n-Dodecane
n-Cetane
Olefins
Ethene
Propene
Butene
Penterte
Hexene
Heptene
Octene
Nonene
Decene
Alkylbenzenes
Benzene
Methytbenzene
Ethylbenzene
Propytbenzene
Butylbenzene
Other fuels
Gasoline
Diesel T-T
JP8jet fuel
Methanol
Ethanol
Nitromethane
Phenol
Hydrogen
kJ
kg
LIQUID H 2
IN PRODUCTS
GAS H 2
IN PRODUCTS
Formula
CH 4
CjH,
CjHg
QH]o
C 5 H, 2
QH 14
C 8 H 18
C10H22
C 12^26
C16H34
C 2 H 4
QH 6
C 4 H 3
C 5 H 10
C,H 14
C S H 56
C9H 1S
CioH 2 o
C 7 H 8
C 9 H 12
CioH 14
C 7 H 17
Cl4.4^24.9
C13H23.8
CH 3 OH
C2H 5 OH
CH 3 N0 2
C 6 H 5 OH
H 2
Liq. HC
Gas HC
-49 973
-49 130
-48 643
-48 308
-48 071
-47 893
-47 641
-47 470
-47 300
-41 831
-42 437
-42 997
-43 416
-43 748
-48 201
-45 700
-45 707
-22 657
-29 676
-11 618
-32 520
-55 496
-51 875
-50 343
-49 500
-49 011
-4 8 676
-48 436
-48 256
-48 000
-47 828
-47 658
-50 296
-48 917
-48 453
-48 134
-47 937
-47 800
-47 693
-47 612
-47 547
-42 266
-42 847
-43 395
-43 800
-44 123
-48 582
-46 074
-46087
-23 840
-30 596
-12 247
-33 176
-141781
Liq. HC
Gas HC
-45 982
-45 344
-44 983
-44 733
-44 557
-44 425
-44 239
-44 109
-44 000
-40 141
-40 527
-40 924
-41 219
-41 453
-44 506
-42 934
-42 800
-19910
-26 811
-10 537
-31 117
-50 010
-47 484
-46 352
-45 714
-45 351
-45 101
-44 922
-44 788
-44 598
-44 467
-44 358
-47 158
-45 780
-45 316
-44 996
-44 800
-44 662
-44 556
-44 475
-44 410
-40 576
-40 937
-41 322
-41 603
-41 828
-44 886
-43 308
-43 180
-21 093
-27 731
-11 165
-31 774
-119953
EntiIalpy and Internal energy of Combustion; Heat of Reaction
h 583
When all the gaseous constituents can be considered as ideal gases, and the volume
of the liquid and solid constituents is negligible compared to the value of the gaseous con-
stituents, this relation for reduces to
U RP ~ h$P ~ ^("gastous products ~ "gaseous reactanls) (14.17)
Frequently the term 'Seating value" or "heat of reaction" is used. This represents
the heat transferred from the chamber during combustion or reaction at constant tempera-
ture. In the case of a constant pressure or steady- flow process, we conclude from the first
law of thermodynamics that it is equal to the negative of the enthalpy of combustion. For
this reason, this heat transfer is sometimes designated the constant-pressure heating value
for combustion processes.
In the case of a constant-volume process, the heat transfer is equal to the negative of
the internal energy of combustion. This is sometimes designated the constant-volume
heating value in the case of combustion.
When the term heating value is used, the terms "higher" and "lower" heating value
are used. The higher heating value is the heat transfer with liquid water in the products,
and the lower heating value is the heat transfer with vapor water in the products.
EXAMPLE 14.10 Calculate the enthalpy of combustion of propane at 25°C on both a kilomole and kilo-
gram basis under the following conditions.
1. Liquid propane with liquid water in the products.
2. Liquid propane with gaseous water in the products.
3. Gaseous propane with liquid water in the products.
4. Gaseous propane with gaseous water in the products.
This example is designed to show how the enthalpy of combustion can be determined
from enthalpies of formation. The enthalpy of evaporation of propane is 370 kj/kg.
Analysis and Solution
The basic combustion equation is
C 3 H S + 5 2 -» 3 C0 2 + 4H 2
From Table A.10 = -103 900kJ/kmol. Therefore,
fyfkjim = ~ 103 9 °0 ~ 44.097(370) = -120 216 kJ/kmol
1. Liquid propane-liquid water:
= 3(-393 522) + 4(-285 830) - (-120 216)
= -2 203 670 kJ/kmol = "^||~ ~ "49 973 kJ/kg
The higher heating value of liquid propane is 49 973 kJ/kg.
584 H CHAPTER FOURTEEN CHEMICAL REACTIONS
2. Liquid propane-gaseous water:
V, " 3$) CO| I 4(^) Hj0fe) (^)c,H t( o
= 3(-393 522) + 4(-241 826) - (-120 216)
= -2 027 654 kJ/kmol = - 2 ^^ 4 = -45 982 kj/kg
The lower heating value of liquid propane is 45 982 kJ/kg.
3. Gaseous propane-liquid water:
:-: 3(-393 522) !• 4(-285 830) - (-103 900)
- -2 219 986 kJ/kmol = -^HHP - -50 343 kJ/kg
The higher heating value of gaseous propane is 50 343 kJ/kg.
4. Gaseous propane-gaseous water:
= 3(-393 522) + 4(-241 826) - (- 103 900)
- -2 043 970 kJ/kmol - -^|^ - -46 352 kJ/kg
The lower heating value of gaseous propane is 46 352 kJ/kg.
Each of the four values calculated in this example corresponds to the appropriate
value given in Table 14.3.
EXAMPLE 14.11 Calculate the enthalpy of combustion of gaseous propane at 500 K. (At this temperature all
the water formed during combustion will be vapor.) This example will demonstrate how the
enthalpy of combustion of propane varies with temperature. The average constant-pressure
specific heat of propane between 25°C and 500 Kis 2.1 kJ/kg K.
Analysis
The combustion equation is
C 3 H s (g) + 50 2 -> 3C0 3 + 4H 2 0fe)
The enthalpy of combustion is, from Eq. 14.13,
(V)r = 2 + A*) e - 2 n,{h} + A/*),
P R
Adiabatic Flame Temperature M 585
Solution
= $ + q, av (A7)j CjHife) + k 0i (A*) 0j
= -103 900 + 2.1 X 44.097(500 - 298.2) + 5(6095)
= -54 73SkJ/kmoI
= 3(-393 522 + 8297) + 4(-241 826 4- 6896)
= -2 095 395 kJ/kmol '
V M = -2 095 395 - (-54 738) = -2 040 657 kJ/kmol
h - ~ 2 040 657 - ^ 1*7-7 ITfl
^ 44.097 ~ ~ 46 277 ^
This compares with a value of —46 352 at 25°C.
This problem could also have been solved using the given value of the enthalpy of
combustion at 25°C by noting that
= n COi (hj + Ah) C02 + r %0 (h} + AA) Hi0
-{h} + C^ v (AJ)] CiHsCg) - « Oi (A/0 O2
= Aju-, + n COi (Mi) COi + k Hj0 (A^) Hi0
-C^ av (A7) CjHtfe) - « 0j (AA) 0]
hRp m = -46 352 X 44.097 + 3(8297) + 4(6896)
-2.1 X 44.097(500 - 298.2) - 5(6095)
= -2 040 657 kJ/kmol
h - ~ 2 040 657 - AA<yinm
h se m 44 097 ~ ~ 46 277 kJ/kg
14.6 Adiabatic Flame temperature
Consider a given combustion process that takes place adiabatically and with no work or
changes in kinetic or potential energy involved. For such a process the temperature of
the products is referred to as the adiabatic flame temperature. With the assumptions of
no work and no changes in kinetic or potential energy, this is the maximum temperature
that can be achieved for the given reactants because any heat transfer from the reacting
substances and any incomplete combustion would tend to lower the temperature of the
products.
For a given fuel and given pressure and temperature of the reactants, the maxi-
mum adiabatic flame temperature that can be achieved is with a stoichiometric mixture.
The adiabatic flame temperature can be controlled by the amount of excess air that is
used. This is important, for example, in gas turbines, where the maximum permissible
586 H CHAPTER FOURTEEN CHEMICAL REACTIONS
temperature is determined by metallurgical considerations in the turbine, and close con-
trol of the temperature of the products is essential.
Example 14.12 shows how the adiabatic flame temperature may be found. The dis-
sociation that takes place in the combustion products, which has a significant effect on the
adiabatic flame temperature, will be considered in the next chapter.
EXAMPLE 14.12 Liquid octane at 25°C is burned with 400% theoretical air at 25°C in a steady-state
process. Determine the adiabatic flame temperature.
Control volume: Combustion chamber.
Met states: Tknown for fuel and air.
Process: Steady state.
Model: Gases ideal gases, Table A.9; liquid octane, Table A. 10.
Analysis
The reaction is
CgH !8 C0 + 4(12.5)0 2 + 4(12.5)(3.76)N 2 ^
8C0 2 + 9H 2 0fe) + 37.50 2 + 188.0N 2
First law: Since the process is adiabatic,
Hr^ Hp
R P
where Lh e refers to each constituent in the products at the adiabatic flame temperature.
Solution
From Tables A.9 and A. 10,
Hr = E n$f + = Qt/icfiM = -250 105 kJ/kmol fuel
p
= 8(-393 522 + kh co ) + 9(-241 826 + &h Ul0 ) + 37.5 A^ 0j + 188.0 AA Nl
By trial-and-error solution, a temperature of the products is found that satisfies this
equation. Assume that
T P = 900 K
p
= 8(— 393 522 + 28 030) + 9(-241 826 + 21 892) .
+ 37.5(19 249) + 188(18 222)
= -755 769 kJ/kmol fuel
THE THIRD Law of Thermodynamics and Absolute Entropy H 587
Assume that
p
= 8(-393 522 + 33 400) + 9(-241 826 + 25 956)
+ 37.5(22 710) + 188(21 461)
= 62 487 kJ/kmol fuel
Since H P = H R ~ -250 105 kJ/kmol, we find by linear interpolation that the adiabatic
flame temperature is 961.8 K. Because the ideal-gas enthalpy is not really a linear func-
tion of temperature, the true answer will be slightly different from this value.
14.7 The third Law of Thermodynamics
and Absolute Entropy
As we consider a second-law analysis of chemical reactions, we face the same problem we
had with the first law: What base should be used for the entropy of the various substances?
This problem leads directly to a consideration of the third law of thermodynamics.
The third law of thermodynamics was formulated during the early part of the twenti-
eth century. The initial work was done primarily by W. H. Nernst (1864-1941) and Max
Planck (1 858-1947). The third law deals with the entropy of substances at the absolute "zero
of temperature and in essence states that the entropy of a perfect crystal is zero at absolute
zero. From a statistical point of view, this means that the crystal structure has the maximum
degree of order. Furthermore, because the temperature is absolute zero, the thermal energy
is minimum. It also follows that a substance that does not have a perfect crystalline structure
at absolute zero, but instead has a degree of randomness, such as a solid solution or a glassy
solid, has a finite value of entropy at absolute zero. The experimental evidence on which the
third law rests is primarily data on chemical reactions at low temperatures and measure-
ments of heat capacity at temperatures approaching absolute zero. In contrast to the first and
second laws, which lead, respectively, to the properties of internal energy and entropy, the
third law deals only with the question of entropy at absolute zero. However, the implica-
tions of the third law are quite profound, particularly in respect to chemical equilibrium.
The particular relevance of the third law is that it provides an absolute base from
which to measure the entropy of each substance. The entropy relative to this base is
termed the absolute entropy. The increase in entropy between absolute zero and any given
state can be found either from calorimetric data or by procedures based on statistical ther-
modynamics. The calorimetric method gives precise measurements of specific-heat data
over the temperature range, as well as of the energy associated with phase transforma-
tions. These measurements are in agreement with the calculations based on statistical
thermodynamics and observed molecular data.
Table A.10 gives the absolute entropy at 25°C and 0.1-MPa pressure for a number of
substances. Table A.9 gives the absolute entropy for a number of gases at 0.1-MPa pressure
and various temperatures. For gases the numbers in all these tables are the hypothetical
588 B Chapter fourteen chemical reactions
ideal-gas values. The pressure P° of 0.1 MPa is termed the standard-state pressure, and the
absolute entropy as given in these tables is designated s°. The temperature is designated in
kelvins with a subscript such as s° mo .
If the value of the absolute entropy is known at the standard-state pressure of 0. 1 MPa
and a given temperature, it is a straightforward procedure to calculate the entropy change
from this state (whether hypothetical ideal gas or real substance) to another desired state fol-
lowing the procedure described in Section 13.11 . If the substance is listed in Table A.8, then
5 w = 5S-51n^+(3 v -5Sp) (14.18)
In this expression, the first term on the right side is the value from Table A.9, the second
is the ideal-gas term to account for a change in pressure from P° to P, and the third is the
term that corrects for real-substance behavior, as given in the generalized entropy chart in
the Appendix. If the real-substance behavior is to be evaluated from an equation of state
or thermodynamic table of properties, the term for the change in pressure should be made
to a low pressure P* t at which ideal-gas behavior is a reasonable assumption, but it is also
listed in the tables. Then
If the substance is not one of those listed in Table A.9, and the absolute entropy is known
only at one temperature T Q , as given in Table A. 10 for example, then it will be necessary
to calculate S£ from
( i4 - 2 °>
and then proceed with the calculation of Eq. 14.17 or 14.19.
If Eq. 14.18 is being used to calculate the absolute entropy of a substance in a re-
gion in which the ideal-gas model is a valid representation of the behavior of that sub-
stance, then the last term on the right-side of Eq. 14.18 simply drops out of the
calculation.
For calculation of the absolute entropy of a mixture of ideal gases at T, P, the mix-
ture entropy is given in terms of the component partial entropies as
where
Sf=^ T -R\n^-R\n yi = ^--R ln*£ (14.22)
For a real-gas mixture, a correction can be added to the ideal-gas entropy calculated from
Eqs. 14:21 and 14.22 by using a pseudocritical method such as was discussed in Section
13.11. The corrected expression is
in which the second term on the right side is the correction term from the generalized en-
tropy chart.
Second -Law Analysis of reacting systems M 589
14.8 Second-law Analysis
of reacting Systems •
The concepts of reversible work, irreversibility, and availability (exergy) were introduced
in Chapter 10. These concepts included both the first and second laws of thermodynamics.
We proceed now to develop this matter further, and we will be particularly concerned
with determining the maximum work (availability) that can be done through a combustion
process and with examining the irreversibilities associated with such processes.
The reversible work for a steady-state process in which there is no heat transfer with
reservoirs other than the surroundings, and also in the absence of changes in kinetic and
potential energy is, from Eq. 10.9 on a total mass basis,
W«* = 2 mjh, " m - 2 mlK - T$s e )
Applying this equation to a steady-state process that involves a chemical reaction,
and introducing the symbols from this chapter, we have
W™ = 2 >#? + AA - JoS), - 2 nM + Ah ~ T s) e (14.24)
x P
Similarly, the irreversibility for such a process can be written as
/= JF rev - W= y 2n e T s e - - Q cv (14.25)
P R
The availability, iff, for a steady-flow process, in the absence of kinetic and potential
energy changes, is given by Eq. 10.19 as
ifi = (h- T s) - (h ~ T s )
We further note that if a steady-state chemical reaction takes place in such a manner
that both the reactants and products are in temperature equilibrium with the surroundings,
the Gibbs function (g = h — Ts), defined in Eq. 13.14, becomes a significant variable. For
such a process, in the absence of changes in kinetic and potential energy, the reversible
work is given by the relation
^ v = 2 ttiii ~ 2« e & = -AG (14.26)
J! P
in which
AC? = AH - T&S (14.27)
We should keep in mind that Eq. 14.26 is a special case and that the reversible work is
given by Eq. 14.24 if the reactants and products are not in temperature equilibrium with
the surroundings.
Let us now consider the question of the maximum work that can be done during a
chemical reaction. For example, consider 1 kmol of hydrocarbon fuel and the necessary
air for complete combustion, each at 0.1-MPa pressure and 25°C, the pressure and tem-
perature of the surroundings. What is the maximum work that can be done as this fuel re-
acts with the air? From the considerations covered in Chapter 10, we conclude that the
maximum work would be done if this chemical reaction took place reversibly and the
590 H chapter Fourteen chemical reactions
products were finally in pressure and temperature equilibrium with the surroundings. We
conclude that this reversible work could be calculated from the relation in Eq. 14.26,
R P
However, since the final state is in equilibrium with the surroundings, we could
consider this amount of work to be the availability of the fuel and air.
EXAMPLE 14.13 Ethene (g) at 25°C and 0.1-MPa pressure is burned with 400% theoretical air at 25°C
and 0.1-MPa pressure. Assume that this reaction takes place reversibly at 25°C and that
the products leave at 25°C and 0.1-MPa pressure. To simplify this problem further, as-
sume that the oxygen and nitrogen are separated before the reaction takes place (each at
0.1 MPa, 25°C), that the constituents in the products are separated, and that each is at
25°C and 0.1 MPa. Thus, the reaction takes place as shown in Fig. 14.4. For purposes of
comparison between this and the two subsequent examples, we consider all the water in
the products to be a gas (a hypothetical situation in this example and Example 14.15).
Determine the reversible work for this process (that is, the work that would be
done if this chemical action took place reversibly and isothermally).
ConU-ol volume: Combustion chamber.
Inlet states: P, '/'known for each gas.
Exit states: P, 7/ known for each gas.
Model: All ideal gases, Tables A.9 and A.10.
Sketch: Figure 14.4.
Analysis
The equation for this chemical reaction is
C 2 H 4 (g) + 3(4)0 2 + 3(4)(3.76)N 2 ^2C0 2 + 2H 2 Ofe) + 90 2 + 45.1N 2 ■
The reversible work for this process is equal to the decrease in Gibbs function during
this reaction, Eq. 14.26. Since each component is at the standard-state pressure P° , we
write Eqs. 14.26 and 14.27 as
FT ev =-AG°, AG = Afl° - TA5
We also note that the 45.1 N 2 cancels out of both sides in these expressions, as does 9 of
thel20 2 .
Each at
T = 25°C
p = 0.1 MPa
FIGURE 14.4 Sketch
for Example 14.13.
V
co 2
-SB- '>
Each at
T = 25°C
p = 0.1 MPa'
second-Law analysis of reacting systems H 591
Solution
Using values from Tables A.8 and A.9 at 25°C,
AW° - 2lij COi •(• 2h% m ■ h%. t]l> - 3h} 02
- 2(-393 522) + 2( -241 826) - (+52 467) - 3(0)
- -1 323 163 kJ/kmol fuel
AS — 2sq 0i + 2su i0(s) — ^CjH, ~ 3.? 0j
= 2(213.795) + 2(188.843) - (219.330) - 3(205.148)
= -29.51 6 kJ/kmol fuel
AG - -I 323 163 - 298.1 5( -29.516)
- - I 314 363 kJ/kmol C 2 n 4
» /,ev - -AG - 1 314 363 kJ/kmol C 2 H 4
= = 46851 kJ/kg
Therefore, we might say that when 1 kg of ethene is at 25°C, and the standard-
state pressure, 0.1 MPa } it has an availability of 46 85 1 kJ.
Thus, it would seem logical to rate the efficiency of a device designed to do work by
utilizing a combustion process, such as an internal-combustion engine or a steam power
plant, as the ratio of the actual work to the reversible work, or in Example 14.13, the de-
crease in Gibbs function for the chemical reaction, instead of comparing the actual work
to the heating value, as is commonly done. This is, in fact, the basic principle of the
second-law efficiency, which was introduced in connection with availability analysis in
Chapter 10. As noted from Example 14.13, the difference between the decrease in Gibbs
function and the heating value is small, which is typical for hydrocarbon fuels. The differ-
ence in the two types of efficiencies will, therefore, not usually be large. We must always
be careful, however, when discussing efficiencies, to note the definition of the efficiency
under consideration.
It is of particular interest to. study the irreversibility that takes place during a com-
bustion process. The following examples illustrate this matter. We consider the same hy-
drocarbon fuel that was used in Example 14.13 — ethene (g) at 25°C and 0.1 MPa. We
determined its availability and found it to be 46 851 kj/kg. Now let us burn this fuel with
400% theoretical air in a steady-state adiabatic process. We can determine the irreversibil-
ity of this process in two ways. The first way is to calculate the increase in entropy for the
process. Since the process is adiabatic, the increase in entropy is due entirely to the irre-
versibilities for the process, and we can find the irreversibility from Eq. 14.25. We can
also calculate the availabilities of the products of combustion at the adiabatic flame tem-
perature, and note that they are less than the availability of the fuel and air before the
combustion process. The difference is the irreversibility that occurs during the combus-
tion process.
1
592 m CHAPTER FOURTEEN CHEMICAL REACTIONS
EXAMPLE 14.14 Consider the same combustion process as in Example 14. 13, but let it take place adiabat-
ically. Assume that each constituent in the products is at 0.1-MPa pressure and at the
adiabatic flame temperature. This combustion process is shown schematically in Fig.
14.5. The temperature of the surroundings is 25°C. -u
For this combustion process, determine (1) the increase in entropy during combus-
tion and (2) the availability of the products of combustion.
Control volume: Combustion chamber.
Inlet states: P, T known for each gas.
Exit states: P known for each gas.
Sketch: Figure 14.5.
Model: All ideal gases, Table A.9; Table A.10 for ethene. ;
Analysis
The combustion equation is
C 2 H 4 (g)+ 1202 + 12(3.76)N 2 ->2C0 2 + 2H 2 0(g) + 90 2 +'45.1N 2
The adiabatic flame temperature is detennined first.
First law:
th - ///>
Solution
52 467 = 2(-393 522 + A£ C o 2 ) + 2 <~ 241 826 + A Vxg>) + 9 ^ + 45.lAft N;
By a trial-and-error solution we find the adiabatic flame temperature to be
1016 K. . . .
We now proceed to find the change in entropy during this adiabatic combustion
process.
it
= 219.330 + 12(205.147) + 45.1(191.610)
= 11 322.705 kJ/kmol (fuel) K
S P = 2 (»^)ioi6 - (^cc, + Hpm + + 45.1^) 10 i6
/'
= 2(270.194) + 2(233.355) + 9(244.135) + 45.1(228.691)
13 518.277 kJ/kmol (fuel) K
S P -S R = 2195.572 kJ/kmol (fuel) K
Second-Law Analysis of Reacting Systems 11 593
' Each at
T = 25°C
P = 0.1 MPa
FIGURE 14.5 Sketch
for Example 14.14.
v////////////////////^,
Each at the
adiabatic ffame
' temperature and
P = 0.1 MPa
Since this is an adiabatic process, the increase in entropy indicates the irre-
versibility of the adiabatic combustion process. This irreversibility can be found from
Eq. 14.25.
\ p R
- 298.15 X 2195.572 = 654 610kJ/kmot
= f|~™ = 23 334kj/kgfuel
Therefore, the availability after the combustion process is
tjf F = 46 851 - 23 334 = 23 517 kJ/kg
The availability of the products can also be found from the relation
p
Since in this problem the products are separated, and each is at 0.1-MPa pres-
sure and the adiabatic flame temperature of 1016 K, this equation can be evaluated,
yielding
to- - 2».K*2 - *8) - Tic? 1 *8)]
p
= 2(34 271) + 2(26 618) + 9(23 268) + 45.1(21 985)
— 298.15[2(270.194 - 213.795) + 2(233.355 - 188.834)
+ 9(244.135 - 205.147) + 45.1(228.691 - 191.610)]
= 659 746 kJ/kmol - 23 5 17 kJ/kg
In other words, if every process after the adiabatic combustion process is reversible, the
maximum amount of work that could be done is 23 517 kJ/kg fuel. This compares to a
value of 46 851 kJ/kg for the reversible isothermal reaction. This means that if we had
an engine with the indicated adiabatic combustion process, and if all other processes
were completely reversible, the efficiency would be about 50%.
What is the effect of having the combustion process of Example 14.14 take place
with pure oxygen instead of air? In this case, the first law is the same, except with no
594 Wi CHAPTER FOURTEEN CHEMICAL REACTIONS
nitrogen term in the reactants or products. As a result, the adiabatic flame temperature
will be much higher, without having to heat up the 45.1 kmol of nitrogen. By trial and
error, this temperature is found to be 2800 K. The entropy of the reactants is now
2681.094 kJ/kmol fuel K, and that of the products at 2800 K is found to be 3760.695
kJ/kmol fuel K. Therefore, the process irreversibility from Eq. 14.25 is, converting to a
mass basis, 1 1474 kJ/kg fuel, such that the availability of the products is 35 377 kJ/kg
fuel, significantly higher than for combustion with air in Example 14.14. This is be-
cause the products in this case are at a much higher temperature.
In the two prior examples we assumed, for purposes of simplifying the calcula-
tion, that the constituents in the reactants and products were separated, and each was at
0.1-MPa pressure. This of course is not a realistic problem. In the following example,
Example 14.13 is repeated with the assumption that the reactants and products each
consist of a mixture at 0.1-MPa pressure.
Example 14.15
Consider the same combustion process as in Example 14.13, but assume that the re-
actants consist of a mixture at 0.1-MPa pressure and 25°C and that the products also
consist of a mixture at 0.1 MPa and 25°C. Thus, the combustion process is as shown
in Fig. 14.6.
Determine the work that would be done if this combustion process took place re-
versibly and in pressure and temperature equilibrium with the surroundings.
Control volume: Combustion chamber.
Inlet state: P, T known.
Exit state: P, Tknown.
Sketch: Figure 14,6.
Model: Reactants— ideal-gas mixture, Table A.9. Products— ideal-gas
mixture, Table A.9.
Analysis
The combustion equation, as noted previously, is
QH^fg) + 3(4)0 2 + 3(4)(3.76)N 2 -> 2C0 2 + 2H 2 0(g) + 9 2 + 45.1N 2
Reactants
P = OA MPa, T = 25°C
Wc.v. Qcv.
n
Products
P = 0.1 MPa, T = 25°C
FIGURE 14.6 Sketch
for Example 14.15.
Second-law Analysis of Reacting Systems H 595
In this case we must find the entropy of each substance as it exists in the mixture—
that is, at its partial pressure and the given temperature of 25°C. Because the absolute en-
tropies given in Tables A.9 and A.10 are at 0.1-MPa pressure and 25°Q the entropy of
each constituent in the mixture can be found, using Eq. 14.20 from the relation
where 5* = partial entropy of the constituent in the mixture
j° =absoluteentropyatthesametemperatureand0.1-MPapressure
P — pressure of the mixture
P° = 0.1 -MPa pressure
y = mole fraction of the constituent
Since P° and the pressure of the mixture are both 0.1 MPa, the partial entropy of
each constituent can be found by the relation
S* = ^ - R]ny = ^ + Rhijj
Solution
For the reactants:
n
1/y
R In 1/y
C 2 H 4
1
58.1
33.774
219.330
253.104
o 2
12
4.842
13.114
205.147
218.261
45.1
1.288
2.104
191.610
193.714
58.1
ie products:
n
Vy
R In Vy
S*
co 2
2
29.05
28.011
213.795
241.806
H 2
2
29.05
28.011
188.834
216.845
2
9
6.456
15.506
205.147
220.653
N 2
45.1
1.288
2.104
191.610
193.714
58.1
596 B CHAPTER FOURTEEN CHEMICAL REACTIONS
With the assumption of ideal-gas behavior, the enthalpy of each constituent is equal to
the enthalpy of formation at 25°C. The values of entropy are as just calculated. There-
fore, from Eq. 14.24,
-298.15(5^ + 12SS, + - - 2S£ ofe) - 9S& - 45.1S&)
- 52 467 • 2(- 393 522) - 2(-241 826)
29!U5[253.104 -I 12(218.264) + 45.1(193.714) // y
-2(241.806) - 2(216.845) - 9(220.653) - 45.1(193.714)]
- ! 332 378 kJ/kmol
Note that this value is essentially the same as the value that was obtained in Exam-
ple 14.13, when the reactants and products were each separated and at 0.1 -MPa pressure.
14.9 Fuel Cells
The previous examples raise the question of the possibility of a reversible chemical reac-
tion. Some reactions can be made to approach reversibility by having them take place m
an electrolytic ceil, as described in Chapter 1. When a potential exactly equal to the elec-
tromotive force of the ceil is applied, no reaction takes place. When the applied potential
is increased slightly, the reaction proceeds in one direction, and if the applied potential is
decreased slightly, the reaction proceeds in the opposite direction. The work done is the
electrical energy supplied or delivered.
Consider a reversible reaction occurring at constant temperature equal to that of its
environment. The work output of the fuel cell ts
where AG is the change in Gibbs function for the overall chemical reaction. We also real-
ize that the work is given in terms of the charged electrons flowing through an electrical
potential % as
in which n t is the number of kilomoles of electrons flowing through the external circuit and
N a e = 6.022 136 X 10 26 elec/kmol X 1.602 177 X 10~ 22 kJ/elec V
= 96 485 kJ/kmoiV
FUEL CELLS B 597
Thus, for a given reaction, the maximum (reversible reaction) electrical potential %° of a
fuel cell at a given temperature is
mMz (14 - 28)
EXAMPLE 14.16 Calculate the reversible electromotive force (EMF) at 25°C for the hydrogen-oxygen
fuel cell described in Section 1,2.
Solution
The anode side reaction was stated to be
2II 2 — > 411 ' + 4e
and the cathode side reaction is
4II + (• 4e r 2 -»2II 2
Therefore, the overall reaction is, in kilomoles,
2H 2 ! 2 - >2H 2 ()
for which 4 kmol of electrons flow through the external circuit. Let us assume that each
component is at its standard-state pressure of 0.1 MPa and that the water formed is liq-
uid. Then
- 2(-285 830) 2(0) - 1(0) - -571 660 kJ
Atf - 24 Aj! 2s\ - si
= 2(69.950) - 2(130.678) - 1(205.148) - -326.604 kJ/K
AG - 571 660 - 298.15( 326.604) - -474 283 kJ .
Therefore, from Eq. 14.28,
-(-474 283)
%° ■-- ■ --■ -- 1 229 V
96 485 X 4 ' C J v
In Example 14.16, we found the shift in the Gibbs function and the reversible EMF
at 25°C. In practice, however, many fuel cells operate at an elevated temperature where
the water leaves as a gas and not as a liquid; thus, it carries away more energy. The com-
]
598 M CHAPTER FOURTEEN CHEMICAL REACTIONS
FIGURE 14.7
Hydrogen-oxygen fuel
cell ideal EMF as a
function of temperature.
1.2
& 1.1
1.0
Water liquid
J L
_L
J I i
300 400 500 600 700 800 900 1000 1100
Temperature (K)
putations can be done for a range of temperatures, leading to lower EMF as the tempera-
ture increases. This behavior is shown in Fig. 14.7.
A variety of fuel cells are being investigated for use in stationary as well as mobile
power plants. The low-temperature fuel cells use hydrogen as the fuel, whereas the higher
temperature cells can use methane and carbon monoxide that are then internally reformed
into hydrogen and carbon dioxide. The most important fuel cells are listed in Table 14.4
with their main characteristics.
The low-temperature fuel cells are very sensitive to being poisoned by CO gas so
that they require an external reformer and purifier to deliver hydrogen gas. The higher
temperature fuel cells can reform natural gas, mainly methane, but also ethane and
propane as shown in Table 14.2, into hydrogen gas and carbon monoxide inside the cell.
The latest research is being done with gasified coal as a fuel and operating the cell at
higher pressures like 15 atmospheres. As the fuel cell has exhaust gas with a small
amount of fuel in it, additional combustion can occur and then combine the fuel cell with
a gas turbine or steam power plant to utilize the exhaust gas energy. These combined
cycle power plants strive to have an efficiency of up to 60%.
Table 14.4
Fuel Cell
PEC
PAC
MCC
soc
Polymer
Phosphoric
Molten
Solid
Electrolyte
Acid
Carbonate
Oxide
T
80°C
200°C
650°C
900°C
Fuel
Hydrogen, H 2
Hydrogen, H 2
CO, Hydrogen
Natural Gas
Carrier
H +
H +
cor
Charge, n 4
2e~ per H 2
2e~ per H 2
2e~ per H 2
8e~ per CH 4
2e~ per CO
Catalyst
Pt
Pt
Ni
Zr0 2
Poison
CO
CO
Evaluation of Actual Combustion Processes H 599
14.10 Evaluation of Actual
Combustion Processes
A number of different parameters can be defined for evaluating the performance of an actual
combustion process, depending on the nature of the process and the system considered. In
the combustion chamber of a gas-turbine, for example, the objective is to raise the tempera-
ture of the products to a given temperature (usually the maximum temperature the metals in
the turbine can withstand). If we had a combustion process that achieved complete combus-
tion and that was adiabatic, the temperature of the products would be the adiabatic flame
temperature. Let us designate the fuel-air ratio needed to reach a given temperature under
these conditions as the ideal fuel-air ratio. In the actual combustion chamber, the combus-
tion will be incomplete to some extent, and there will be some heat transfer to the surround-
ings. Therefore, more fuel will be required to reach the given temperature, and this we
designate as the actual fuel-air ratio. The combustion efficiency, t) cosrib is defined here as
FA. ideal
T?comb = ~pl (14.29)
-^-"actual
On the other hand, in the furnace of a steam generator (boiler), the purpose is to
transfer the maximum possible amount of heat to the steam (water). In practice, the effi-
ciency of a steam generator is defined as the ratio of the heat transferred to the steam to
the higher heating value of the fuel. For a coal this is the heating value as measured in a
bomb calorimeter, which is the constant-volume heating value, and it corresponds to the
internal energy of combustion. We observe a minor inconsistency, since the boiler in-
volves a flow process, and the change in enthalpy is the significant factor. In most cases,
however, the error thus introduced is less than the experimental error involved in measur-
ing the heating value, and the efficiency of a steam generator is defined by the relation
_ heat transferred to steam/kg fuel
1Wn se ™ " higher heating value of the fuel C14 ' 30}
In an internal-combustion engine the purpose is to do work. The logical way to
evaluate the performance of an internal-combustion engine would be to compare the ac-
tual work done to the maximum work that would be done by a reversible change of state
from the reactants to the products. This, as we noted previously, is called the second-law
efficiency.
In practice, however, the efficiency of an internal-combustion engine is defined as the
ratio of the actual work to the negative of the enthalpy of combustion of the fuel (that is, the
constant-pressure heating value). This ratio is usually called the thermal efficiency,
v = = w (14.31)
— hgp^ heating value
The overall efficiency of a gas-turbine or steam power plant is defined in the same
way. It should be pointed out that in an internal-combustion engine or fuel-burning steam
power plant, the fact that the combustion is itself irreversible is a significant factor in the
relatively low thermal efficiency of these devices.
One other factor should be pointed out regarding efficiency. We have noted that the
enthalpy of combustion of a hydrocarbon fuel varies considerably with the phase of the water
in the products, which leads to the concept of higher and lower heating values. Therefore,
when we consider the thermal efficiency of an engine, the heating value used to determine
600 M chapter Fourteen chemical reactions
this efficiency must be borne in mind. Two engines made by different manufacturers may
have identical performance, but if one manufacturer bases his or her efficiency on the higher
heating value and the other on the lower heating value, the latter will be able to claim a
higher thermal efficiency. This claim is not significant, of course, as the performance is the
same; this would be revealed by consideration of how the efficiency was defined.
The whole matter of the efficiencies of devices that undergo combustion processes
is treated in detail in textbooks dealing with particular applications; our discussion is in-
tended only as an introduction to the subject. Two examples are given, however, to illus-
trate these remarks.
EXAMPLE 14.17 The combustion chamber of a gas-turbine uses a liquid hydrocarbon fuel that has an ap-
proximate composition of CsH 18 . During testing, the following data are obtained.
T ■
-*■ air
= 400K
T
1 products
- hook
v •
"atr
- 100 m/s
^products
= 150 m/s
= 50°C
actual
= 0.0211 kg fuel/kg air
Calculate the combustion efficiency for this process.
Control volume: Combustion chamber.
Inlet states: T known for air and fuel.
Exit state: T known.
Model: Air and products— ideal gas, Table A. 8 . Fuel— Table A. 9.
Analysis
For the ideal chemical reaction the heat transfer is zero. Therefore, writing the first law
for a control volume that includes the combustion chamber, we have
H R + KE R = Hp + KE P
- [hj + C p (50 - 25)]^ + «c, (AA +
+ 3 ,^ + *£)
+ K -12,)(A^^) 0i + 3.76^ + ^) Ki
EVALUATION OF ACTUAL COMBUSTION PROCESSES H 601
Solution
Therefore,
H R + KE R — -250 105 4- 2.23 X 114.23(50 - 25)
32 X (100) r
3034 +
+ 3.76;/ 0z
2971 +
2 X 1000 ^
28.02 X (100) 2
2 X 1000
-243 737 4- 14 892h Qj
Hp + KEp — 8
-393 522 + 38 891 +
44.01 X (150) 2
+ 9
-241 826 4- 30 147 +
2 X 1000
18.02 X(150) 2
+ <> 0j - 12.5)
26 218 +
2 X 1000
32 X (150) 2 "
+ 3,76h 0i
24 758 +
2 X 1000
28,02 X (150) 2
2 X 1000
= -5 068 599 + 120 853« D
-243 737 + 14 892« 0i = -5 068 599 + 120 853h 0j
« 0j = 45.53 kmol 2 /kmol fuel
kmol air/kmol fuel 4.76(45.53) = 216.72
F Ad&i = 216 jjt^L 97 = 0,0182 k S fuel/kg air
Vcomb = q 021 1 X 100 = 86,2 P ercent
EXAMPLE 14.18 In a certain steam power plant, 325 000 kg of water per hour enters the boiler at a pres-
sure of 12.5 MPa and a temperature of 200°C. Steam leaves the boiler at 9 MPa, 500°C.
The power output of the turbine is 81 000 kW. Coal is used at the rate of 26 700 kg/h
and has a higher heating value of 33 250 kJ/kg. Determine the efficiency of the steam
generator and the overall thermal efficiency of the plant.
In power plants, the efficiency of the boiler and the overall efficiency of the plant
are based on the higher heating value of the fuel.
Solution
The efficiency of the boiler is defined by Eq. 14.30 as
_ heat transferred to H 2 0/kg fuel
^eamg^tor - - higher hearing value
602 H CHAPTER FOURTEEN CHEMICAL REACTIONS
Therefore,
= 325 000(3386.1 - 857.1) x Q = „ 6%
Vstezm generator 26 700 X 33 250
The thermal efficiency is defined by Eq. 14.31,
= w - 81 000 X 3600 x 10Q = 32 g%
77(11 heating value 26 700 X 33 250
>UMMARY
An introduction to combustion of hydrocarbon fuels and chemical reactions in general is
given. A simple oxidation of a hydrocarbon fuel with pure oxygen or air burns the hydro-
gen to water and the carbon to carbon dioxide. We apply the continuity equation for each
kind of atom to balance the stoichiometric coefficients of the species in the reactants and
products. The reactant mixture composition is described by the air-fuel ratio on a mass or
mole basis or percent theoretical air or equivalence ratio, according to the practice of the
particular area of use. The product of a given fuel for a stoichiometric mixture and com-
plete combustion is unique, whereas actual combustion can lead to incomplete combus-
tion and more complex products described by measurements on a dry or wet basis. As
water is part of the products, they have a dew point, so it is possible to see water condens-
ing out from the products as they are cooled.
Because of the chemical changes from the reactants to the products, we need to
measure energy from an absolute reference. Chemically pure substances (not compounds
like CO) in their ground state (graphite for carbon, not diamond form) are assigned a
value of for the formation enthalpy at reference temperature and pressure (25°C, 100
kPa). Stable compounds have a negative formation enthalpy, and unstable compounds
have a positive formation enthalpy. The shift in the enthalpy from the reactants to the
products is the enthalpy of combustion, which is also the negative of the heating value
HV. When a combustion process takes place without any heat transfer, the resulting prod-
uct temperature is the adiabatic flame temperature. The enthalpy of combustion, the heat-
ing value (lower or higher), and the adiabatic flame temperature depend on the mixture
(fuel and A/F ratio) and the reactant' s supply temperature. When a single unique number
for these properties is used, it is understood to be for a stoichiometric mixture at the refer-
ence conditions.
Similarly to the enthalpy, an absolute value of entropy is needed for the application
of the second law. The absolute entropy is zero for a perfect crystal at K, which is the
third law of thermodynamics. The combustion process is an irreversible process, and
there is thus a loss of availability (exergy) associated with it. This irreversibility is in-
creased by mixtures different from stoichiometric and by dilution of the oxygen (i.e., ni-
trogen in air), which lowers the adiabatic flame temperature. From the concept of flow
exergy, we apply the second law to find the reversible work given by the change in Gibbs
function. A process that has less irreversibility than combustion at high temperature is the
chemical conversion in a fuel cell where we approach a chemical equilibrium process
(covered in detail in the following chapter). Here the energy release is directly converted
key Concepts and formulas H 603
into an electrical power output, a system that is currently under intense study and devel-
opment for future energy conversion -systems.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to:
• Write the combustion equation for the stoichiometric reaction of any fuel.
* Balance the stoichiometric coefficients for a reaction with a set of products mea-
sured on a dry basis,
* Handle the combustion of fuel mixtures as well as moist air oxidicer.
* Apply the energy equation with absolute values of enthalpy or internal energy.
• Use the proper tables for high-temperature products,
• Deal with condensation of water in low-temperature products of combustion.
• Calculate the adiabatic flame temperature for a given set of reactants.
• Know the difference between the enthalpy of formation and the enthalpy of
combustion.
• Know the definition of the higher and lower heating values.
* Apply the second law to a combustion problem and find irreversibilities.
■ Calculate the change in Gibbs function and the reversible work.
* Know how a fuel cell operates and how to find its electrical potential.
* Know some basic definition of combustion efficiencies.
Key Concepts
and formulas
Reaction
Stoichiometric ratio
Stoichiometric coefficients
Stoichiometric reaction
Air-fuel ratio
Equivalence ratio
Enthalpy of formation
Enthalpy of combustion
Heating value HV
Internal energy of combustion
Adiabatic flame temperature
Reversible work
Gibbs function
Irreversibility
Fuel + Oxidizer Products
hydrocarbon + air =i> carbon dioxide + water 4-
nitrogen
No excess fuel, no excess oxygen
Factors to balance atoms between reactants and products
Q^+v 02 CO 2 + 3.76N 2 )
=> v COl C0 2 + v Hi0 H 2 + v N N 2
= x + y/4; v COi = x; v HjQ = y/2; v N; - 3 .76v 02
AF = Hl^L = AF M*L
FA
.AF,
AF
hf, zero for chemically pure substance, ground state
= Hp - H R
Urp = Up ~ U R ~ hup - RT(n p - n R ) if ideal gases
H P = H R if flow; Up = U R if constant volume
}r ev =G R ~G P = -AG = ~(&H- TAS)
This requires that any Q is transferred at the local T.
G = H-TS
i = W»-Y!= T£ g Jm = V gea
j=W m ~W= TfjSgJh = TiSU for 1 kmole fuel
604 M CHAPTER FOURTEEN CHEMICAL REACT tONS
Concept-Study Guide problems
14.1 How many kmoles of air are needed to burn 14.11
1 kmol of carbon?
14.2 If I burn 1 kmol of hydrogen H 2 with 6 kmol air, 14.12
what is the A/F ratio on a mole basis and what is
the percent theoretical air? 14.13
14.3 Why would I sometimes need A/F on a mole
basis? on a mass basis? 14,14
14.4 Why is there no significant difference between the 14.15
number of moles of reactants versus products in
combustion of hydrocarbon fuels with air? 14A6
14.5 For the 110% theoretical air in Eq. 14.8, what is
the equivalence ratio? Is that mixture rich or lean?
14.6 Why are products measured on a dry basis? 14.17
14.7 What is the dew point of hydrogen burned with
stoichiometric pure oxygen? air? 14.18
14.8 How does the dew point change as the equiva-
lence ratio goes from 0.9 to 1 to 1 . 1? 14,19
14.9 In most cases, combustion products are exhausted
above the dew point. Why? 14.20
14.10 Why does combustion contribute to global
warming?
What is the enthalpy of formation for oxygen as
2 ?ifO?forC0 2 ?
How is a fuel enthalpy of combustion connected
to its enthalpy of formation?
What is the higher and lower heating value HHV,
LHV ofn-butane?
What is the value of h fg for n-octane?
Why do some fuels not have entries for liquid fuel
in Table 14.3?
Does it make a difference for the enthalpy of
combustion whether I burn with pure oxygen or
air? What about the adiabatic flame temperature?
What happens to the adiabatic flame temperature
if I burn rich or lean?
Is the irreversibility in a combustion process sig-
nificant? Why is that?
If the A/F ratio is larger than stoichiometric, is it
more or less reversible?
What makes the fuel cell attractive from a power-
generating point of view?
Homework problems
Fuels and the Combustion Process
14.21 Calculate the theoretical air-fuel ratio on a mass
and mole basis for the combustion of ethanol,
CjHjOH.
14.22 A certain fuel oil has the composition C i0 H 22 . If
this fuel is burned with 150% theoretical air, what
is the composition of the products of combustion?
14.23 Methane is burned with 200% theoretical air.
Find the composition and the dew point of the
products.
14.24 In a combustion process with decane, C I& H 22 , and
air, the dry product mole fractions are 83.61% N 2)
4.91% 2l 10.56% C0 2 , and 0.92% CO. Find the
equivalence ratio and the percent theoretical air of
the reactants.
14.25 Natural gas B from Table 14.2 is burned with
20% excess air. Determine the composition of the
products.
14.26 A Pennsylvania coal contains 74.2% C, 5.1% H,
6.7% O (dry basis, mass percent) plus ash and
small percentages of N and S. This coal is fed into
a gasifier along with oxygen and steam, as shown
in Fig. P14.26. The exiting product gas composi-
tion is measured on a mole basis to: 39.9% CO,
30.8% H 2 , 11.4% C0 2 , 16.4% H 2 plus small
percentages of CH 4 , N 2) and H 2 S. How many kilo-
grams of coal are required to produce 100 kmol
of product gas? How much oxygen and steam are
required?
Oxygen
Product gas
FIGURE P14.26
14,27 Repeat Problem 14.26 for a certain Utah coal that
contains, according to the coal analysis, 68.2% C,
4.8% H, and 15.7% O on a mass basis. The exit-
ing product gas contains 30.9% CO, 26.7% H 2)
15.9% C0 2 , and 25.7% H 2 on a mole basis.
HOMEWORK PROBLEMS II 605
14.28 For complete stoichiometric combustion of gaso-
line, C 7 H 17 , determine the fuel molecular weight,
the combustion products, and the mass of carbon
dioxide produced per kg of fuel burned.
14.29 A sample of pine bark has the following ultimate
analysis on a dry basis, percent by mass; 5.6% H,
53.4% C, 0.1% S, 0.1% N, 37.9% O, and 2,9%
ash. This bark will be used as a fuel by burning it
with 1 00% theoretical air in a furnace. Determine
the air-fuel ratio on a mass basis,
14.30 Liquid propane is burned with dry air. A volumet-
ric analysis of the products of combustion yields
the following volume percent composition on a
dry basis: 8.6% C0 2 , 0.6% CO, 7.2% 2 , and
83.6% N 2 . Determine the percent of theoretical ab-
used in this combustion process.
14.31 A fuel, C x H y , is burned with dry air, and the prod-
uct composition is measured on a dry mole basis
to be: 9.6% C0 2 , 7.3% 2 , and 83.1% N 2 . Find
the fuel composition (x/y) and the percent theoret-
ical air used.
14.32 For the combustion of methane, 150% theoretical
air is used at 25°C, 100 kPa, and relative humidity
of 70%. Find the composition and dew point of
the products.
14.33 Many coals from the western United States have a
high moisture content. Consider the following
sample of Wyoming coal, for which the ultimate
analysis on an as-received basis is, by mass:
Component Moisture H C S N O Ash
%mass 28.9 3.5 48.6 0.5 0.7 12.0 5.8
This coal is burned in the steam generator of a
large power plant with 150% theoretical air. De-
termine the air-fuel ratio on a mass basis.
14.34 Pentane is burned with 120% theoretical air in a
constant-pressure process at 100 kPa. The products ■
are cooled to ambient temperature, 20°C. How
much mass of water is condensed per kilogram of
fuel? Repeat the answer, assuming that the air used
in the combustion has a relative humidity of 90%.
14.35 The coal gasifier in an integrated gasification
combined cycle (IGCC) power plant produces a
gas mixture with the following volumetric percent
composition:
Product CH 4 H 2 CO C0 2 N 2 H 2 H 2 S NH 3
%voI. 0.3 29.6 41.0 10.0 0.8 17.0 1.1 0.2
This gas is cooled to 40°C, 3 MPa, and the H 2 S
• and NH 3 are removed in water scrubbers. Assum-
ing that the resulting mixture, which is sent to the
combustors, is saturated with water, determine the
mixture composition and the theoretical air-fuel
ratio in the combustors.
14.36 The hot exhaust gas from an internal combustion
engine is analyzed and found to have the follow-
ing percent composition on a volumetric basis at
the engine exhaust manifold: 10% C0 2 , 2% CO,
13% H 2 0, 3% 2 , and 72% N 2 . This gas is fed to
an exhaust gas reactor and mixed with a certain
amount of air to eliminate the carbon monoxide,
as shown in Fig. P14.36. It has been determined
that a mole fraction of 10% oxygen in the mixture
at state 3 will ensure that no CO remains. What
must be the ratio of flows entering the reactor?
Exhaust gas
Exhaust out
FIGURE P14.36
14.37 Butane is burned with dry air at 40°C, 100 kPa, with
AF = 26 on a mass basis. For complete combustion
find the equivalence ratio, % theoretical air, and the
dew point of the products. How much water (kg/kg
fuel) is condensed out, if any, when the products are
cooled down to ambient temperature 40°C?
14.38 Methanol, CH 3 OH, is burned with 200% theoreti-
cal air in an engine, and the products are brought
to 100 kPa, 30°C. How much water is condensed
per kilogram of fuel?
14.39 The output gas mixture of a certain air-blown coal
gasifier has the composition of producer gas as
listed in Table 14.2. Consider the combustion of
this gas with 120% theoretical air at 100 kPa pres-
sure. Determine the dew point of the products and
find how many kilograms of water will be con-
densed per kilogram of fuel if the products are
cooled 10°C below the dew-point temperature.
14.40 In an engine, liquid octane and ethanol, mole ra-
tion 9:1, and stoichiometric air are taken in at 298
K, 100 kPa. After complete combustion, the prod-
ucts run out of the exhaust system where they are
L
606 M Chapter fourteen Chemical reactions
cooled to 10°C. Find the dew point of the prod-
ucts and the mass of water condensed per kilo-
gram of fuel mixture.
Energy Equation, Enthalpy of Formation
14.41 A rigid vessel initially contains 2 kmol of carbon
and 2 kmol of oxygen at 25°C, 200 kPa. Combus-
tion occurs, and the resulting products consist of 1
kmol of carbon dioxide, 1 kmol of carbon monox-
ide, and excess oxygen at a temperature of 1000
K. Determine the final pressure in the vessel and
the heat transfer from the vessel during the
process.
14.42 In a test of rocket propeliant performance, liquid
hydrazine (N 2 H 4 ) at 100 kPa, 25°C, and oxygen
gas at 100 kPa, 25°C, are fed to a combustion
chamber in the ratio of 0.5 kg 2 /kg N 2 H 4 . The
heat transfer from the chamber to the surround-
ings is estimated to be 1 00 kJ/kg N 2 H 4 . Determine
the temperature of the products exiting the cham-
ber. Assume that only H 2 0, H 2 , and N 2 are pre-
sent. The enthalpy of formation of liquid
hydrazine is +50 417 kJ/kmol.
14.43 The combustion of heptane C 7 H 16 takes place in a
steady-flow burner where fuel and air are added
as gases at P 0t T . The mixture has 125% theoreti-
cal air, and the products pass through a heat ex-
changer where they are cooled to 600 K. Find the
heat transfer from the heat exchanger per kmol of
heptane burned.
14.44 Butane gas and 200% theoretical air, both at
25°C, enter a steady-flow combustor. The prod-
ucts of combustion exit at 1000 K. Calculate the
heat transfer from the combustor per kmol of bu-
tane bumed.
14.45 One alternative to using petroleum or natural gas
as fuels is ethanol (C 2 H 5 OH), which is commonly
produced from grain by fermentation. Consider a '
combustion process in which liquid ethanol is
burned with 120% theoretical air in a steady-flow
process. The reactants enter the combustion
chamber at 25°C, and the products exit at 60°C,
100 kPa. Calculate the heat transfer per kilomole
of ethanol.
14.46 Do the previous problem with the ethanol fuel de-
livered as a vapor.
14.47 Another alternative to using petroleum or natural
gas as fuels is methanol, (CH 3 OFJ), which can be
produced from coal. Both methanol and ethanol
have been used in automotive engines. Repeat
Problem 14.45 using liquid methanol as the fuel
instead of ethanol.
14.48 Another alternative fuel to be seriously consid-
ered is hydrogen. It can be produced from water
by various techniques that are under extensive
study. Its biggest problems at the present time are
cost, storage, and safety, repeat Problem 14.45
using hydrogen gas as the fuel instead of ethanol.
14.49 In a new high-efficiency furnace, natural gas, as-
sumed to be 90% methane and 10% ethane (by
volume) and 110% theoretical air each enter at
25°C, 100 kPa, and the products (assumed to be
100% gaseous) exit the furnace at 40°C, 100 kPa.
What is the heat transfer for this process? Com-
pare this to an older furnace where the products
exit at 250°C, 100 kPa.
14.50 Repeat the previous problem but take into account
the actual phase behavior of the products exiting
the furnace.
14.51 Pentene, C 5 H I0 , is bumed with pure oxygen in a
steady-flow process. The products at one point are
brought to 700 K and used in a heat exchanger,
where they are cooled to 25°C. Find the specific
heat transfer in the heat exchanger.
14.52 Methane, CH 4 , is bumed in a steady-flow process
with two different oxidizers: Case A; Pure oxy-
gen, 2 , and case B: A mixture of 2 + xAr. The
reactants are supplied at T , P and the products
for both cases should be at 1800 K. Find the re-
quired equivalence ratio in case A and the amount
of argon, x, for a stoichiometric ratio in case B.
14.53 A closed, insulated container is charged with a
stoichiometric ratio of oxygen and hydrogen at
25°C and 150 kPa. After combustion, liquid water
at 25°C is sprayed in such a way that the final
temperature is 1200 K. What is the final pressure?
14.54 Gaseous propane mixes with air, both supplied at
500 K, 0. 1 MPa. The mixture goes into a combus-
tion chamber, and products of combustion exit at
1300 K, 0.1 MPa. The products analyzed on a dry
basis are 11.42% C0 2 , 0.79% CO, 2.68% 2 , and
85.11% N 2 on a volume basis. Find the equiva-
lence ratio and the heat transfer per kmol of fuel.
HOMEWORK. PROBLEMS B 607
Enthalpy of Combustion and Heating Value
14.55 Liquid pentane is burned with dry air, and the
products are measured on a dry basis as 10.1%
C0 2 , 0.2% CO, 5.9% 2 remainder N 2 . Find the
enthalpy of formation for the fuel and the actual
equivalence ratio.
14.56 Phenol has an entry in Table 14.3, but it does not
have a corresponding value of the enthalpy of for-
mation in Table A. 10. Can you calculate it?
14.57 Do Problem 14.43 using Table 14.3 instead of
Table A. 10 for the solution.
14.58 Wet biomass waste from a food-processing
plant is fed to a catalytic reactor, where in a
steady-flow process it is converted into a low-
energy fuel gas suitable for firing the processing
plant boilers. The fuel gas has a composition of
50% methane, 45% carbon dioxide, and 5% hy-
drogen on a volumetric basis. Determine the
lower heating value of this fuel gas mixture per
unit volume.
14.59 Determine the lower heating value of the gas gen-
erated from coal, as described in Problem 14.35.
Do not include the components removed by the
water scrubbers.
14.60 Do Problem 14.45 using Table 14.3 instead of
Table A. 10 for the solution.
14.61 Propylbenzene, C 9 H 12 , is listed in Table 14.3, but
not in Table A.9. No molecular weight is listed in
the book. Find the molecular weight, the enthalpy
of formation for the liquid fuel, and the enthalpy
of evaporation.
14.62 Determine the higher heating value of the sample
Wyoming coal as specified in Problem 14.33.
14.63 Do Problem 14.47 using Table 14.3 instead of
Table A.10 for the solution.
14.64 A burner receives a mixture of two fuels with,
mass fraction 40% n-butane and 60% methanol,
both vapor. The fuel is burned with stoichiometric
air. Find the product composition and the lower
heating value of this fuel mixture (kJ/kg fuel
mix).
14.65 Consider natural gas A and natural gas D, both of
which are listed in Table 14.2. Calculate the en-
thalpy of combustion of each gas at 25°C, assum-
ing that the products include vapor water. Repeat
the answer for liquid water in the products.
14.66 Blast furnace gas in a steel mill is available at
. 250°C to be burned for the generation of steam.
The composition of this gas is, on a volumetric
basis,
Component CH 4 H 2 CO
Percent by volume 0.1 2.4 23.3
C0 2 N 2
14.4 56.4
H 2
3.4
Find the lower heating value (kJ/m 3 ) of this gas at
250°C and ambient pressure.
14.67 Natural gas, we assume methane, is burned with
200% theoretical air, shown in Fig. P14.67, and
the reactants are supplied as gases at the reference
temperature and pressure. The products are flow-
ing through a heat exchanger where they give off
energy to some water flowing in at 20°C, 500 kPa,
and out at 700°C, 500 kPa. The products exit at
400 K to the clumney. How much energy per
kmole fuel can the products deliver, and how
many kg water per kg fuel can they heat?
Fuel
FIGURE P14.67
14.68 Gasoline, C 7 H 17 , is burned in a steady-state burner
with stoichiometric air at P , T Q , shown in Fig.
P14.68. The gasoline is flowing as a liquid at T Q to
a carburetor where it is mixed with air to produce
a fuel air-gas mixture at T . The carburetor takes
some heat transfer from the hot products to do
the heating. After the combustion, the products
60O K
<*H —
H. Excb.
Fuel
Air
Carb.
Combustor
FIGURE P14.68
608 H CHAPTER FOURTEEN CHEMICAL REACTIONS
go through a heat exchanger, which they leave at
600 K. The gasoline consumption is 10 kg per
hour. How much power is given out in the heat
exchanger, and how much power does the carbu-
retor need?
14.69 In an engine a mixture of liquid octane and
ethanol, mole ratio 9 : 1, and stoichiometric air are
taken in at T Q , P . In the engine, the enthalpy of
combustion is used so that 30% goes out as work,
30% goes out as heat loss, and the rest goes out
the exhaust. Find the work and heat transfer per
kilogram of fuel mixture and also the exhaust
temperature.
14.70 Liquid nttromethane is added to the air in a carbu-
retor to make a stoichiometric mixture where both
fuel and air are added at 298 K, 100 kPa. After
combustion, a constant-pressure heat exchanger
brings the products to 600 K before being ex-
hausted. Assume the nitrogen in the fuel becomes
N 2 gas. Find the total heat transfer per kmol fuel
in the whole process.
Adiabatic Flame Temperature
14.71 Hydrogen gas is burned with pure oxygen in a
steady-flow burner, shown in Fig. PI 4.71, where
both reactants are supplied in a stoichiometric
ratio at the reference pressure and temperature.
What is the adiabatic flame temperature?
FIGURE P14.71
14.72 In a rocket, hydrogen is burned with air, both-
reactants supplied as gases at P » T o- com -
bustion is adiabatic, and the mixture is stoichio-
metric (100% theoretical air). Find the products'
dew point and the adiabatic flame temperature
(-2500 K).
14.73 Carbon is burned with air in a furnace with 150%
theoretical air, and both reactants are supplied at
the reference pressure and temperature. What is
the adiabatic flame temperature?
14.74 A stoichiometric mixture of benzene, C 6 H 6 , and
air is mixed from the reactants flowing at 25°C,
100 kPa, Find the adiabatic flame temperature.
What is the error if constant-specific heat at T for
the products from Table A.5 is used?
14.75 Hydrogen gas is burned with 200% theoretical air
in a steady-flow burner where both reactants are
supplied at the reference pressure and tempera-
ture. What is the adiabatic flame temperature?
14.76 A gas-furbine burns natural gas (assume methane)
where the air is supplied to the combustor at 1000
kPa, 500 K, and the fuel is at 298 K, 1000 kPa.
What is the equivalence ratio and the percent the-
oretical air if the adiabatic flame temperature
should be limited to 1800 K?
14.77 Liquid n -butane at T 0> is sprayed into a gas tur-
bine, as in Fig. P14.77, with primary air flowing
at 1.0 MPa, 400 K, in a stoichiometric ratio. After
complete combustion, the products are at the adi-
abatic flame temperature, which is too high, so
secondary air at 1.0 MPa, 400 K, is added, with
the resulting mixture being at 1400 K. Show that
T iA > 1400 K and find the ratio of secondary to
primary airflow.
Combustor
Tad
— j — 1>
1400 K
1
— I— »
Air 2
FIGURE P14.77
14.78 Butane gas at 25°C is mixed with 150% theoreti-
cal air at 600 K and is burned in an adiabatic
steady-flow combustor. What is the temperature
of the products exiting the combustor?
14.79 Natural gas, we assume methane, is burned with
200% theoretical air, and the reactants are sup-
Exhaust
f
FIGURE P14.79
Homework problems El 609
plied as gases at the reference temperature and
pressure. The products are flowing through a
heat exchanger and then out the exhaust, as in
Fig. P14.79. What is the adiabatic flame tem-
perature right after combustion before the heat
exchanger?
14.80 Liquid butane at 25°C is mixed with 150% theo-
retical air at 600 K and is burned in a steady-flow
burner. Use the enthalpy of combustion from
Table 14.3 to find the adiabatic flame temperature
out of the burner.
14.81 Acetylene gas at 25°C, 100 kPa, is fed to the head
of a cutting torch. Calculate the adiabatic flame
temperature if the acetylene is burned with
a. 100% theoretical air at 25°C.
b. 100% theoretical oxygen at 25°C.
14.82 Ethene, C 2 H 4) burns with 150% theoretical air in a
steady-flow constant-pressure process with reac-
tants entering at P , T Q , Find the adiabatic flame
temperature.
14.83 Solid carbon is burned with stoichiometric air in
a steady- flow process. The reactants at T , P
are heated in a preheater to T 2 = 500 K, as
shown in Fig. PI 4. 83, with the energy given by
the product gases before flowing to a second
heat exchanger, which they leave at T Q . Find the
temperature of the products T 4 , and the heat
transfer per kmol of fuel (4 to 5) in the second
heat exchanger.
FIGURE P14.83
14,84 Gaseous ethanol, C2H5OFL is burned with pure
oxygen in a constant-volume combustion bomb.
The reactants are charged in a stoichiometric ratio
at the reference condition. Assume no heat trans-
. fer and find the final temperature (>5000 K).
14.85 The enthalpy of formation of magnesium oxide,
MgO(s), is -601 827 kj/kmol at 25°C. The melt-
ing point of magnesium oxide is approximately
3000 K, and the increase in enthalpy between 298
and 3000 K is 128 449 kJ/kmol, The enthalpy of
sublimation at 3000 K is estimated at 418 000
kJ/kmol, and the specific heat of magnesium
oxide vapor above 3000 K is estimated at 37,24
kJ/kmol K.
a. Determine the enthalpy of combustion per
kilogram of magnesium.
b. Estimate the adiabatic flame temperature when
magnesium is burned with theoretical oxygen.
Second Law for the Combustion Process
14.86 Calculate the irreversibility for the process de-
scribed in Problem 14.41.
14.87 Methane is burned with air, both of which are
supplied at the reference conditions. There is
enough excess air to give a flame temperature of
1800 K. What are the percent theoretical air and
the irreversibility in the process?
14.88 Consider the combustion of hydrogen with pure
oxygen in a stoichiometric ratio under steady-
flow adiabatic conditions. The reactants enter sep-
arately at 298 K, 100 kPa, and the product(s) exit
at a pressure of 100 kPa, What is the exit temper-
ature, and what is the irreversibility?
14.89 Pentane gas at 25°C, 150 kPa, enters an insulated
steady-flow combustion chamber. Sufficient ex-
cess air to hold the combustion products tempera-
ture to 1800 K enters separately at 500 K, 150
kPa. Calculate the percent theoretical air required
and the irreversibility of the process per kmol of
pentane burned.
14.90 Consider the combustion of methanol, CH 3 OFf,
with 25% excess air. The combustion products
are passed through a heat exchanger and exit at
200 kPa, 400 K. Calculate the absolute entropy of
the products exiting the heat exchanger assuming
all the water is vapor.
14.91 Consider the combustion of methanol, CH 3 OFf,
with 25% excess air. The combustion products
are passed through a heat exchanger and exit at
200 kPa, 40°C. Calculate the absolute entropy of
610 H Chapter Fourteen Chemical Reactions
the products exiting the heat exchanger per kilo-
mole of methanol burned, using proper amounts
of liquid and vapor water,
14.92 An inventor claims to have built a device that will
take 0.001 kg/s of water from the faucet at 10°C,
100 kPa, and produce separate streams of hydro-
gen and oxygen gas, each at 400 K, 175 kPa. It is
stated that this device operates in a 25°C room on
10-kW electrical power input. How do you evalu-
ate this claim?
14.93 Two kilomoles of ammonia are burned in a
steady-flow process with x kmol of oxygen. The
products, consisting of H 2 0, N 2) and the excess
2 , exit at 200°C, 7 MPa.
a. Calculate x if half the water in the products is
condensed.
b. Calculate the absolute entropy of the products
at the exit conditions.
14.94 Graphite, C, at P , T Q is burned with air coming in
at P 0i 500 K, in a ratio so the products exit at P 0>
1200 K. Find the equivalence ratio, the percent
theoretical air, and the total irreversibility.
14.95 A flow of hydrogen gas is mixed with a flow of
oxygen in a stoichiometric ratio, both at 298 K
and 50 kPa. The mixture burns without any heat
transfer in complete combustion. Find the adia-
batic flame temperature and the amount of en-
tropy generated per kmole hydrogen in the
process.
14.96 A closed, rigid container is charged with propene,
C 3 H 6 , and 150% theoretical air at 100 kPa, 298 K.
The mixture is ignited and bums with complete
combustion. Heat is transferred to a reservoir at
500 K so the final temperature of the products is
700 K. Find the final pressure, the heat transfer
per kmol fuel, and the total entropy generated per
kmol fuel in the process.
Problems Involving Generalized Charts
or Real Mixtures
14.97 Repeat Problem 14.42, but assume that satu-
rated-liquid oxygen at 90 K is used instead of
25°C oxygen gas in the combustion process. Use
the generalized charts to determine the proper-
ties of liquid oxygen.
14.98 Hydrogen peroxide, H 2 2 , enters a gas generator
at 25°C, 500 kPa, at the rate of 0.1 kg/s and is
decomposed to steam and oxygen exiting at 800
K, 500 kPa, The resulting mixture is expanded
through a turbine to atmospheric pressure, 100
kPa, as shown in Fig. P 14.98. Determine the
power output of the turbine and the heat transfer
rate in the gas generator. The enthalpy of forma-
tion of liquid H 2 2 is - 1 87 583 kJ/kmoi.
FIGURE P14.93
14.99 Liquid butane at 25°C is mixed with 150% theo-
retical air at 600 K and is burned in an adiabatic
steady-state combustor. Use the generalized
charts for the liquid fuel and find the tempera-
ture of the products exiting the combustor.
14.100 Saturated liquid butane enters an insulated con-
stant-pressure combustion chamber at 25°C, and
x times theoretical oxygen gas enters at the same
P and T. The combustion products exit at 3400
K. With complete combustion find x, What is
the pressure at the chamber exit? What is the ir-
reversibility of the process?
14.101 A gas mixture of 50% ethane and 50% propane
by volume enters a combustion chamber at 350
K, 10 MPa. Determine the enthalpy per kilomole
of this mixture relative to the thermochemical
base of enthalpy using Kay's rule.
14.102 A mixture of 80% ethane and 20% methane on a
mole basis is throttled from 10 MPa, 65°C, to
100 kPa and is fed to a combustion chamber
where it undergoes complete combustion with
air, which enters at 1 00 kPa, 600 K. The amount
of air is such that the products of combustion
exit at 100 kPa, 1200 K. Assume that the com-
bustion process is adiabatic and that all compo-
nents behave as ideal gases except the fuel
mixture, which behaves according to the gener-
alized charts, with Kay's rule for the pseudocrit-
HOMEWORK PROBLEMS M 611
ical constants. Determine the percentage of theo-
retical air used in the process and the dew-point
temperature of the products.
14.103 Liquid hexane enters a combustion chamber at
31°C, 200 kPa, at the rate of 1 kmol/s. 200% the-
oretical air enters separately at 500 K, 200 kPa,
and the combustion products exit at 1000 K,
200 kPa, The specific heat of ideal-gas hexane
is CpQ = 143 kJ/kmol K. Calculate the rate of ir-
reversibility of the process.
Fuel Cells
14.104 In Example 14.16, a basic hydrogen-oxygen
fuel cell reaction was analyzed at 25°C, 100 kPa.
Repeat this calculation, assuming that the fuel
cell operates on air at 25 °C, 100 kPa, instead of
on pure oxygen at this state.
14.105 Assume that the basic hydrogen-oxygen fuel
cell operates at 600 K instead of 298 K as in Ex-
ample 14.16. Find the change in the Gibbs func-
tion and the reversible EMF it can generate.
14.106 Consider a methane-oxygen fuel cell in which
the reaction at the anode is
CH 4 + 2H 2 C0 2 + 8e~ + 8H +
The electrons produced by the reaction flow
through the external load, and the positive ions
migrate through the electrolyte to the cathode,
where the reaction is
Se~ + 8H + + 20 2 ^4H 2
Calculate the reversible work and the reversible
EMF for the fuel cell operating at 25°C, 100 kPa.
14.107 Redo the previous problem, but assume that the
fuel cell operates at 1200 K instead of at room
temperature.
Combustion Efficiency
14.108 Consider the steady- state combustion of propane
at 25°C with air at 400 K. The products exit the
combustion chamber at 1200 K. It may be as-
sumed that the combustion efficiency is 90%
and that 95% of the carbon in the propane burns
to form carbon dioxide; the remaining 5% forms
carbon monoxide. Determine the ideal fuel-air
ratio and the heat transfer from the combustion
chamber.
14.109 A gasoline engine is converted to run on
propane as shown in Fig. P14.109. Assume the
propane enters the engine at 25°C 3 at the rate 40
kg/h. Only 90% theoretical air enters at 25°C,
such that 90% of the C burns to form C0 2 and
1 0% of the C bums to form CO. The combustion
products, also including H 2 0, H 2 , and N 2 , exit
the exhaust pipe at 1000 K. Heat loss from the
engine (primarily to the cooling water) is 120
kW. What is the power output of the engine?
What is the thermal efficiency?
C 3 H 8 gas
90% theo air
1 ^
Internal .
combustion
engine
Combustion
-£>
FIGURE P14.109
14.110 A small air-cooled gasoline engine is tested,
and the output is found to be 1.0 kW. The tem-
perature of the products is measured as 600 K.
The products are analyzed on a dry volumetric
basis, with the result; 11.4% C0 2 , 2.9% CO,
1.6% 3j and 84.1% N 2 . The fuel may be' con-
sidered to be liquid octane. The fuel and air
enter the engine at 25°C, and the flow rate of
fuel to the engine is 1.5 X 10~ 4 kg/s. Deter-
mine the rate of heat transfer from the engine
and its thermal efficiency.
14.111 A gasoline engine uses liquid octane and air,
both supplied at P 0t T , in a stoichiometric
ratio. The products (complete combustion)
flow out of the exhaust valve at 1100 K. As-
sume that the heat loss carried away by the
cooling water, at 100°C, is equal to the work
output. Find the efficiency of the engine ex-
pressed as (work/lower heating value) and the
second-law efficiency.
Review Problems
14.112 Ethene, C 2 H 4 , and propane, C 3 H S , in a 1 : 1 mole
ratio as gases are burned with 120% theoretical
air in a gas turbine. Fuel is added at 25°C,
1 MPa, and the air comes from the atmosphere,
25°C, 100 kPa, through a compressor to 1 MPa
612 H Chapter Fourteen Chemical reactions
and mixed with the fuel. The turbine work is
such that the exit temperature is 800 K with an
exit pressure of 100 kPa. Find the mixture tem-
perature before combustion, and also the work,
assuming an adiabatic turbine.
14.113 Carbon monoxide, CO, is burned with 150%
theoretical air, and both gases are supplied
at 150 kPa and 600 K. Find the reference
enthalpy of reaction and the adiabatic flame
temperature.
14.114 Consider the gas mixture fed to the combustors
in the integrated gasification combined cycle
power plant, as described in Problem 14.35. If
the adiabatic flame temperature should be lim-
ited to 1500 K, what percent theoretical air
should be used in the combustors?
14.115 A study is to be made using liquid ammonia as
the fuel in a gas-turbine engine. Consider the
compression and combustion processes of this
engine.
a. Air enters the compressor at 100 kPa, 25°C,
and is compressed to 1600 kPa, where the
isentropic compressor efficiency is 87%. De-
termine the exit temperature and the work
input per kilomole.
b. Two kilomoles of liquid ammonia at 25°C
and x times theoretical air from the compres-
sor enter the combustion chamber. What is x
if the adiabatic flame temperature is to be
fixed at 1600 K?
14.116 A rigid container is charged with butene, C 4 H S ,
and air in a stoichiometric ratio at P Qt T Q . The
charge burns in a short time with no heat trans-
fer to state 2. The products then cool with time
to 1200 K, state 3. Find the final pressure, P 3 ,
the total heat transfer, x Qi, and the temperature
immediately after combustion, T 2 .
14.117 The turbine in Problem 14.112 is adiabatic. Is it
reversible, irreversible, or impossible?
14.118 Consider the combustion process described in
Problem 14.102.
a. Calculate the absolute entropy of the fuel
mixture before it is throttled into the combus-
tion chamber.
b. Calculate the irreversibility for the overall
process.
14.119 Natural gas (approximate it as methane) at a rate
of 0.3 kg/s is burned with 250% theoretical air in
a combustor at 1 MPa where the reactants are
supplied at T Q . Steam at 1 MPa, 450°C, at a rate
of 2.5 kg/s is added to the products before they
enter an adiabatic turbine with an exhaust pres-
sure of 150 kPa. Determine the turbine inlet
temperature and the turbine work assuming the
turbine is reversible.
14.120 Consider one cylinder of a spark-ignition, internal-
combustion engine. Before the compression
stroke, the cylinder is filled with a mixture of air
and methane. Assume that 110% theoretical air
has been used and that the state before compres-
sion is 100 kPa, 25°C. The compression ratio of
the engine is 9 to 1.
a. Determine the pressure and temperature after
'compression, assuming a reversible adiabatic
process.
b. Assume that complete combustion takes
place while the piston is at top dead center (at
minimum volume) in an adiabatic process.
Determine the temperature and pressure after
combustion, and the increase in entropy dur-
ing the combustion process.
c. What is the irreversibility for this process?
14.121 Liquid acetylene, C 2 H 2 , is stored in a high-
pressure storage tank at ambient temperature,
25°C. The liquid is fed to an insulated combus-
tor/steam boiler at the steady rate of 1 kg/s, along
with 140% theoretical oxygen, 2 , which enters at
Steam
FIGURE P14.121
500 K, as shown in Fig. P14.121. The combustion
products exit the unit at 500 kPa, 350 K. Liquid
water enters the boiler at 10°C, at the rate of 15
kg/s, and superheated steam exits at 200 fcPa.
a. Calculate the absolute entropy, per kmol, of
liquid acetylene at the storage tank state.
English Unit Problems M 613
b. Determine the phase(s) of the combustion
products exiting the combustor boiler unit,
and the amount of each, if more than one.
c. Determine the temperature of the steam at the
boiler exit.
English Unit Problems
Concept Problems
14.122E What is the enthalpy of formation for oxygen
as 2 ? if O? for C0 2 ?
14.123E What is the higher heating value (HHV) of
n-butane?
Fuels and the Combustion Process
14.124E Pentane is burned with 120% theoretical air in
a constant-pressure process at 14.7 lbf/in. 2 . The
products are cooled to ambient temperature,
70 F. How much mass of water is condensed
per pound-mass of fuel? Repeat the problem,
assuming that the air used in the combustion
has a relative humidity of 90%.
14.125 E The output gas mixture of a certain air-blown
coal gasifier has the composition of producer
gas as listed in Table 14.2. Consider the com-
bustion of this gas with 120% theoretical air at
14.7 lbf/in. 2 pressure. Find the dew point of the
products and the mass of water condensed per
pound-mass of fuel if the products are cooled
20 F below the dew-point temperature?
Energy and Enthalpy of Formation
14.126E A rigid vessel initially contains 2-pound moles
of carbon and 2-pound moles of oxygen at
77 F, 30 lbf/in. 2 . Combustion occurs, and the
resulting products consist of a 1 -pound mole of
carbon dioxide, 1-pound mole of carbon
monoxide, and excess oxygen at a temperature
of 1800 R. Determine the final pressure in the
vessel and the heat transfer from the vessel
during the process.
14.127E In a test of rocket propellant performance, liq-
uid hydrazine (N 2 H 4 ) at 14.7 lbf/in. 2 , 77 F, and
oxygen gas at 14.7 lbf/in. 2 , 77 F, are fed to a
combustion chamber in the ratio of 0.5 Ibm
2 /lbm N 2 H 4 . The heat transfer from the cham-
ber to the surroundings is estimated to be 45
Btu/lbm N 2 H 4 . Determine the temperature of
the products exiting the chamber. Assume that
only H 2 0, H 2 , and N 2 are present. The enthalpy
of formation of liquid hydrazine is +21 647
Btu/lb mole.
14.128E One alternative to using petroleum or natural
gas as fuels is ethanol (C,H 5 OH), which is
commonly produced from grain by fermenta-
tion. Consider a combustion process in which
liquid ethanol is burned with 120% theoretical
air in a steady-flow process. The reactants
enter the combustion chamber at 77 F, and the
products exit at 140 F, 14.7 lbf/in. 2 . Calculate
the heat transfer per pound mole of ethanol,
using the enthalpy of formation of ethanol gas
plus the generalized tables or charts.
14.129E In a new high-efficiency furnace, natural gas,
assumed to be 90% methane and 10% ethane
(by volume) and 110% theoretical air each
enter at 77 F, 14.7 lbf/in. 2 , and the products (as-
sumed to be 100% gaseous) exit the furnace at
1 00 F, 14.7 lbf/in. 2 . What is the heat transfer for
this process? Compare this to an older furnace
where the products exit at 450 F, 14.7 lbf/in. 2 .
14.130E Repeat the previous problem, but take into ac-
count the actual phase behavior of the products
exiting the furnace.
14.131E Pentene, C 5 H 10 is burned with pure oxygen in a
steady-state process. The products at one point
are brought to 1300 R and used in a heat ex-
changer, where they are cooled to 77 F. Find
the specific heat transfer in the heat exchanger.
14.132E Methane, CFf 4 , is burned in a steady-state
process with two different oxidizers: case A—
pure oxygen, 2 and case B — a mixture of
614 H CHAPTER FOURTEEN CHEMCAL REACTIONS
2 + xAr. The reactants are supplied at T Ql P Q ,
and the products are at 3200 R in both cases.
Find the required equivalence ratio in case A
and the amount of argon, x, for a stoichiomet-
ric ratio in case B.
14.133E A closed, insulated container is charged with a
stoichiometric ratio of oxygen and hydrogen at
77 F and 20 lbf/in. 2 . After combustion, liquid
water at 77 F is sprayed in such a way that the
final temperature is 2100 R. What is the final
pressure?
Enthalpy of Combustion and Heating Value
14.134E A burner receives a mixture of two fuels with
mass fraction 40% n-butane and 60%
methanol, both vapor. The fuel is burned with
stoichiometric air. Find the product composi-
tion and the lower heating value of this fuel
mixture (Btu/lbm fuel mix).
14.135E Blast furnace gas in a steel mill is available at
500 F to be burned for the generation of steam.
The composition of this gas is, on a volumetric
basis,
Component CH 4 H 2 CO C0 2 N 2 H 2
Percent by volume 0.1 2.4 23.3 14.4 56.4 3.4
Find the lower heating value (Btu/ft 3 ) of this
gas atSOOFandiV
Adiabatic Flame Temperature
14.136E Hydrogen gas is burned with pure oxygen in a
steady-flow burner where both reactants are
supplied in a stoichiometric ratio at the refer-
ence pressure and temperature. What is the
adiabatic flame temperature?
14.137E Carbon is burned with air in a furnace with
150% theoretical air, and both reactants are
supplied at the reference pressure and tempera- .
ture. What is the adiabatic flame temperature?
14.138E Butane gas at 77 F is mixed with 150% theoreti-
cal air at 1000 R and is burned in an adiabatic
steady-state combustor. What is the temperature
of the products exiting the combustor?
14.139E Liquid H-butane at T , is sprayed into a gas tur-
bine with primary air flowing at 150 lbf/in. 2 ,
700 R in a stoichiometric ratio. After complete
combustion, the products are at the adiabatic
flame temperature, which is too high. Therefore,
secondary air at 150 Ibffin. 2 , 700 R is added, see
Fig. P14.77, with the resulting mixture being at
2500 R. Show that T 2d > 2500 R and find the
ratio of secondary to primary airflow.
14.140E Acetylene gas at 77 F, 14.7 lbf/in. 2 is fed to the
head of a cutting torch. Calculate the adiabatic
flame temperature if the acetylene is burned
with 100% theoretical air at 77 F. Repeat the
answer for 100% theoretical oxygen at 77 F.
14.141E Ethene, C 2 H,, burns with 150% theoretical ak-
in a steady-state constant-pressure process,
with reactants entering at P 0> T . Find the adia-
batic flame temperature,
14.142E Solid carbon is burned with stoichiometric air in
a steady-state process, as shown in Fig. P14.83.
The reactants at T , P are heated in a preheater
to T 2 - 900 R with the energy given by the prod-
ucts before flowing to a second heat exchanger,
which they leave at T . Find the temperature of
the products T 4 and the heat transfer per lb mol
of fuel (4 to 5) in the second heat exchanger.
Second Law for the Combustion Process
14.143E Methane is burned with air, both of which are
supplied at the reference conditions. There is
enough excess air to give a flame temperature
of 3200 R. What are the percent theoretical air
and the irreversibility in the process?
14.144E Two-pound moles of ammonia are burned in a
steady-state process with x lb mol of oxygen.
The products, consisting of H 2 0, N 2 , and the
excess 2 , exit at 400 F, 1000 lbf/in. 2 .
a. Calculate x if half the water in the products
is condensed.
b. Calculate the absolute entropy of the prod-
ucts at the exit conditions.
14.145E Graphite, C, at P Ql T is burned with air coming
in at P 0) 900 R, in a ratio so the products exit at
P 0t 2200 R. Find the equivalence ratio, the per-
cent theoretical air, and the total irreversibility.
Problems Involving Generalized Charts
or Real Mixtures
14.146E Repeat Problem 14.127E, but assume that satu-
rated-liquid oxygen at 170 R is used instead of
77 F oxygen gas in the combustion process.
COMPUTER, DESIGN, AND OPEN-ENDED PROBLEMS M 615
Use the generalized charts to determine the
properties of liquid oxygen.
14.147E Hydrogen peroxide, H 2 2 , enters a gas genera-
tor at 77 F, 75 lbt7in. 2 , at the rate of 0.2 lbm/s
and is decomposed to steam and oxygen exit-
ing at 1500 R, 75 lbf/in. 2 . The resulting mix-
ture is expanded through a turbine to
atmospheric pressure, 14.7 lbf/in. 2 , as shown in
Fig. P 14.98. Detennine the power output of the
turbine and the heat-transfer rate in the gas
generator. The enthalpy of formation of liquid
H 2 2 is -80 541 Btu/lb mol.
Fuel Cells, Efficiency, and Review
14.148E In Example 14.16, a basic hydrogen-oxygen
fuel cell reaction was analyzed at 25 °C, 100
kPa. Repeat this calculation, assuming that the
fuel cell operates on air at 77 F, 14.7 lbf/in. 2 ,
instead of on pure oxygen at this state.
14.149E A small, air-cooled gasoline engine is tested,
and the output is found to be 2.0 hp. The tem-
perature of the products is measured and found
to be 730 F. The products are analyzed on a
dry volumetric basis, with the following result:
1 1 .4% C0 2 , 2.9% CO, 1 .6% 2l and 84. 1 % N 2 .
The fuel may be considered to be liquid oc-
tane. The fuel and air enter the engine at 77 F,
and the flow rate of fuel to the engine is 1.8
Ibm/h. Determine the rate of heat transfer from
the engine and its thermal efficiency,
14.150E A gasoline engine uses liquid octane and air,
both supplied at P 0i T , in a stoichiometric
ratio. The products (complete combustion)
flow out of the exhaust valve at 2000 R. As-
sume that the heat loss carried away by the
cooling water, at 200 F, is equal to the work
output. Find the efficiency of the engine ex-
pressed as (work/lower heating value) and the
second-law efficiency.
14.151E Ethene, C2H4, and propane, C 3 H g , in a 1 : 1 mole
ratio as gases are burned with 120% theoretical
air in a gas turbine. Fuel is added at 77 F, 150
lbf/in. 2 , and the air comes from the atmosphere,
77 F, 15 lbf/in. 2 through a compressor to 150
lbf/in. 2 and mixed with the fuel. The turbine
work is such that the exit temperature is 1500 R
with an exit pressure of 14.7 lbf/in. 2 . Find the
mixture temperature before combustion, and
also the work, assuming an adiabatic turbine.
Computer, Design, and Open-Ended Problems
14.152 Write a program to solve the general case of
Problem 14.31 for any hydrocarbon fuel C x B y ,
where x and y are input parameters. We wish to
calculate the percentage of theoretical air for any
given percentages of combustion products.
14.153 Write a program to solve the general case of Prob-
lem 14.26 for different percentages of the compo-
nents given in the ultimate analysis of the coal.
14.154 Use the software program for the ideal-gas prop-
erties to do Problem 14.83.
14.155 Write a program to study the effect of the per-
centage of theoretical air on the adiabatic flame
temperature for a (variable) hydrocarbon fuel.
Assume reactants enter the combustion chamber
at 25°C, and complete combustion. Use con-
stant-specific heat of the various products of
combustion and let the fuel composition and its
enthalpy of formation be program inputs.
14.156 Power plants may use off-peak power to com-
press air into a large storage facility (see Prob-
lem 9.69). The compressed air is then used as
the air supply to a gas-turbine system where it is
burned with some fuel, usually natural gas. The
system is then used to produce power at peak
load times. Investigate such a setup and estimate
the power generated with conditions given in
Problem 9.69 and combustion with 200-300%
theoretical air and exhaust to the atmosphere.
14.157 A car that runs on natural gas has it stored in a
heavy tank with a maximum pressure of 3600
psi (25 MPa). Size the tank for a range of 300
miles (500 km) assuming a car engine that has a
30% efficiency requiring about 25 hp (20 kW) to
drive the car at 55 mi/li (90 km/h).
14.158 The Cheng cycle, shown in Fig. PI 2. 176, is
powered by the combustion of natural gas (es-
sentially methane) being burned with 250-300%
theoretical air. In the case with a single water-
condensing heat exchanger, where T 6 = 40°C
and $ 6 = 100%, is any makeup water needed at
616 M Chapter Fourteen Chemical reactions
state 8, or is there a surplus? Does the humidity
in the compressed atmospheric air at state 1
make any difference? Study the problem over a
range of air-fuel ratios.
14.159 The cogeneratiiig power plant shown in Problem
11.65 bums 170 kg/s air with natural gas, CH 4 .
The set up is shown in Fig. P14.159 where a
fraction of the airflow out of the compressor
with pressure ratio 15.8: 1 is used to preheat the
feedwater in the steam cycle. The fuel flow rate
is 3.2 kg/s. Make an analysis of the system de-
termining the total heat transfer to the steam
cycle from the turbine exhaust gases, the heat
transfer in the preheater, and the gas turbine
inlet temperature.
FIGURE P14.159
14.160 Consider the combustor in the Cheng cycle (see
Problems 12.176 and 14.119E). Atmospheric air
is compressed to 1.25 MPa, state 1. It is burned
with natural gas, CH 4 , with the products leaving
at state 2. The fuel should add a total of about 15
MW to the cycle, with an airflow of 12 kg/s. For
a compressor with an intercooler estimate the
temperatures T u T 2 , and the fuel flow rate.
14.161 Study the coal gasification process that will pro-
duce methane, CH 4 , or methanol CH 3 OH. What is
involved in such a process? Compare the heating
values of the gas products with those of the origi-
nal coal. Discuss the merits of this conversion.
14.162 Ethanol, C^HjOH, can be produced from com or
biomass. Investigate the process and the chemi-
cal reactions that occur. For different raw mate-
rials estimate the amount of ethanol that can be
obtained per mass of the raw material.
14.163 A Diesel engine is used as a stationary power
plant in remote locations such as a ship, oil
drilling rig, or farm. Assume diesel fuel is used
with 300% theoretical air in a 1000-hp diesel en-
gine. Estimate the amount of fuel used, the effi-
ciency, and the potential use of the exhaust
gases for heating of rooms or water. Investigate
if other fuels can be used.
14.164 When a power plant burns coal or some blends
of oil, the combustion process can generate pol-
lutants as $>O x and NO x Investigate the use of
scrubbers to remove these. Explain the processes
that take place and the effect on the power plant
operation (energy, exhaust pressures, etc.).
14.165 For a number of fuels listed in Table 14.3 esti-
mate their adiabatic flame temperature when
they are burned with 200% theoretical air. As-
sume a power-generating device like a gasoline
or diesel engine or a gas-turbine with reasonable
choices for their operating conditions. Find the
power that can be generated as a fraction of the
enthalpy of combustion. Does a ranking of
the fuels follow the magnitude of the enthalpy of
combustion?
introduction to phase and
Chemical equilibrium
Up to this point, we have assumed that we are dealing either with systems that are in
equilibrium or with those in which the deviation from equilibrium is infinitesimal, as in a
quasi-equilibrium or reversible process. For irreversible processes we made no attempt to
describe the state of the system during the process but dealt only with the initial and final
states of the system, in the case of a control mass, or the inlet and exit states as well in the
case of a control volume. For any case, we either considered the system to be in equilib-
rium throughout or at least made the assumption of local equilibrium.
In this chapter we examine the criteria for equilibrium and from them derive certain re-
lations that will enable us, under certain conditions, to determine the properties of a system
when it is in equilibrium. The specific case we will consider is that involving chemical equi-
librium in a single phase (homogeneous equilibrium) as well as certain related topics.
15.1 REQUIREMENTS FOR EQUILIBRIUM
As a general requirement for equilibrium, we postulate that a system is in equilibrium
when there is no possibility that it can do any work when it is isolated from its surround-
ings. In applying this criterion, it is helpful to divide the system into two or more subsys-
tems, and consider the possibility of doing work by any conceivable interaction between
these two subsystems. For example, in Fig. 15.1 a system has been divided into two sys-
tems and an engine, of any conceivable variety, placed between these subsystems. A sys-
tem may be so defined as to include the immediate surroundings. In this case, we can let
the immediate surroundings be a subsystem and thus consider the general case of the
equilibrium between a system and its surroundings.
The first requirement for equilibrium is that the two.subsystems have the same tem-
perature, for otherwise we could operate a heat engine between the two systems and do
work. Thus, we conclude that one requirement for equilibrium is that a system must be at
a uniform temperature to be in equilibrium. It is also evident that there must be no unbal-
anced mechanical forces between the two systems, or else one could operate a turbine or
piston engine between the two systems and do work.
We would like to establish general criteria for equilibrium that would apply to all
simple compressible substances, including those that undergo chemical reactions. We
■-. will find that the Gibbs function is a particularly significant property in defining the crite-
ria for equilibrium.
Let us first consider a qualitative example to illustrate this point. Consider a natural
. gas well that is 1 km deep, and let us assume that the temperature of the gas is constant
617
618 H chapter Fifteen introduction to phase and chemical equilibrium
FIGURE 15,1 Two
subsystems that
communicate through an
engine.
Subsystem
Engine
<=>
Subsystem
2
w
throughout the gas wel!. Suppose we have analyzed the composition of the gas at the top
of the well, and we would like to know the composition of the gas at the bottom of the
well. Furthermore, let us assume that equilibrium conditions prevail in the well. If this is
true we would expect that an engine such as that shown in Fig. 15.2 (which operates on
the basis of the pressure and composition change with elevation and does not involve
combustion) would not be capable of doing any work.
If we consider a steady-state process for a control volume around this engine, the re-
versible work for the change of state from i to e is given by Eq. 10.9 on a total mass basis
W rev = in, (a, + ^ + gZ l - 7*,*) - ib, ^ e + ^ + gZ e - ToS t
Furthermore, since T t — T e — T = constant, this reduces to the form of the Gibbs
function g = A - Ts t Eq. 13,14, and the reversible work is
/ V? \ / V 2
However,
Then we can write
V? V 2
W m = 0, m t = m e and y = -y
Si + S Z i = Se + S Z e
and the requirement for equilibrium in the well between two levels that are a distance dZ
apart would be
dg T + gdZ T =0
FIGURE 15.2
Illustration showing the
relation between
reversible work and the
criteria for equilibrium.
4 z
Gas
well
I
Mass flow =
Rev.
eng.
W mv =0
Equilibrium between two phases of a pure substance H 619
FIGURE 15.3
Illustration of the
requirement for chemical
equilibrium.
Equilibrium point
In contrast to a deep gas well, most of the systems that we consider are of such size that
AZ is negligibly small, and therefore we consider the pressure to be uniform throughout.
This leads to the general statement of equilibrium that applies to simple compressible
substances that may undergo a change in chemical composition, namely, that at equilibrium
dG T>P = Q (15.1)
In the case of a chemical reaction, it is helpful to think of the equilibrium state as
the state in which the Gibbs function is a minimum. For example, consider a control mass
consisting initially of n A moles of substance^ and n B moles of substance B, which react in
accordance with the relation
Let the reaction take place at constant pressure and temperature. If we plot G for
this control mass as a function of n A , the number of moles of A present, we would have a
curve as shown in Fig. 15.3. At the minimum point on the curve, dG TP = 0, and this will
be the equilibrium composition for this system at the given temperature and pressure. The
subject of chemical equilibrium will be developed further in Section 15.4.
15.2 Equilibrium Between Two
Phases of a Pure Substance
As another example of this requirement for equilibrium, let us consider the equilibrium be-
tween two phases of a pure substance. Consider a control mass consisting of two phases of a
pure substance at equilibrium. We know that under these conditions the two phases are at the
same pressure and temperature. Consider the change of state associated with a transfer ofdn
moles from phase 1 to phase 2 while the temperature and pressure remain constant. That is,
dn x = ~dn 2
The Gibbs function of this control mass is given by
G=f{T,P ) n\n 2 )
where «' and n 7 designate the number of moles in each phase. Therefore,
CG = (f ) + (f ) OP + fef ) rf „. + ( 4\ *•
\*t*Jpjtj V^/ryy \(J?i7r^ \dn 2 jT^
620 a CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
By definition,
Therefore, at constant temperature and pressure,
Now at equilibrium (Eq. 15.1)
dG TP —
Therefore, at equilibrium, we have
That is, under equilibrium conditions, the Gibbs function of each phase of a pure sub-
stance is equal Let us check this by determining the Gibbs function of saturated liquid
(water) and saturated vapor (steam) at 300 kPa. From the steam tables:
For the liquid:
gf =ft f -Ts f = 561.47 - 406.7 X 1.6718 = -118.4 kJ/kg
For the vapor;
g g = h s ~ Ts g = 2725.3 - 406.7 X 6.9919 = —118.4 kJ/kg
Equation 15.2 can also be derived by applying the relation
Tds = dh-v dP
to the change of phase that takes place at constant pressure and temperature. For this
process this relation can be integrated as follows:
dh
f
7\s g - s f ) = {h g - h f )
The Clapeyron equation, which was derived in Section 13.1, can be derived by an
alternate method by considering the fact that the Gibbs functions of two phases in equilib-
rium are equal. In Chapter 13 we considered the relation (Eq. 13.15) for a simple com-
pressible substance:
dg = vdP~sdT
Consider a control mass that consists of a saturated liquid and a saturated vapor in
equilibrium, and let this system undergo a change of pressure dP. The corresponding
change in temperature, as determined from the vapor-pressure curve, is dT. Both phases
will undergo the change in Gibbs function, dg, but since the phases always have the same
value of the Gibbs function when they are in equilibrium, it follows that
dg f =dg g
Equilibrium between Two Phases of a Pure Substance 88 621
But, from Eq. 13.15,
dg=vdP- sdT
it follows that
dg f = v f dP~s f dT
dg z = v g dP~ s g dT
Since
it follows that
v f dP ~s f dT^ v g dP - s s dT
dP{v g -v f )^dT{s g ~s f ) (15.3)
dP _ s Jg „ h fg
dT v& Tv /g
In summary, when different phases of a pure substance are in equilibrium, each
phase has the same value of the Gibbs function per unit mass. This fact is relevant to dif-
ferent solid phases of a pure substance and is important in metallurgical applications of
thermodynamics. Example 15.1 illustrates this principle.
EXAMPLE 15.1 What pressure is required to make diamonds from graphite at a temperature of 25°C?
The following data are given for a temperature of 25°C and a pressure of 0. 1 MPa.
Graphite
Diamond
s
V
Pr
0.000 444 m 3 /kg
0.304 X 10" 6 1/MPa
2867.8 kJ/mol
0.000 284 m 3 /kg
0.016 X 10" 5 1/MPa
Analysis and Solution
The basic principle in the solution is that graphite and diamond can exist in equilibrium
when they have the same value of the Gibbs function". At 0.1 MPa pressure the Gibbs
function of the diamond is greater than that of the graphite. However, the rate of increase
in Gibbs function with pressure is greater for the graphite than for the diamond, therefore,
at some pressure they can exist in equilibrium. Our problem is to find this pressure.
We have already considered the relation
dg=vdP~sdT
Since we are considering a process that takes place at constant temperature, this
reduces to
dg T =vdP T (a)
622 M Chapter fifteen introduction to phase and chemical equilibrium
Now at any pressure P and the given temperature, the specific volume can be found
from the following relation, which utilizes isothermal compressibility factor.
- v° - f vp, dF (b)
Jp=0A
The superscript will be used in this example to indicate the properties at a pres-
sure of 0.1 MPa and a temperature of 25°C.
The specific volume changes only slightly with pressure, so that v « v°. Also, we
assume that j3 r is constant and that we are considering a very high pressure. With these
assumptions this equation can be integrated to give
v = v°-v a IS T P=v <i (\ -PjF)
We can now substitute this into Eq. (a) to give the relation
(P 2 - P 02 )
If we assume that P° <P, this reduces to
For the graphite, g° = and we can write
P 2
For the diamond, g° has a definite value and we have
gD = g Q D + v° D P-(Pt)d-
But at equilibrium the Gibbs function of the graphite and diamond are equal
Sg = Zd
Therefore,
V G
P - 03 r ) G
P 2
P 2
(v° G ~ v° D )P - [vKfirh ~ v D (f3 T ) D ] ^ = g° D
(0.000 444 - 0.000 284)P
(c)
■(E)
2867 8
-(0.000 444 X 0.304 X 10~ 6 - 0.000284 X 0.016 X 10-^/2 = 12 Q11 x I000
Metastable Equilibrium H 623
Solving this for P we find
P = 1493 MPa
That is, at 1493 MPa, 25°C, graphite and diamond can exist in equilibrium, and the pos-
sibility exists for conversion from graphite to diamonds.
15.3 METASTABLE EQUILIBRIUM
Although the limited scope of this book precludes an extensive treatment of metastable
equilibrium, a brief introduction to the subject is presented in this section. Let us first con-
sider an example of metastable equilibrium.
Consider a slightly superheated vapor, such as steam, expanding in a convergent-
divergent nozzle, as shown in Fig. 1 5.4. Assuming the process is reversible and adiabatic,
the steam will follow path l-a on the T-s diagram, and at point a we would expect con-
densation to occur. However, if point a is reached in the divergent section of the nozzle, it
is observed that no condensation occurs until point b is reached, and at this point the con-
densation occurs very abruptly in what is referred to as a condensation shock. Between
points a and b the steam exists as a vapor, but the temperature is below the saturation tem-
perature for the given pressure. This is known as a metastable state. The possibility of a
metastable state exists with any phase transformation. The dotted lines on the equilibrium
diagram shown in Fig. 15.5 represent possible metastable states for solid-liquid-vapor
equilibrium.
The nature of a metastable state is often pictured schematically by the kind ofdia-
gram shown in Fig. 15.6. The ball is in a stable position (the "metastable state") for small
Point where condensation
would begin if equilibrium prevailed
Point where condensation
occurs very abruptiy
supersaturation in a
nozzle.
FIGURE 15.4
Illustration of the
phenomenon of
T
624 ■
CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
FIGURE 15.5
Metastable states for
solid-liquid-vapor
equilibrium.
FIGURE 15.6 \
Schematic diagram \.
illustrating a metastable
displacements, but with a large displacement it moves to a new equilibrium position. The
steam expanding in the nozzle is in a metastable state between a and b. This means that
droplets smaller than a certain critical size will reevaporate, and only when droplets larger
than this critical size have formed (this corresponds to moving the ball out of the depres-
sion) will the new equilibrium state appear.
Components
A,B,C,D
in chemical
equilibrium
FIGURE 15.7
Schematic diagram for
consideration of chemical
equilibrium.
15.4 Chemical equilibrium
We now turn our attention to chemical equilibrium and consider first a chemical reaction
involving only one phase. This is referred to as a homogeneous chemical reaction. It may
be helpful to visualize this as a gaseous phase, but the basic considerations apply to any
phase.
Consider a vessel, Fig. 15.7, that contains four compounds, A, B, C, and D, which
are in chemical equilibrium at a given pressure and temperature. For example, these might
consist of C0 2 , H 3 , CO, and H 2 in equilibrium. Let the number of moles of each compo-
nent be designated n A , n B , n g and n D . Furthermore, let the chemical reaction that takes
place between these four constituents be
VjA + v B B ^ v c C + VpD
(15.4)
where the v's are the stoichiometric coefficients. It should be emphasized that there is a
very definite relation between the v's (the stoichiometric coefficients), whereas the n's
(the number of moles present) for any constituent can be varied simply by varying the
amount of that component in the reaction vessel.
Let us now consider how the requirement for equilibrium, namely, that dG TF = at
equilibrium, applies to a homogeneous chemical reaction. Let us assume that the four
Chemical equilibrium H 625
components are in chemical equilibrium and then assume that from this equilibrium state,
while the temperature and pressure remain constant, the reaction proceeds an infinitesimal
amount toward the right as Eq. 15.4 is written. This results in a decrease in the moles of A
and B and an increase in the moles of C and D. Let us designate the degree of reaction by
e } and define the degree of reaction by the relations
dn A = —v A ds
dn B — —v B de
dn c — +v c ds
dn D =+v D d£ (15.5)
That is, the change in the number of moles of any component during a chemical re-
action is given by the product of the stoichiometric coefficients (the u's) and the degree of
reaction.
Let us evaluate the change in the Gibbs function associated with this chemical reac-
tion that proceeds to the right in the amount de. In doing so we use, as would be expected,
the Gibbs function of each component in the mixture— the partial molal Gibbs function
(or its equivalent, the chemical potential):
dG TP = G c dn c + G D dn D + G A dn A + G B dn B
Substituting Eq. 15,5, we have
dG ZP = (v c G c + v D G D - v A G A - v B G B ) de (15.6)
We now need to develop expressions for the partial molal Gibbs functions in
terms of properties that we are able to calculate. From the definition of Gibbs function
Eq. 13.14,
G = H — TS
For a mixture of two components A and B, we differentiate this equation with re-
spect to n A at constant T } P, and n Bi which results in
\ dn A}T?,n s \^a)tA^ \ dn A/T?,n a
All three of these quantities satisfy the definition of partial molal properties according to
Eq. 13.68, such that
G A =H A ~ TS A (15.7)
For an ideal-gas mixture, enthalpy is not a function of pressure, and
H A = h ATP = h ATI * (15.8)
Entropy is, however, a function of pressure, so that the partial entropy of A can be ex-
pressed by Eq. 14.22 in terms of the standard-state value,
$a ~ s ATP J = Y ll P
In
P° J (15.9)
626 H Chapter Fifteen introduction to Phase and Chemical Equilibrium
Now, substituting Eqs. 15,8 and 15.9 into Eq. 15.7,
(15.10)
Equation 15.10 is an expression for the partial Gibbs function of a component in a mixture
in terms of a specific reference value, the pure-substance standard-state Gibbs function at
the same temperature, and a function of the temperature, pressure, and composition of the
mixture. This expression can be applied to each of the components in Eq. 15.6, resulting in
dG T p - \ v c
+ v D
3, + RTln
(¥)])
ds (15.11)
Let us define AG as follows:
AG = v^ c + Vd & - - (15.12)
That is, AG is the change in the Gibbs function that would occur if the chemical re-
action given by Eq. 15.4 (which involves the stoichiometric amounts of each component)
proceeded completely from left to right, with the reactants^ and B initially separated and
at temperature T and the standard-state pressure and the products C and D finally sepa-
rated and at temperature T and the standard-state pressure. Note also that AG 3 for a given
reaction is a function of only the temperature. This will be most important to bear in mind
as we proceed with our developments of homogeneous chemical equilibrium. Let us now
digress from our development to consider an example involving the calculation of AG .
EXAMPLE 15.2 Determine the value of AG for the reaction 2H 2 ^ 2H 2 + 2 at 25°C and at 2000 K,
with the water in the gaseous phase.
Solution
At any given temperature, the standard-state Gibbs function change of Eq. 15.12 can be
calculated from the relation
AG° = Afl°- rA5°
At25°C,
^ = 2/^ + ^-2^)
= 2(0) + 1(0) - 2(-241 826) = 483 652 kJ
AS = 24, + 4^25^)
= 2(130.678) + 1(205.148) - 2(188.834) - 88.836 kJ/K
Therefore, at 25°C,
AG = 483 652 - 298.15(88.836) = 457 166 kJ
CHEJ.OCAL EQUtLlBRJU v \f H 627
At 2000 K,
^ = Wtm ~ f?mhk + (Soo - h° 29 s) 0i - 2$ + h° 2m - h%\
= 2(52 942) + (59 176) - 2(-24 1 826 + 72 788)
= 503 136 kJ
AS = 2($ m ) ni + ($ m ) 0i ~ 2(s^ 000 ) HiO
' = 2(188.419) + (268.748) - 2(264.769)
= 1 16.048 U/K
Therefore,
AG = 503 136 -2000 X 116.048 = 271 040 kJ
Returning now to our development, substituting Eq. 15.12 into Eq. 15.11 and rear-
ranging we can write
cIGtp —
AG 4- RTln
At equilibrium dG ZP = 0. Therefore, since ds is arbitrary,
In
For convenience, we define the equilibrium constant K as
AG
RT
\nK= -
AG
RT
(15.13)
(15.14)
(15.15)
which we note must be a function of temperature only for a given reaction, since AG is
given by Eq. 15. 12 in terms of the properties of the pure substances at a given temperature
and the standard-state pressure.
Combining Eqs. 15.14 and 15.15, we have
(15,16)
which is the chemical equilibrium equation corresponding to the reaction equation, Eq. 15.4.
From the equilibrium constant definition in Eqs. 15.15 and 15.16 we can draw a few
conclusions. If the shift in the Gibbs function is large and positive, In K is large and nega-
tive leading to a very small value of K. At a given P in Eq. 15.16 this leads to relatively
small values of the RHS (component C andD) concentrations relative to the LHS compo-
nent concentrations; the reaction is shifted to the left. The opposite is the case of a shift in
the Gibbs function that is large and negative, giving a large value of K and the reaction is
shifted to the right as shown in Fig. 15.8. If the shift in Gibbs function is zero, then In JSTis
zero, and K is exactly equal to 1, the reaction is in the middle with all concentrations of
the same order of magnitude, unless the stoichiometric coefficients are extreme.
628
CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
FIGURE 15.8 The
shift in the reaction with
the change in Gibbs
function.
■Bight'
K»1
Centered.:
K~ 1
Left
K«1
AG
-f*- — —
RT
The other trends we can see are the influences of the temperature and pressure. For
a higher temperature, but same shift in Gibbs function, the absolute value of In K is
smaller, which means K is closer to 1 and the reaction is more centered. For low tempera-
tures the reaction is shifted toward the side with the smallest Gibbs function G . The
pressure has an influence only if the power in Eq. 15.16 is different from zero. That is so
when the number of moles on the RHS (v c + v D ) is different from the number of moles
on the LHS (y A + v B ). Assuming we have more moles on the RHS, then we see that the
power is positive. So if the pressure is larger than the reference pressure, the whole pres-
sure factor is larger than 1, which reduces the RHS concentrations as K is fixed for a given
temperature. You can argue all the other combinations, and the result is that a higher pres-
sure pushes the reaction toward the side with fewer moles, and a lower pressure pushes
the reaction toward the side with more moles. The reaction tries to counteract the exter-
nally imposed pressure variation.
EXAMPLE 15.3 Determine the equilibrium constant K, expressed as in K, for the reaction 2H 2 ^
21 1 2 + 2 at 25°C and at 2000 K.
Solution
We have already found, in Example 15.2, AG for this reaction at these two tempera-
tures. Therefore, at 25°C,
(lni0 293 = — =
J T 8.3145 X 298.15
^2J4^= -184.42
At 2000 K, we have
_ A^ooo -271040 - _ .jggpp-
(lnK) 2O0o - -=^- - 83145 x 2000
Table A.1 1 gives the values of the equilibrium constant for a number of reactions.
Note again that for each reaction the value of the equilibrium constant is determined from
the properties of each of the pure constituents at the standard-state pressure and is a func-
tion of temperature only.
For other reaction equations, the chemical equilibrium constant can be calculated as
in Example 15.3. Sometimes you can write a reaction scheme as a linear combination of
Chemical equilibrium H 629
the elementary reactions that are already tabulated, as for example in Table A. 11 . Assume
we can write a reaction HI as a linear combination of reaction I and reaction II, which
means
LHS m = a LHS £ + b LHS U (15.17)
RHSm - a RHS t + b RHS ir
From the definition of the shift in the Gibbs function, Eq, 15.12, it follows that
AGfji = <j?h rhs " GfiiLHs = a AG? + b AGft
Then from the definition of the equilibrium constant in Eq. 15.15 we get
, „ AGm AG° AGjj
In Aiii = — = — = — a ~ c> — — — a In At + b In a,,
AT RT RT
K m = K° l K b u (15.18)
EXAMPLE 15.4 Show that the equilibrium constant for the reaction called the water-gas reaction
HIH 2 + C0 2 ^H 2 + CO
canbecalculatedfromvalueslistedinTableA.il.
Solution
Using the reaction equations from Table A.l 1 , -
I 2C0 2 ^2CO + 2
II 2H 2 0^2H 2 + 2
It is seen that
m = £i-4n = £(i-n)
so that
A',,, , „
where K m is calculated from the Table A. 1 1 values
lni: m = |(InA' I --ln^: II )
We now consider a number of examples that illustrate the procedure for deteimin-
ing the equilibrium composition for a homogeneous reaction and the influence of certain
variables on the equilibrium composition.
630 CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
EXAMPLE 15.5 One kilomole of carbon at 25°C and 0.1 MPa pressure reacts with 1 kmol of oxygen at
25°C and 0.1 MPa pressure to form an equilibrium mixture of C0 2 , CO, and 2 at 3000
K, 0.1 MPa pressure, in a steady-state process. Determine the equilibrium composition
and the heat transfer for this process.
Control volume: Combustion chamber.
Inlet states: P, J known for carbon and for oxygen.
Exit state: P, 7' known.
Process: Steady-state.
Sketch: Figure 15.9.
Model: Table A. 1 for carbon; ideal gases, Tables A .9 and A. 1 0.
Analysis and Solution
It is convenient to view the overall process as though it occurs in two separate steps, a
combustion process followed by a heating and dissociation of the combustion product
carbon dioxide, as indicated in Fig. 15.9. This two-step process is represented as
Combustion: C + 2 — » C0 2
Dissociation reaction: 2C0 2 ^ 2CO + 2
That is, the energy released by the combustion of C and 2 heats the C0 2 formed to high
temperature, which causes dissociation of part of the C0 2 to CO and 2 . Thus, the over-
all reaction can be written
C + 2 -*■ aC0 2 + bCO + d0 2
where the unknown coefficients a, b, and d must be found by solution of the equilibrium -
equation associated with the dissociation reaction. Once this is accomplished, we can
write the first law for a control volume around the combustion chamber to calculate the
heat transfer.
• Control surface around
combustion chamber
c + o 2
25°C
(reactants)
Energy transfer
-6> Combustion
1
Heating and
dissociation
FIGURE T5.9 Sketch
for Example 15.5.
' -Q
To surroundings
ofwiK- Equilibrium mixture
cKMJK aC o 2 + bCO+d0 2
(products)
Chemical Equilibrium H 631
From the combustion equation we find that the initial composition for the dissoci-
ation reaction is 1 kmol C0 2 . Therefore, letting 2z be the number of kilomoles of C0 2
dissociated, we find
2C0 2 ^2CO + 2
Initial: 1
Change: -2z +2z +z
At equilibrium: (1 — 2z) 2z z
Therefore, the overall reaction is
C + 2 -» (1 - 2z)C0 2 + 2zCO + z0 2
and the total number of kilomoles at equilibrium is
n = (1 - 2z) + 2z + z= 1 +z
The equilibrium mole fractions are
_ 1 - 2z „ _ 2z _ z
yc0 > ~ 1 + z yco ~ 1 + z y °> l+z
From Table A.ll we find that the value of the equilibrium constant at 3000 K for the
dissociation reaction considered here is
In £=-2.217 £=0.1089
Substituting these quantities along with P = 0.1 MPa into Eq. 15.16, we have the equi-
librium equation,
2z
K = 0.1089 = ^ = . (1)
W ( 1 - 2z X2
1 + z
or, in more convenient form,
K 0.1089 / 2z
PIP 1 * 1 \1 - 2z/ \1 + z
To obtain the physically meaningful root of this mathematical relation, we note
that the number of moles of each component must be greater than zero. Thus, the root of
interest to us must lie in the range
0<z<0.5
Solving the equilibrium equation by trial and error, we find
z = 0.2189
Therefore, the overall process is
C + 2 ^ 0.5622CO 2 + 0.4378CO + 0.2189O 2
632 M Chapter Fifteen Introduction to Phase and Chemical Equilibrium
where the equilibrium mole fractions are
, ... 0.2189 _ n r?0 ,
^-1.2189 ai7%
The heat transfer from the combustion chamber to the surroundings can be calcu-
lated using the enthalpies of formation and Table A.9. For this process
H R - (lj) c I (h}) 0i -o + o-o
The equilibrium products leave the chamber at 3000 K. Therefore,
Hp ~ n CoSf§ ^~ ^3000 — ^29g)cOj
+ n CC>{hf + ^3000 — ^29s)cO
+ "O^ 1 / + A 3000 ~ ^298)o 2
- 0.5622(~393 522 + 152 853)
+ 0.4378(- 110 527 + 93 504)
+ 0.2189(98 013)
= -121 302 kJ •
Substituting into the first law gives
Qcv. - Hp If x
= - 121 302 kJ/kmol C burned
EXAMPLE 15,6 One kilomole of carbon at 25°C reacts with 2 kmol of oxygen at 25°C to form an equi-
librium mixture of C0 2 , CO, and 2 at 3000 K, 0.1 MPa pressure. Determine the equi-
librium composition.
Confrol volume: Combustion chamber.
Inlet states: T known for carbon and for oxygen.
Exit state: P, T known.
Process: Steady-state,
Model: Ideal-gas mixture at equilibrium.
Chemical equilibrium H 633
Analysis and Solution
The overall process can be imagined to occur in two steps as in the previous example.
The combustion process is
C + 20 2 -> C0 2 + 2
and the subsequent dissociation reaction is
2C0 2 ^± 2CO + 2
Initial: 1 1
Change: — 2z +2z +z
At equilibrium: (1 - 2z) 2z (I+z)
"We find that in this case the overall process is
C + 20 2 -> (1 - 2z)C0 2 + 2zCO + (1 + z)0 2
and the total number of ktlomoles at equilibrium is
n = (1 - 2z) + 2z + (1 + z) = 2 + z
The mole fractions are
_ 1 - 2z 2z _ 1 +z
yco>- 2 + z y™~ 2 + z y °*~2 + z
The equilibrium constant for the reaction 2C0 2 ^ 2CO + 2 at 3000 K. was found in
Example 15.5 to be 0.1089. Therefore, with these expressions, quantities, and P = 0.1
MPa substituted, the equilibrium equation is
2z W 1 + z
K = o. 1 089 = Z = V / V ^ (1)
2+z
K 0.1089 / 2z Y 1+z
P/P° 1 V ~ 2z J \ 2 +z
We note that in order for the number of kilomoles of each component to be greater than
zero,
< z < 0.5
Solving the equilibrium equation for z, we find
z - 0.1553
so that the overall process is
C + 20 2 -» 0.6894CO 2 + 0.3106CO + 1.15530 2
634 H Chapter fifteen introduction to phase and Chemical equilibrium
When we compare this result with that of Example 15.5, we notice that there is more
C0 2 and less CO. The presence of additional oxygen shifts the dissociation reaction
more to the left side.
The mole fractions of the components in the equilibrium mixture are
0.6894
2.1553
0.3106
2.1553
1.1553
2.1553
= 0.320
= 0.144
- 0.536
The heat transferred from the chamber in this process could be found by the same proce-
dure followed in Example 15.5, considering the overall process.
15.5 Simultaneous Reactions
In developing the equilibrium equation and equilibrium constant expressions of Sec-
tion 15.4, it was assumed that there was only a single chemical reaction equation re-
lating the substances present in the system. To demonstrate the more general situation
in which there is more than one chemical reaction, we will now analyze a case involv-
ing two simultaneous reactions by a procedure analogous to that followed in Section
15.4. These results are then readily extended to systems involving several simultaneous
reactions.
Consider a mixture of substances A,B,C, D, L, M t and TV as indicated in Fig. 15.10.
These substances are assumed to exist at a condition of chemical equilibrium at tempera-
ture T and pressure P, and are related by the two independent reactions
(1) v AX A + v B B ^±v c C + v D D (15.19)
(2) v A2 A 4- v L L ^± v M M + v N N (1 5.20)
a
Components
A, B, C, D,
/., M, N
In chemical
equilibrium
FIGURE 15.10
Sketch demonstrating
simultaneous reactions.
We have considered the situation where one of the components (substance A) is in-
volved in each of the reactions in order to demonstrate the effect of this condition on the
resulting equations. As in the previous section, the changes in amounts of the components
are related by the various stoichiometric coefficients (which are not the same as the num-
ber of moles of each substance present in the vessel). We also realize that the coefficients
v Al and v A1 are not necessarily the same. That is, substance A does not in general take part
in each of the reactions to the same extent.
Development of the requirement for equilibrium is completely analogous to that of
Section 15.4, We consider that each reaction proceeds an infinitesimal amount toward the
right side. This results in a decrease in the number of moles of A, B, and L, and an increase
in the moles of C, £>, M, and N. Letting the degrees of reaction be e x and s 2 for reactions
Simultaneous Reactions H 635
I and 2, respectively, the changes in the number of moles are, for infinitesimal shifts from
the equilibrium composition,
dn A = ~v Al ds l ~ v Ai de 2
dn B — — v B ds x
dn L - ~v L ds 2
dn c = +v c ds\
dn D = +v D dsi
dn M - +v M de 2
dn N = +v N ds 2 (15.21)
The change in Gibbs function for the mixture in the vessel at constant temperature
and pressure is
dG TiP = G A dn A + G B dn B + G c dn c -f G D dn D + G L dn L + G M dn M + G M dn N
Substituting the expressions of Eq. 15.21 and collecting terms,
dG TtP ~ (v c G c + v D G D - v A G A - v B G B ) ds l
HvmG m + v N G H -v A p A - v L G L ) de 2 (15.22)
It is convenient to again express each of the partial molal Gibbs functions in terms of
G t = $ + RTin
Equation 15.22 written in this form becomes
P
v c +v D -v AI -
+ Jag? + rt\r
y W { p
y v A a yl L \p°
ds-,
(15.23)
In this equation the standard-state change in Gibbs function for each reaction is defined as
AG? = v^ c + v^ D - v A $ A - (15.24)
AG§" = vtfftt + - v A2 T A - v$ L (15.25)
Equation 15.23 expresses the change in Gibbs function of the system at constant T,
P, for infinitesimal degrees of reaction of both reactions 1 and 2, Eqs. 15.19 and 15.20.
The requirement for equilibrium is that dG TP = 0. Therefore, since reactions 1 and 2 are
independent, de x and de 2 can be independently varied. It follows that at equilibrium each
of the bracketed terms of Eq. 15.23 must be zero. Defining equilibrium constants for the
two reactions by
CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
and
\nK 2
AGS
(15.27)
RT
we find that, at equilibrium
ft ft
(15.28)
and
K 2
.A
(15.29)
These expressions for the equilibrium composition of the mixture must be solved simulta-
neously. The following example demonstrates and clarifies this procedure.
EXAMPLE 15.7 One kilomole of water vapor is heated to 3000 K, 0. 1 MPa pressure. Determine the equi-
librium composition, assuming that H 2 0, H 2 , 2> and OH are present.
Control volume: Heat exchanger.
Exit slate: P,rknown.
Model: Ideal-gas mixture at equilibrium.
Analysis and Solution
There are two independent reactions relating the four components of the mixture at equi-
librium. These can be written as
Let 2a be the number of kilomoles of water dissociating according to reaction 1 during
the heating, and let 2b be the number of kilomoles of water dissociating according to re-
action 2. Since the initial composition is 1 kmol water, the changes according to the two
reactions are
(1) 2H 2 0^2H 2 + 2
(2) 2H 2 O^H 2 + 2 0H
(l)2H 2 0^2H 2 + 2
Change: ~2a + 2a + a
(2)2H 2 0^ H 2 +2 OH
Change: ~2b + b + 2b
Simultaneous Reactions B 637
Therefore, the number of kilomoles of each component at equilibrium is its initial num-
ber plus the change, so that at equilibrium
n Hi Q — 1 — 2a — lb
n Uj = 2a + b
"p, = a
n = 1 + a + b
The overall chemical reaction that occurs during the heating process can be written
H 2 -» (1 - 2a - 26)H 2 + (2a + b)H 2 + a0 2 + 2bOK
The right-hand side of this expression is the equilibrium composition of the system.
Since the number of kilomoles of each substance must necessarily be greater than zero,
we find that the possible values of a and b are restricted to
a £
b>0
(a + £)<0.5
The two equilibrium equations are, assuming that the mixture behaves as an ideal gas,
A
\ 1+2-2
Since the mole fraction of each component is the ratio of the number of kilomoles of the
component to the total number of kilomoles of the mixture, these equations can be writ-
ten in the form
\2/
2a + b
l+a + bj\l+a + b
\-2a~2b
l+a + b
2a + b \ 2 ( a \l P
and
1 - 2a - 2b j \l + a + b/\p°
2a + b V 2b (
1 + a + b \l + a + b
\~2a~2b
1 +a + b
2a + b V 2b VI P
1 + a + bj\ 1 - 2a - 2b \p°
638 CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
giving two equations in the two unknowns a and b, since P = 0.1 MPa and the values of
K u K 2 are known. From Table A. 1 1 at 3000 K, we find
Ki ~ 0.002 062 K 2 ~ 0.002 893
Therefore, the equations can be solved simultaneously for a and b. The values satisfying
the equations are
a 0.0534 b - 0.0551
Substituting these values into the expressions for the number of kilomoles of each com-
ponent and of the mixture, we find the equilibrium mole fractions to be
Mo " :: 0.7063
^ = 0.1461
y 0i - 0.0482
y 0K = 0.0994
The methods used in this section can readily be extended to equilibrium systems
having more than two independent reactions. In each case, the number of simultaneous
equilibrium equations is equal to the number of independent reactions. The solution of a
large set of nonlinear simultaneous equations naturally becomes quite difficult, however,
and is not easily accomplished by hand calculations. These problems are normally solved
using iterative procedures on a computer.
15.6 Ionization
In the final section of this chapter, we consider the equilibrium of systems that are made
up of ionized gases, or plasmas, a field that has been studied and applied increasingly in
recent years. In previous sections, we discussed chemical equilibrium, with a particular
emphasis on molecular dissociation, as for example the reaction
N 2 ^2N
which occurs to an appreciable extent for most molecules only at high temperature, of the
order of magnitude 3000 to 10 000 K. At still higher temperatures, such as those found in
electric arcs, the gas becomes ionized. That is, some of the atoms lose an electron, accord-
ing to the reaction
N ^ N + + e~
where N + denotes a singly ionized nitrogen atom, one that has lost one electron and conse-
quently has a positive charge, and e~ represents the free electron. As the temperature rises
still higher, many of the ionized atoms lose another electron, according to the reaction
and thus becomes doubly ionized. As the temperature continues to rise, the process continues
until a temperature is reached at which all the electrons have been stripped from the nucleus.
Ionization M 639
Ionization generally is appreciable only at high temperature. However, dissociation
and ionization both tend to occur to greater extents at low pressure, and consequently dis-
sociation and ionization may be appreciable in such environments as the upper atmo-
sphere, even at moderate temperature. Other effects such as radiation will also cause
ionization, but these effects are not considered here.
The problems of analyzing the composition in a plasma become much more diffi-
cult than for an ordinary chemical reaction, for in an electric field the free electrons in the
mixture do not exchange energy with the positive ions and neutral atoms at the same rate
that they do with the field. Consequently, in a plasma in an electric field, the electron gas
is not at exactly the same temperature as the heavy particles. However, for moderate
fields, assuming a condition of thermal equilibrium in the plasma is a reasonable approxi-
mation, at least for preliminary calculations. Under this condition, we can treat the ioniza-
tion equilibrium in exactly the same manner as an ordinary chemical equilibrium analysis.
At these extremely high temperatures, we may assume that the plasma behaves as
an ideal-gas mixture of neutral atoms, positive ions, and electron gas. Thus, for the ioniza-
tion of some atomic species A,
A^±A + + e~ (1530)
we may write the ionization equilibrium equation in the form
The ionization-equilibrium constant K is defined in the ordinary manner
In ^=-4^ (15.32)
RT
and is a function of temperature only. The standard-state Gibbs function change for reac-
tion 15.30 is found from
AG» = ( 15 - 33 >
The standard-state Gibbs function for each component at the given plasma tempera-
ture can be calculated using the procedures of statistical thermodynamics, so that
ionization-equilibrium constants can be tabulated as functions of temperature.
Solution of the ionization-equilibrium equation, Eq. 15.31, is then accomplished in
the same manner as for an ordinary chemical-reaction equilibrium.
EXAMPLE 15.8 Calculate the equilibrium composition if argon gas is heated in an arc to 10 000 K, 1
kPa, assuming the plasma to consist of Ar, Ar + , e~. The ionization-equilibrium constant
for the reaction
Ar ^ Ar + + e~
at this temperature is 0.000 42.
Control volume:
Exit state:
Model:
Heating arc,
P, T known.
Ideal-gas mixture at equilibrium.
640 S CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
Analysis and Solution
Consider an initial composition of 1 kmol neutral argon, and let z be the number of kilo-
moles ionized during the heating process. Therefore,
Ar Ar { I e"
Initial: 1
Change: — z +z +z
Equilibrium: (1 z) z z
and
n = (1 — z) + z + z = 1 + z
Since the number of kilomoles of each component must be positive, the variable z is re-
stricted to the range
0<z< 1
The equilibrium mole fractions are
y*
"Ar = 1 - z
" 1 + z
« l+Z
« 1 + z
The equilibrium equation is
K = y^y e - ( P V T1_I _ V 1 z /\ l 1 6 > ( P
I + l-l \\ + z/\l + z
,1+z
so that, at 10 000 K, 1 kPa,
0.000 42 = (j^) (0 ' 01)
Solving,
z = 0.2008
and the composition is found to be
y Ar = 0.6656
yAr+ = 0.1672
3V = 0.1672
Summary H 641
0.01
0.001
2,000 4,000 6,000 8,000 10,000 12,00014,000 16,000 18,000 20,000 22,000 24,000
Temperature, T, K
™ ^lO" 6 (P = 1.2927 kg/m 3 )
FIGURE 15.11 Equilibrium composition of air [W. E. Moeckel and K. C. Weston, NACA TN 4265 (1958)].
Simultaneous reactions, such as simultaneous molecular dissociation and ionization
reactions or multiple ionization reactions, can be analyzed in the same manner as the ordi-
nary simultaneous chemical reactions of Section 15.5. In doing so, we again make the as-
sumption of thermal equilibrium in the plasma, which, as mentioned before, is, in many
cases, a reasonable approximation. Figure 15.11 shows the equilibrium composition of air
at high temperature and low density, and indicates the overlapping regions of the various
dissociation and ionization processes.
NUMMARY
A short introduction is given to equilibrium in general with application to phase equilib-
rium and chemical equilibrium. From previous analysis with the second law we have
found the reversible shaft work as the change in Gibbs function. This is extended to give
the equilibrium state as the one with minimum Gibbs function at a given T, P. This also
applies to two phases in equilibrium, so each phase has the same Gibbs function.
Chemical equilibrium is formulated for a single-equilibrium reaction assuming
the components are all ideal gases. This leads to an equilibrium equation tying together
the mole fractions of the components, the pressure, and the reaction constant. The reac-
tion constant is related to the shift in Gibbs function from the reactants (LHS) to the
products (RHS) at a temperature T. As temperature or pressure changes, the equilibrium
642 m chapter Fifteen introduction to phase and chemical equilibrium
composition will shift according to its sensitivity to T and P. For very large equilibrium
constants, the reaction is shifted toward the RHS, and for very small ones it is shifted
toward the LHS.
In most real systems of interest, there are multiple reactions coming to equilib-
rium simultaneously with a fairly large number of species involved. Often species are
present in the mixture without participation in the reactions causing a dilution, so all
mole fractions are lower than they otherwise would be. As a iast example of a reaction,
we show an ionization process in which one or more electrons can be separated from
an atom.
You should have learned a number of skills and acquired abilities from studying
this chapter that will allow you to:
• Apply the principle of minimum Gibbs function to a phase equilibrium.
• Understand that the concept of equilibrium can include other effects such as eleva-
tion, surface tension, and electrical potentials, as well as the concept of metastable
states.
• Understand that the chemical equilibrium is written for ideal-gas mixtures.
• Understand the meaning of the shift in Gibbs function due to the reaction.
• Know when the absolute pressure has an influence on the composition.
• Know the connection between the reaction scheme and the equilibrium constant.
• Understand that all species are present and influence the mole fractions.
• Know that a dilution with an inert gas has an effect.
• Understand the coupling between the chemical equilibrium and the energy
equation.
• Intuitively know that most problems must be solved by iterations.
• Be able to treat a dissociation added on to a combustion process.
• Be able to treat multiple simultaneous reactions.
• Know what an ionization process is and how to treat it.
Key concepts
and formulas
Gibbs function
Equilibrium
Phase equilibrium
Equilibrium reaction
Change in Gibbs function
Equilibrium constant
g = h - Ts
Minimum g for given T,P^> dG ZP —
Sf=Sg
AG = v^ c + v D gl - v A gl - v^ B evaluate at T, P°
K = g-Atf/AT
K =
Reaction scheme
Dilution
Simultaneous reactions
Reaction scheme HI = aI + MI=> K m — KfK^
reaction the same, y's are smaller
K U K 2 , . . ■ and more /s
Homework Problems S 643
Concept-Study Guide Problems
15.1 Is the concept of equilibrium limited to thermo-
dynamics?
15.2 How does the Gibbs function vary with quality as
you move from liquid to vapor?
15.3 How is a chemical equilibrium process different
from a combustion process?
15.4 Must P and T be held fixed to obtain chemical
equilibrium?
15.5 The change in Gibbs function for a reaction is a
function of which property?
15.6 In a steady-flow burner, Tis not controlled; which
properties are?
15.7 In a closed, rigid combustion bomb, which prop-
erties are held fixed?
15.8 Is the dissociation of water pressure sensitive?
15.9 At 298 K, K = exp(-184) for the water dissocia-
tion, what does that imply?
15.10 For a mixture of 2 and O the pressure is increased
at constant T; what happens to the composition?
15.11 For a mixture of 2 and O the temperature is
increased at constant P\ what happens to the
composition?
15.12 For a mixture of 2 and O I add some argon,
keeping constant T, P; what happens to the moles
ofO?
15.13 In a combustion process, is the adiabatic flame
temperature affected by reactions?
15.14 When dissociations occur after combustion, does
Tgo up or down?
15.15 In equilibrium, the Gibbs function of the reac-
tants and the products is the same; how about
the energy?
15.16 Does a dissociation process require energy, or
does it give out energy?
15.17 If I consider the nonfrozen (composition can
vary) heat capacity, but still assume all compo-
nents are ideal gases, does that C become a func-
tion of temperature? of pressure?
15.18 What is K for the water-gas reaction in Example
15.4 at 1200 K?
15.19 Which atom in air ionizes first as T increases?
What is the explanation?
15.20 At what temperature range does air become a
plasma?
Homework Problems
Equilibrium and Phase Equilibrium
15.21 Carbon dioxide at 15 MPa is injected into the top
of a 5-km-deep well in connection with an en-
hanced oil-recovery process. The fluid column
standing in the well is at a uniform temperature of
40°C. What is the pressure at the bottom of the
well, assuming ideal-gas behavior?
15.22 Consider a 2-km-deep gas well containing a gas
mixture of methane and ethane at a uniform tem-
perature of 30°C. The pressure at the top of the
well is 14 MPa, and the composition on a mole
basis is 90% methane, 10% ethane. Each com-
ponent is in equilibrium (top to bottom) with
dG + g dZ = and assume ideal gas, so for each
component Eq. 15.10 applies. Determine the pres-
sure and composition at the bottom of the well.
15.23 A container has liquid water at 20°C, 100 kPa, in
equilibrium with a mixture of water vapor and dry
air also at 20°C, 100 kPa. How much is the water
vapor pressure, and what is the saturated water
vapor pressure?
15.24 Using the same assumptions as those in develop-
ing Eq. d in Example 15.1, develop an expression
for pressure at the bottom of a deep column of liq-
uid in terms of the isothermal compressibility,
fi T . For liquid water at 20°C, we know that
j3 r = 0.0005 [1/MPa]. Use the result of the first
question to estimate the pressure in the Pacific
Ocean at the depth of 3 fan.
Chemical Equilibrium, Equilibrium Constant
15.25 Calculate the equilibrium constant for the reaction
2 ^ 20 at temperatures of 298 K and 6000 K.
Verify the result with Table A. 1 1 .
15.26 For the dissociation of oxygen, 2 o 20, around
2000 K we want a mathematical expression for
the equilibrium constant K(T). Assume constant
644 M CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
heat capacity, at 2000 K, for 2 and O from Table
A.9 and develop the expression from Eqs. 15.12
and 15.15. 15.38
15.27 Calculate the equilibrium constant for the reaction
H 2 ^ 2H at a temperature of 2000 K, using prop-
erties from Table A.9. Compare the result with
the value listed in Table A. 11.
15.28 Plot to scale the values of In K versus 1/7* for the 15.39
reaction 2C0 2 ^ 2CO + 2 . Write an equation
for In £ as a function of temperature.
15.29 Calculate the equilibrium constant for the reaction 15.40
2C0 2 ^ 2CO + 2 at 3000 K using values from
Table A.9 and compare the result to Table A. 1 1 .
15.30 Consider the dissociation of oxygen, 2 <^> 20,
starting with 1 kmol oxygen at 298 K and heating
it at constant pressure 100 kPa. At which temper-
ature will we reach a concentration of monatomic
oxygen of 10%? 15.41
15.31 Pure oxygen is heated from 25°C to 3200 K in a
steady-state process at a constant pressure of 200
kPa. Find the exit composition and the heat transfer. 15.42
15.32 Nitrogen gas, N 2 , is heated to 4000 K, lOkPa. What
fraction of the N 2 is dissociated to N at this state?
15.33 Hydrogen gas is heated from room temperature to
4000 K, 500 kPa, at which state the diatomic
species has partially dissociated to the monatomic
form. Determine the equilibrium composition at
this state. 15.43
15.34 One kilomole Ar and one kilomole 2 are heated
at a constant pressure of 100 kPa to 3200 K,
where they come to equilibrium. Find the final
mole fractions for Ar, 2) and O.
15.35 Consider the reaction 2C0 2 & 2CO + 2 ob-
tained after heating 1 kmol C0 2 to 3000 K. Find
the equilibrium constant from the shift in Gibbs
function and verify its value with the entry in
Table A. 11. What is the mole fraction of CO at
3000 K, 100 kPa?
15.36 Air (assumed to be 79% nitrogen and 21% oxy-
gen) is heated in a steady-state process at a con-
stant pressure of 100 kPa, and some NO is formed
(disregard dissociations of N 2 and 2 ). At what
temperature will the mole fraction of NO be
0.001?
15.37 The combustion products from burning pentane,
QH l2 , with pure oxygen in a stoichiometric ratio 15.46
exit at 2400 K, 100 kPa. Consider the dissociation
15.44
15.45
of only C0 2 and find the equilibrium mole frac-
tion of CO.
Find the equilibrium constant for the reaction
2NO + 2 ^ 2N0 2 from the elementary reac-
tions in Table A. 11 to answer which of the nitro-
gen oxides, NO or N0 2 , is the more stable at
ambient conditions? What about at 2000 K?
Pure oxygen is heated from 25°C, 100 kPa, to
3200 K in a constant-volume container. Find the
final pressure, composition, and the heat transfer.
A mixture of 1 kmol carbon dioxide, 2 kmol car-
bon monoxide, and 2 kmol oxygen, at 25°C J 150
kPa, is heated in a constant-pressure steady-state
process to 3000 K. Assuming that only these
same substances are present in the exiting chemi-
cal equilibrium mixture, determine the composi-
tion of that mixture.
Repeat the previous problem for an initial mixture
that also includes 2 kmol of nitrogen, which does
not dissociate during the process.
One approach to using hydrocarbon fuels in a fuel
cell is to "reform" the hydrocarbon to obtain hy-
drogen, which is then fed to the fuel cell. As a
part of the analysis of such a procedure, consider
the reaction CH 4 + H 2 ^ 3H 2 + CO. Deter-
mine the equilibrium constant for this reaction at
a temperature of 800 K.
Consider combustion of methane with pure oxy-
gen forming carbon dioxide and water as the
products. Find the equilibrium constant for the re-
action at 1000 K. Use an average heat capacity of
C p = 52 kJ/kmol K for the fuel and Table A.9 for
the other components.
Find the equilibrium constant for the reaction:
2NO + 2 <=> 2N0 2 from the elementary reaction
in Table A. 11 to answer these two questions.
Which of the nitrogen oxides, NO or N0 2) is the
more stable at 25°C, 100 kPa? At what T do we
have an equal amount of each?
The equilibrium reaction with methane as CH 4 ^
C + 2H 2 has In K - -0.3362 at 800 K and In
IC = -4.607 at 600 K. By noting the relation of if
to temperature, show how you would interpolate
In K in (1/7) to find K at 700 K and compare that
to a linear interpolation.
Water from the combustion of hydrogen and pure
oxygen is at 3800 K and 50 kPa. Assume we only
Homework Problems M 645
have H 2 0, 2 and H 2 as gases, find the equilib-
rium composition.
15.47 Complete combustion, of hydrogen and pure oxy-
gen in a stoichiometric ratio at P , T to form
water would result in a computed adiabatic flame
temperature of 4990 K for a steady-state setup.
a. How should the adiabatic flame temperature be
found if the equilibrium reaction 2H 2 + 2 ^
2H 2 is considered? Disregard all other possi-
ble reactions (dissociations) and show the final
equation(s) to be solved.
b. Which other reactions should be considered,
and which components will be present in the
final mixture?
15.48 The van't Hoff equation
RT P
relates the chemical equilibrium constant K to the
enthalpy of reaction A#°. From the value ofK in
Table A. 1 1 for the dissociation of hydrogen at
2000 K and the value of A#° calculated from
Table A.9 at 2000 K, use the van't Hoff equation
to predict the equilibrium constant at 2400 K.
15.49 Gasification of char (primarily carbon) with steam
following coal pyrolysis yields a gas mixture of
1 kmol CO and 1 kmol H 2 . We wish to upgrade
the hydrogen content of this syngas fuel mixture,
so it is fed to an appropriate catalytic reactor along
with 1 kmol of H 2 0. Exiting the reactor is a chem-
ical equilibrium gas mixture of CO, H 2 , H 2 0, and
C0 2 at 600 K, 500 kPa. Determine the equilibrium
composition. Note: See Example 15.4.
15.50 Consider the water gas reaction in Example 15.4.
Find the equilibrium constant at 500, 1000, 1200,
and 1400 K. What can you infer from the result?
15.51 Catalytic gas generators are frequently used to de-
compose a liquid, providing a desired gas mixture
(spacecraft control systems, fuel cell gas supply,
and so forth). Consider feeding pure liquid hy-
drazine, N 2 H 4l to a gas generator, from which exits
a gas mixture of N 2 , H 2 , and NH 3 in chemical equi-
librium at 100°C, 350 kPa. Calculate the mole frac-
tions of the species in the equilibrium mixture.
15.52 A piston/cylinder contains 0.1 kmol hydrogen and
0.1 kmol Ar gas at 25°C, 200 kPa. It is heated in a
constant-pressure process so the mole fraction of
atomic hydrogen is 10%. Find the final tempera-
ture and the heat transfer needed.
15.53 A tank contains 0.1 kmol hydrogen and 0.1 kmol of
argon gas at 25°C, 200 kPa, and the tank keeps con-
stant volume. To what r should it be heated to have
a mole fraction of atomic hydrogen, H, of 10%?
15.54 A gas mixture of 1 kmol carbon monoxide,
1 kmol nitrogen, and 1 kmol oxygen at 25°C, 150
kPa, is heated in a constant-pressure steady-state
process. The exit mixture can be assumed to be in
chemical equilibrium with C0 2) CO, 2 , and N 2
present. The mole fraction of C0 2 at this point is
0.176. Calculate the heat transfer for the process.
15.55 A rigid container initially contains 2 kmol of car-
bon monoxide and 2 kmol of oxygen at 25°C, 100
kPa. The content is then heated to 3000 K, at
which point an equilibrium mixture of C0 2) CO,
and 2 exists. Disregard other possible species
and determine the final pressure, the equilibrium
composition, and the heat transfer for the process.
15.56 A coal gasiher produces a mixture of ICO and
2H 2 that is fed to a catalytic converter to produce
methane. The reaction is CO + 3H 2 <=> CH 4 +
H 2 0. The equilibrium constant at 600 K is K =
1.83 X 10 6 . What is the composition of the exit
flow assuming a pressure of 600 kPa?
15.57 One approach to using hydrocarbon fuels in a fuel
cell is to "reform" the hydrocarbon to obtain hy-
drogen, which is then fed to the fuel cell. As a part
of the analysis of such a procedure, consider the re-
action CHf + H 2 <^> CO + 3H 2 . One kilomole
each of methane and water are fed to a catalytic re-
former. A mixture of CH 4) H 2 0, H 2t and CO exits
in chemical equilibrium at 800 K, 100 kPa; deter-
mine the equilibrium composition of this mixture
using an equilibrium constant of K = 0.0237.
15.58 Use the information in Problem 15.45 to estimate
the enthalpy of reaction, A#°, at 700 K using the
van't Hoff equation (see Problem 15.48) with fi-
nite differences for the derivatives.
15.59 Acetylene gas at 25°C is burned with 140% theo-
retical air, which enters the burner at 25°C, 1 00
kPa, 80% relative humidity. The combustion
products form a mixture of C0 2) Ff 2 0, N 2 , 2t and
NO in chemical equilibrium at 2200 K, 100 kPa.
This mixture is then cooled to 1000 K very
rapidly, so that the composition does not change.
646 H CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
Determine the mole fraction of NO in the prod-
ucts and the heat transfer for the overall process.
15.60 A step in the production of a synthetic liquid fuel
from organic waste matter is the following conver-
sion process: 1 kmol of ethylene gas (converted
from the waste) at 25°C, 5 MPa, and 2 kmol of
steam at 300°C, 5 MPa, enter a catalytic reactor.
An ideal-gas mixture of ethanol, ethylene, and
water in chemical equilibrium leaves the reactor at
700 K, 5 MPa, Determine the composition of the
mixture and the heat transfer for the reactor.
15.61 Methane at 25°C, 100 kPa, is burned with 200%
theoretical oxygen at 400 K, 100 kPa, in an adia-
batic steady-state process, and the products of
combustion exit at 100 kPa. Assume that the only
significant dissociation reaction in the products is
that of carbon dioxide going to carbon monoxide
and oxygen. Determine the equilibrium composi-
tion of the products and also their temperature at
the combustor exit.
15.62 Calculate the irreversibility for the adiabatic com-
bustion process described in the previous problem.
15.63 An important step in the manufacture of chemical
fertilizer is the production of ammonia, according
to the reaction: N 2 + 3H 2 ^ 2NH 3
a. Calculate the equilibrium constant for this re-
action at 150°C
b. For an initial composition of 25% nitrogen,
75% hydrogen, on a mole basis, calculate the
equilibrium composition at 150°C, 5 MPa.
15.64 One kilomole of carbon dioxide, C0 2 , and 1 kmol
of hydrogen, H 2 , at room temperature and 200
kPa is heated to 1200 K, 200 kPa. Use the water
gas reaction to determine the mole fraction of CO.
Neglect dissociations of H 2 and 2 .
15.65 Consider the production of a synthetic fuel
(methanol) from coal. A gas mixture of 50% CO
and 50% H 2 leaves a coal gasifier at 500 K,
1 MPa, and enters a catalytic converter. A gas mix-
ture of methanol, CO, and H 2 in chemical equilib-
rium with the reaction CO + 2H 2 ^ CH 3 OH
leaves the converter at the same temperature and
pressure, where it is known that lnK= —5.119.
a. Calculate the equilibrium composition of the
mixture leaving the converter.
b. Would it be more desirable to operate the con-
verter at ambient pressure?
15.66 Hydrides are rare earth metals, M, that have the
ability to react with hydrogen to form a different
substance MB X with a release of energy. The hy-
drogen can then be released, the reaction re-
versed, by heat addition to the MH X . In this
reaction only the hydrogen is a gas, so the for-
mula developed for the chemical equilibrium is
inappropriate. Show that the proper expression to
be used instead ofEq. 15.14 is
\n(P m /P a ) = &G°/RT
: when the reaction is scaled to I kmol of H 2 .
Simultaneous Reactions
15.67 Water from the combustion of hydrogen and pure
oxygen is at 3800 K and 50 kPa. Assuming we
only have H 2 0, 2 , OH, and H 2 as gases with the
two simple water dissociation reactions active,
find the equilibrium composition.
15.68 Ethane is burned with 150% theoretical air in a
gas-turbine combustor. The products exiting con-
sist of a mixture of C0 2 , H 2 0, 2 , N 2 , and NO in
chemical equilibrium at 1800 K, 1 MPa. Deter-
mine the mole fraction of NO in the products. Is it
reasonable to ignore CO in the products?
15.69 Butane is burned with 200% theoretical air, and
the products of combustion, an equilibrium mix-
ture containing only C0 2 , H 2 0, 2 , N 2 , NO, and
N0 2 , exits from the combustion chamber at 1400
K, 2 MPa. Determine the equilibrium composi-
tion at tins state.
15.70 A mixture of 1 kmol water and 1 kmol oxygen at
400 K is heated to 3000 K, 200 kPa, in a steady-
state process. Determine the equilibrium compo-
sition at the outlet of the heat exchanger,
assuming that the mixture consists of H 2 0, H 2 ,
2 , and OH.
15.71 One kilomole of air (assumed to be 78% nitrogen,
21% oxygen, and 1% argon) at room temperature
is heated to 4000 K, 200 kPa. Find the equilib-
rium composition at this state, assuming that only
N 2 , 2 , NO, O, and Ar are present.
15.72 One kilomole of water vapor at 100 kPa, 400 K, is
heated to 3000 K in a constant-pressure flow
process. Determine the final composition, assum-
ing that H 2 0, H 2 , H, 2) and OH are present at
equilibrium.
Homework Problems S 647
15.73 Acetylene gas and * times theoretical air (x > 1)
at room temperature and 500 kPa are burned at
constant pressure in an adiabatic flow process.
The flame temperature is 2600 K, and the com-
bustion products are assumed to consist of N 2> 2 ,
C0 2 , H 2 0, CO, and NO. Determine the value of x.
Ionization
15.74 At 10 000 K the ionization reaction for Ar is:
Ar ^> Ar + + e~ with equilibrium constant of
K - 4.2 X 10" 4 . What should the pressure be for
a mole concentration of argon ions (Ar + ) of 1 0%?
15.75 Operation of an MHD converter requires an elec-
trically conducting gas. A helium gas "seeded"
with 1.0 mole percent cesium, as shown in Fig.
P15.75, is used where the cesium is partly ionized
(Cs ^± Cs + + O by heating the mixture to 1800
K, 1 MPa> in a nuclear reactor to provide free elec-
trons. No helium is ionized in this process, so that
the mixture entering the converter consists of He,
Cs, Cs + , and e~ . Determine the mole fraction of
electrons in the mixture at 1 800 K, where In K =
1.402 for the cesium ionization reaction described.
Electrical
power out
1%Cs ' ' ! 1
Mixture
(He, Cs, Cs*, e~)
FIGURE P15.75
15.76 One kilomole of argon gas at room temperature is •
heated to 20 000 K, 100 kPa. Assume that the
plasma in this condition consists of an equilib-
rium mixture of Ar, Ar + , Ar ++ , and e~ according
to the simultaneous reactions
(1) Ar^±Ar + + e" (2)Ar + ^Ar ++ + e~
The ionization equilibrium constants for these re-
actions at 20 000 K have been calculated from
spectroscopic data as In K x = 3.11 and In K 2 =
—4.92. Determine the equilibrium composition of
the plasma.
15.77 At 10 000 K the two ionization reactions for N
' and Ar as
(1) Ar «■ Ar + + e~ (2) N <=> N + + e~
have equilibrium constants of K Y = 4.2 X 10~ 4
an&K 2 = 6.3 X 10~ 4 , respectively. If we start out
with 1 kmol Ar and 0,5 kmol N 2 , what is the equi-
librium composition at a pressure of 10 kPa?
15.78 Plot to scale the equilibrium composition of nitro-
gen at 1 kPa over the temperature range 5000 K
to 15 000 K, assuming thatN 2 , N, N + , and e~ are
present. For the ionization reaction N ^± N + -f-
e", the ionization equilibrium constant K has been
calculated from spectroscopic data as
T[K] 10 000 12 000 14 000 16 000
100K 6.26 XlO" 2 1.51 15.1 92
Review Problems
15.79 Repeat Problem 15.21 using the generalized
charts, instead of ideal-gas behavior.
15.80 In a test of a gas-turbine combustor, saturated-
liquid methane at 1 15 K is burned with excess air
to hold the adiabatic flame temperature to 1600 K.
It is assumed that the products consist of a mix-
ture of C0 2 , H 2 0, N 2 , 2) and NO in chemical
equilibrium. Determine the percent excess air
used in the combustion and the percentage of NO
in the products.
15.81 A space heating unit in Alaska uses propane com-
bustion as the heat supply. Liquid propane comes
from an outside tank at ~-44°C, and the air supply
is also taken in from the outside at — 44°C, The
airflow regulator is misadjusted, such that only
90% of the theoretical air enters the combustion
chamber, resulting in incomplete combustion.
The products exit at 1000 K as a chemical equi-
librium gas mixture, including only C0 2 , CO,
H 2 0, H 2j and N 2 . Find the composition of the
products. Hint: Use the water gas reaction in Ex-
ample 15.4.
15.82 Consider the following coal gasifier proposed
for supplying a syngas fuel to a gas-turbine
power plant. Fifty kilograms per second of dry
coal (represented as 48 kg C plus 2 kg H) enter
the gasifier, along with 4.76 kmol/s of air and
2 kmol/s of steam. The output stream from this
unit is a gas mixture containing H 2 , CO, N 2 ,
648 H chapter Fifteen introduction to phase and chemical equilibrium
CH 4) and C0 2 in chemical equilibrium at 900 ¥L,
1 MPa.
a. Set up the reaction and equilibrium equations)
for this system, and calculate the appropriate
equilibrium constant(s).
b. Determine the composition of the gas mixture
leaving the gasifier.
15.83 One kitomole of liquid oxygen, 2 , at 93 K, andx
kmol of gaseous hydrogen, H 2 , at 25°C, are fed to
a combustion chamber, x is greater than 2, such
that there is excess hydrogen for the combustion
process. There is a heat loss from the chamber of
1000 kJ per kmol of reactants. Products exit the
chamber at chemical equilibrium at 3800 K, 400
kPa, and are assumed to include only H 2 0, H 2)
andO.
a. Determine the equilibrium composition of the
products and also x, the amount of H 2 entering
the combustion chamber.
b. Should another substance(s) have been in-
cluded in part (a) as being present in the prod-
ucts? Justify your answer.
15.84 Saturated liquid butane (note: use generalized
charts) enters an insulated constant-pressure com-
bustion chamber at 25°C, and x times theoretical
English unit problems
15.89E Carbon dioxide at 2200 lbf/in. 2 is injected
into the top of a 3 -mi-deep well in connec-
tion with an enhanced oil recovery process.
The fluid column standing in the well is at a
uniform temperature of 100 F. What is the
pressure at the bottom of the well assuming
ideal-gas behavior?
15.90E Calculate the equilibrium constant for the re-
action 2 ^ 20 at temperatures of 537 R and
10 800 R.
15.91E Pure oxygen is heated from 77 F to 5300 F in a
steady-state process at a constant pressure of
30 lbf/in. 2 . Find the exit composition and the
heat transfer.
15.92 E Air (assumed to be 79% nitrogen and 21%
oxygen) is heated in a steady-state process at a
constant pressure of 14.7 lbf/in. 2 , and some NO
is formed. At what temperature will the mole
fraction of NO be 0.001?
oxygen gas enters at the same pressure and tem-
perature. The combustion products exit at 3400 K.
Assuming that the products are a chemical equi-
librium gas mixture that includes CO, what is x?
15.85 Derive the van't Hoff equation given in Problem
15.48, using Eqs. 15.12 and 15.15. Note: The
rf(g/7) at constant P a for each component can be
expressed using the relations in Eqs. 13.18 and
13.19.
15.86 A coal gasifier produces a mixture of ICO and
2H 2 which is then fed to a catalytic converter to
produce methane. A chemical-equilibrium gas
mixture containing CH 4 , CO, H 2 , and H 2 exits
the reactor at 600 K, 600 kPa. Determine the mole
fraction of methane in the mixture.
15.87 Dry air is heated from 25°C to 4000 K in a 100-
kPa constant-pressure process. List the possible
reactions that may take place and determine the
equilibrium composition. Find the required heat
transfer.
15.88 Methane is burned with theoretical oxygen in a
steady-state process, and the products exit the
combustion chamber at 3200 K, 700 kPa. Calcu-
late the equilibrium composition at this state, as-
suming that only C0 2) CO, H 2 0, H 2l 2 , and OH
are present.
15.93E The combustion products from burning pen-
tane, C 5 H 12 , with pure oxygen in a stoichiomet-
ric ratio exit at 4400 R. Consider the
dissociation of only C0 2 and find the equilib-
rium mole fraction of CO.
15.94E Pure oxygen is heated from 77 F, 14.7 lbf/in. 2 , to
5300 F in a constant-volume container. Find the
final pressure, composition, and the heat transfer.
15.95E The equilibrium reaction with methane as
CH 4 ?± C + 2H 2 has In K = -0.3362 at 1440 R
and In K = -4.607 at 1080 R. By noting the re-
lation of K to temperature, show how you would
interpolate In K in (1/7) to find K at 1260 R and
compare that to a linear interpolation.
15.96E A gas mixture of 1 pound mol carbon monox-
ide, 1 pound mol nitrogen, and 1 pound mol
oxygen at 77 F, 20 lbf/in. 2 , is heated in a con-
stant-pressure flow process. The exit mixture
can be assumed to be in chemical equilibrium
Computer, design, and open-ended problems H 649
with C0 2) CO, 2j and N 2 present. The mole
fraction of C0 2 at this point is 0.176. Calculate
the heat transfer for the process.
15.97E Use the information in Problem 15.95E to esti-
mate the enthalpy of reaction, A// , at 1260 R
using the van't Hoff equation (see Problem
15.48) with finite differences for the derivatives.
15.98E Acetylene gas at 77 F is burned with 140%
theoretical air, which enters the burner at 77 F,
14.7 lbf/in. 2 , 80% relative humidity. The com-
bustion products form a mixture of C0 2 , H 2 0,
N 2) 2> and NO in chemical equilibrium at
3500 F, 14.7 Ibf/in. 2 . This mixture is then
cooled to 1340 F very rapidly, so that the com-
position does not change. Determine the mole
fraction of NO in the products and the heat
transfer for the overall process.
15.99E An important step in the manufacture of chem-
ical fertilizer is the production of ammonia, ac-
cording to the reaction N 2 + 3H 2 ^ 2NH 3
a. Calculate the equilibrium constant for this
reaction at 300 F.
b. For an initial composition of 25% nitrogen,
75% hydrogen, on a mole basis, calculate
the equilibrium composition at 300 F, 750
lbf/in. 2 .
15.100E Ethane is burned with 150% theoretical air in a
gas-turbine combustor. The products exiting
consist of a mixture of C0 2 , H 2 0, 2 , N 2 , and
NO in chemical equilibrium at 2800 F, 150
lbf/in. 2 . Determine the mole fraction of NO in
the products. Is it reasonable to ignore CO in
the products?
15.101E One-pound mole of air (assumed to be 78% ni-
trogen, 21% oxygen, and 1% argon) at room
temperature is heated to 7200 R, 30 lbf/in. 2 .
Find the equilibrium composition at this state,
assuming that only N 2 , 2 , NO, O, and Ar are
present.
15.102E One-pound mole of water vapor at 14.7
lbf/in. 2 , 720 R, is heated to 5400 R in a con-
stant-pressure flow process. Determine the
final composition, assuming that H 2 0, H 2 , H,
2 , and OH are present at equilibrium.
15.103E Acetylene gas and x times theoretical air (x >
1) at room temperature and 75 lbf/in. 2 are
burned at constant pressure in an adiabatic
flow process. The flame temperature is 4600 R,
and the combustion products are assumed to
consist of N 2 , 2 , C0 2 , H 2 0, CO, and NO. De-
termine the value of x.
15.104E Methane is burned with theoretical oxygen in a
steady-state process, and the products exit the
combustion chamber at 5300 F, 100 lbf/in. 2 .
Calculate the equilibrium composition at this
state, assuming that only C0 2 , CO, H 2 0, H 2 ,
2 , and OH are present.
15.1 05E In a test of a gas-turbine combustor, saturated-
liquid methane at 210 R is to be burned with
excess air to hold the adiabatic flame tempera-
ture to 2880 R. It is assumed that the products
consist of a mixture of C0 2 , H 2 0, N 2> 2 , and
NO in chemical equilibrium. Determine the
percent excess air used in the combustion, and
the percentage of NO in the products.
15.106E Dry air is heated from 77 F to 7200 R in a 14.7
lbf/in. 2 constant-pressure process. List the pos-
sible reactions that may take place and deter-
mine the equilibrium composition. Find the
required heat transfer.
Computer, design, and Open-ended problems
15.107 Write a program to solve the general case of mixture of C0 2 , CO, and 2 . We wish to deter-
Problem 15.60, in wirich the relative amount of mine the flame temperature for various combina-
steam input and the reactor temperature and tions of b and the pressure P, assuming constant
pressure are program input variables and use specific heat for the components from Table A.5.
constant specific heats. 15tl n 9 Study the chemical reactions that take place
15.108 Write a program to solve the following problem. when CFC-type refrigerants are released into the
One kmol of carbon at 25°C is burned with b atmosphere. The chlorine may create com-
kmol of oxygen in a constant-pressure adiabatic pounds as HCI and CI0N0 2 that react with the
process. The products consist of an equilibrium ozone 3 .
650 H CHAPTER FIFTEEN INTRODUCTION TO PHASE AND CHEMICAL EQUILIBRIUM
15.110 Examine the chemical equilibrium that takes place
in an engine where CO and various nitrogen oxy-
gen compounds summarized as NO x may be
formed. Study the processes for a range of air-fuel
ratios and temperatures for typical fuels. Are there
important reactions not listed in the book?
15.111 A number of products may be produced from
the conversion of organic waste that can be used
as fuel (see Problem 15.60). Study the subject
and make a list of the major products that are
formed and the conditions at which they are
formed in desirable concentrations.
15.112 The hydrides as explained in Problem 15.66 can
store large amounts of hydrogen. The penalty
for the storage is that energy must be supplied
when the hydrogen is released. Investigate the
literature for quantitative information about the
quantities and energy involved in such a hydro-
gen storage.
15.113 The hydrides explained in Problem 15.66 can be
used in a chemical heat pump. The energy in-
volved in the chemical reaction can be added
and removed at different temperatures. For some
hydrides these temperatures are low enough to
make them feasible for heat pumps for heat up-
grade, refrigerators, and air conditioners. Inves-
tigate the literature for such applications and
give some typical values for these systems,
15.114 Power plants and engines have high peak tem-
peratures in the combustion products where NO
is produced. The equilibrium NO level at the
high temperature is frozen at that level during
the rapid drop in temperature with the expan-
sion. The final exhaust therefore contains NO at
a level much higher than the equilibrium value
at the exhaust temperature. Study the NO level
at equilibrium when natural gas, CH 4 , is burned
adiabatically with air (at 7* ) in various ratios.
15.115 Excess air or steam addition is often used to
lower the peak temperature in combustion to
limit formation of pollutants like NO. Study the
steam addition to the combustion of natural gas
as in the Cheng cycle (see Problem 12.176), as-
suming the steam is added before the combus-
tion. How does this affect the peak temperature
and the NO concentration?
Contents of Appendix
SI Units: Single-State Properties
653
Table A. I Conversion Factors, 653
Table A.2 Critical Constants, 656
Table A3 Properties of Selected Solids at 25°C, 657
Table A.4 Properties of Some Liquids at 25°C, 657
Table A.5 Properties of Various Ideal Gases at 25°C, 100 kPa (SI Units), 658
Table A.6 Constant-Pressure Specific Heats of Various Ideal Gases, 659
Table A7.1 Ideal-Gas Properties of Air, Standard Entropy at 0,1-MPa (1-bar)
Pressure, 660
Table A7.2 The Isentropic Relative Pressure and Relative Volume Functions, 661
Table A.8 Ideal-Gas Properties of Various Substances, Entropies at O.i-MPa
(1-bar) Pressure, Mass Basis, 662
Table A.9 Ideal-Gas Properties of Various Substances (SI Units), Entropies at
O.i-MPa (1-bar) Pressure, Mole Basis, 664
Table A. 1 Enthalpy of Formation and Absolute Entropy of Various Substances
at 25°C, 100 kPa Pressure, 670
Table A. 11 Logarithms to the Base e of the Equilibrium Constant^, 671
SI UNITS: THERMODYNAMIC TABLES 673
Table B.l Thermodynamic Properties of Water, 674
Table B.2 Thermodynamic Properties of Ammonia, 692
Table B.3 Thermodynamic Properties of R-12, 698
Table B.4 Thermodynamic Properties of R-22, 702
Table B.5 Thermodynamic Properties of R-134a, 708
Table B.6 Thermodynamic Properties of Nitrogen, 714
Table B.7 Thermodynamic Properties of Methane, 718
Table D.l Equations of State, 725
Table D.2 Empirical Constants for Benedict-Webb-Rubin Equation, 726
Table D.3 The Lee-Kesler Equation of State, 727
Table D.4 Saturated Liquid- Vapor Compressibilities Lee-Kesler Simple
Fluid, 727
Table D.5 Acentric Factor for Some Substances, 727
IDEAL-GAS SPECIFIC HEAT
723
EQUATIONS OF STATE
725
651
652 M Contents of Appendix
Figure D.l Lee-Kesler Simple Fluid Compressibility Factor, 728
Figure D.2 Lee-Kesler Simple Fluid Enthalpy Departure, 729
Figure D.3 Lee-Kesler Simple Fluid Entropy Departure, 730
E FIGURES 73 ~
Figure E.l Temperature-Entropy Diagram for Water, 732
Figure E.2 Pressure-Enthalpy Diagram for Ammonia, 733
Figure E.3 Pressure-Enthalpy Diagram for Oxygen, 734
Figure E.4 Psychrometric Chart, 735
F ENGLISH UNIT TABLES 73
Table F.l Critical Constants (English Units), 738
Table F.2 Properties of Selected Solids at 77 F, 739
Table F.3 Properties of Some Liquids at 77 F, 739
Table F.4 Properties of Various Ideal Gases at 77 F, 1 arm (English Units), 740
Table F.5 Ideal-Gas Properties of Air (English Units), Standard Entropy at
1 atm = 101.325 kPa = 14.696 lbf/in. 2 , 741
Table F.6 Ideal-Gas Properties of Various Substances (English Units), Entropies
at 1 atm Pressure, 742
Table F.7 Thermodynamic Properties of Water, 748
Table F.8 Thermodynamic Properties of Ammonia, 760
Table F.9 Thermodynamic Properties of R-22, 766
Table F.10 Thermodynamic Properties of R- 13 4a, 772
Table F.l 1 Enthalpy of Formation, and Absolute Entropy of Various Substances
at 77 F, 1 atm Pressure, 778
Appendix A
SI Units: Single-
State properties
Table A.l
Conversion Factors
Area (A)
1 mm 2 = 1.0 X 10" s m 2
1cm 2 = 1.0 X l(r 4 m 2 = 0.1550iiL 2
Im 2 = 10.7639 ft 2
L ft 2 = 144 in. 2
lin. 2 = 6.4516cm 2 = 6.4516 X 10" 4 m 2
1 ft 2 = 0.092 903 m 2
Conductivity (k)
1 W/m-K = 1 J/s-m-K
= 0.577 789 Bhi/h-ft-°R
1 Btu/h-ft-R = 1.730 735 W/m-K
Density {p)
1 kg/m 3 = 0.06242797 lbm/ft 3
1 g/cm 3 = I000kg/iii 3
1 g/cm 3 = 1 kg/L
1 lbm/ft 3 = 16.018 46 kg/m 3
Energy (£, U)
1 J =1 N-m = 1 kg-m 2 /s 2
1 J = 0.737 562 lbf-ft
leal (Int.) = 4.1868 J
1 erg = 1.0 X 10~ 7 J
1 eV = 1.602 177 33 X 10~ 19 J
1 lbf-ft = 1.355 818 J
= 1.28507 X 10~ 3 Btu
1 BtuQnt.) = 1.055 056 kJ
= 778.1693 lbf-ft
Force (F)
1 N = 0.224809 ]bf
1 kp = 9.80665 N (1 kgf)
1 Ibf = 4.448 222 N
Gravitation
g = 9.80665 m/s 2
g = 32.17405 ft/s 2
Heat capacity (C p , C Bi C), specific entropy (s)
1 kJ/kg-K = 0.238 846 Bta/lbm-°R
1 Btu/lbm-°R = 4.1868 kJ/kg-K
Heat flux (per unit area)
1 W/m 2 = 0.316 998 Btu/h-ft 2
1 Btu/h-ft 2 = 3.15459 W/m 2
653
654 £3 appendix A Si Units: single-state properties
TABLE A.l {continued)
Conversion Factors
Heat-transfer coefficient (/i)
1 W/m 2 -K = 0.176 11 Btu/h-fP^R
l Btu/h-rr- K — 5,67826 W/m -K
Length (I)
1 nun = 0.001 m = 0.1 cm
l ft = 12 in.
1 cm = 0.01 m = 10 mm = 0.3970 in.
1 in. = 2.54 cm = 0.0254 m
1 m = 3.28084 ft = 39.370 in.
1 ft = 0.3048 m
1 km = 0.621 371 mi
lmi= 1.609344 km
1 mi = 1609.3 m (US statute)
1 yd = 0.9144 m
Mass (//()
1 kg = 2.204 623 Ibm
1 Ibm = 0.453 592 Kg
1 tonne = 1000 kg
1 slug = 14.5939 kg
1 grain = 6.47989 X 10 _5 kg
1 ton = 2000 Ibm
Moment (torque, T)
1 N-m = 0.737 562 Ibf-ft
1 lbf-ft = 1.355 8I8N-m
Momentum (mV)
1 kg-m/s = 7.232 94 lbm-ft/s
1 lbm-ft/s = 0.138 256 kg-m/s
= 0.224809 lbf-s
Power (Q, )V)
1 W = IJ/s = 1 N-m/s
1 lbf-ft/s = 1.355 818 W
= 0.737 562 lbf-ft/s
= 4.626 24 Btu/h
I kW = 3412.14 Btu/h
lBtu/s = 1.055 056 kW
1 hp (metric) = 0.735 499 kW
1 hp (UK) = 0.7457 kW
= 550 lbf-ft/s
= 2544.43 Btu/h
1 ton of
1 ton of
refrigeration = 3.516 85 kW
refrigeration = 12 000 Btu/h
Pressure (F)
1 Pa =1 N/m 2 = 1 kg/m-s 2
1 Ibf/in 2 = 6.894 757 kPa
1 bar = 1.0 X 10 5 Pa = 100 kPa
lata = 101.325 kPa
1 atm = 14.695 94 lbf/in. 2
= 1.01325 bar
= 29.921 in. Hg [32°F]
= 760 mm Hg [0°C]
= 33.899 5 ft H 2 [4°C]
= 10.332 56 mH 2 0[4°C]
1 ton =1 mm Hg [0°C]
1 mmHg [0°C] = 0.133 322 kPa
lin.Hg[0°C] =0.49115 lbf/in. 2
1 m H 2 [4°C] = 9.806 38 kPa
1 in. H 2 [4°C] = 0.036126 lbf/in 2
Specific energy (e, u)
1 kj/kg = 0.42992 Btu/lbm
1 Btu/lbm = 2.326 kJ/kg
= 334.55 lbf-ft/lbm
1 lbf-ft/lbm = 2.98907 X 10~ 3 kJ/kg
= 1.28507 X 10 -3 Btu/lbm
Appendix A SI Units: Single-State Properties ® 655
TABLE A.l {continued)
Conversion Factors
Specific kinetic energy {; V 2 )
I m 2 /s 2 = 0.001 kj/kg
1 kJ/kg = 1000 m 2 /s 2
1 fr7s 2 = 3.9941 X 10 -5 Bhi/lbm
1 Btu/lbm = 25037 frVs 2
Specific potential energy (Zg)
1 m sstd ~~ y.oiroto x ju KJ/kg
.= 4.21607 X 10~ 3 Btu/lbm
lft-&u= l.Olbf-ft/ibm
= 0.001285 Btu/lbm
= 0.002989 U/kg
Specific volume (v)
1 cm 3 /g = 0.001 m 3 /kg
1 cm 3 /g = 1 17kg
lm 3 /kg = 16.018 46 ftVlbm
1 ftVlbm = 0.062 428 m 3 /kg
Temperature (?)
1 K = 1*C = 1.8 R= 1.8 F
TC = TK — 273.15
= (TF - 32)/1.8
TK = TR/I.8
1 R = (5/9) K
TF = TR — 459.67
= 1.8 TC + 32
TR = 1.8 TK
Universal Gas Constant
R-=N k = 8.31451 kJ/kmoI-K
= 1.98589 kcal/kmol-K
= 82.0578 atm-L/kmol-K
R = 1.98589 Btu/lbmol-R
= 1545.36 lbf-ft/lbmol-R
= 0.73024 atm-ft 3 /lbmol-R
= 10.7317 (lbtfin.^tf/lbmol-R
Velcoity (V)
1 m/s = 3.6 km/h
= 3.28084 ft/s
= 2.23694 mi/h
1 km/h = 0.27778 m/s
= 0.91134 ft/s
= 0.62137 mi/h
lft/s = 0.681818 mi/h
= 0.3048 m/s
= 1.09728 km/h
1 mi/h = 1.46667 ft/s
= 0.44704 m/s
= 1.609344 km/h
Volume (K)
lm' = 35.3147 ft 3
1 L = 1 dm 3 = 0.001 m 3
1 Gal (US) = 3.785 412 L
= 3.785412 X*10~ 3 m 3
1 ft 3 = 2.831 685 X 10- 2 m 3
1 in. 3 = 1.6387 X 10" 5 m 3
1 Gal (UK) = 4.546 090 L
1 Gal (US) = 231.00 in. 3
656 H APPENDIX A SI UNITS: SINGLE- STATE PROPERTIES
TABLE A.2
Critical Constants
Substance
Formula
Molec.
Mass
Temp.
(K)
Press.
(MPa)
Vol.
(m 3 /kg)
Ammonia
Argon
Bromine
Carbon dioxide
Carbon monoxide
Chlorine
Fluorine
Helium
Hydrogen (normal)
Krypton
Neon
Nitric oxide
Nitrogen
Nitrogen dioxide
Nitrous oxide
Oxygen
Sulfur dioxide
Water
Xenon
Acetylene
Benzene
«-Butane
Chlorodifluoroethane (142b)
Chlorodifluoromethane (22)
Dichloroftuoroethane (141)
Dichlorotrifluoroethane (123)
Difiuoroethane (152a) '
Difiuoromethane (32)
Ethane
Ethyl alcohol
Ethylene
rt-Heptane
«-Hexane
Methane
Methyl alcohol
tt-Octane
Pentafluoroethane (125)
M-Pentane
Propane
Propene
Tetrafluoroethane (134a)
NH 3
Ar
Br 2
C0 2
CO
a 2
He
H 2
Kr
Ne
NO
N 2
N0 2
N 2
o 2
so 2
H 2
Xe
C 2 H 2
C 6 H 6
C 4 H l0
CH 3 CC1F 2
CHC1F 2
CH 3 CC1 2 F
CHC1 2 CF 3
CHF 2 CHj
CF 2 H 2
CjH 6
C2H5OH
Q2H4
-C7FE16
QH H
CH 4
CH3OH
C B H lg
CHF 2 CF 3
C 5 H l2
CjH s
CF 3 CH 2 F
17.031
39.948
159.808
44.01
; 28.01
70.906
37.997
4.003
2.016
83.80
20,183
30.006
28.013
46.006
44.013
31.999
64.063
18.015
131.30
26.038
78.114
58.124
100.495
86.469
116.95
152.93
66.05
52.024
30.070
46.069
28.054
100.205
86.178
16.043
32.042
114.232
120.022
72.151
44.094
42.081
102.03
405.5
150.8
588
304.1
: 132.9
416.9
144.3
5.19
33.2
209.4
44.4
180
126.2
431
309.6
154.6
430.8
647.3
289.7
308.3
562.2
425.2
410.3
369.3
481.5
456.9
386.4
351.3
305.4
513.9
282.4
540.3
507.5
190.4
512.6
568.8
339.2
469.7
369.8
364.9
374.2
11.35
4.87
10.30
7.38
3.50
7.98
5.22
0.227
1.30
5.50
2.76
6.48
3.39
10.1
7.24
5.04
7.88
22.12
5.84
6.14
4.89
3.80
4.25
4.97
4.54
3.66
4.52
5.78
4.88
6.14
5.04
2.74
3.01
4.60
8.09
2.49
3.62
3.37
4.25
4.60
4.06
0.00426
0.00188
0.000796
0.00212
0.00333
0.00175
0.00174
0.0143
0.0323
0.00109
0.00206
0.00192
0.0032
0.00365
0.00221
0.00229
0.00191
0.00317
0.000902
0.00433
0.00332
0.00439
0.00230
0.00191
0.00215
0.00182
0.00272
0.00236
0.00493
0.00363
0.00465
0.00431
0.00429
0.00615
0.00368
0.00431
. 0.00176
0.00421
0.00454
0.00430
0.00197
APPENDIX A SI UNITS: StNGLE- STATE PROPERTIES M 657
TABLE A.3
Table a.4
Properties of Selected Solids at 25°C
Properties of Some Liquids at 25°C*
P
C p
P
Substance
(kg/m 3 )
(KJ/Kg-K)
Substance
(kg/m 3 )
(kJ/kg-K)
Asphalt
2120
0.92
AironoiiiR
A fid
Brick, common
1800
0.84
879
1 7?
Carbon j diamond
3250
0.51
Hul£lH6
556
2A1
Carbon, graphite
2000-2500
0.61
CC1,
1584
0.83
Coal ■.'■:■;■-'/;' /■ 7 ■
1200-1500
1.26
co 2
680
2.9
Concrete
2200
0.8S
Ethanol
783
2.46
Glass, plate
2500
0.80
Gasoline
750
2.08
Glass, wool
200
0.66
Glycerine
1260
2.42
Granite
2750
0.89
Kerosene
815
2.0
Ice (0°C) ■
917
2.04
Methanol
787
2.55
Paper
700
1.2
H-octane
692
2.23
Plexiglass
1180
1.44
Oil engine
885
1.9
Polystyrene
920
2.3
Oil light
910
1.8
Polyvinyl chloride
1380
0.96
Propane
510
2.54
Rubber, soft
1100
1.67
R-12
1310
0.97
Sand, dry
1500
0.8
R-22
1190
1.26
Salt, rock
2100-2500
0.92
R-32
961
1.94
Silicon
2330
0.70
R-125
1191
1.41
Snow, firm
560
2.1
R-i34a
1206
1.43
Wood, hard (oak)
720
1.26
Water
997
4.18
Wood, soft (pine)
510
1.38
Liquid metals
Wool
100
1.72
Bismuth, Bi
10040
0.14
Metals
Lead, Pb
10660
0.16
Aluminum
2700
0.90
Mercury, Hg
13580
0.14
Brass, 60-40
8400
0.38
NaK (56/44)
887
1.13
Copper, commercial
8300
0.42
Potassium, K
828
0.81
Gold
19300
0.13
Sodium, Na
929
1.38
Iron, cast
. . 7272
0.42
Tin, Sn
6950
0.24
Iron, 304 St Steel
7820
0.46
Zinc, Zn
6570
0.50
Lead
11340
0.13
Magnesium, 2% Mn
1778
1.00
*Orr ra!l if higher.
Nickel, 10% Cr
S666
0.44
Silver, 99.9% Ag
10524
0.24
Sodium
971
1.21
Tin
7304
0.22
Tungsten
19300
0.13
Zinc
7144
0,39
658 M APPENDIX A SI UNITS: SINGLE-STATE PROPERTIES
Table a.5
Properties of Various Ideal Gases at 25°C, 100 kPa* (SI Units)
Chemical
Molecular
R
P
c
Gas
1Y1H5S
fkJ/ke-Kl
fke/m 3 )
fkJ/kg-K)
(kJ/kg-K)
c v
Steam
H 2
18.015
0.4615
0.0231
1.872
1.410
1.327
Acetylene
C 2 H 2
26.038
0.3193
1.05
1.699
1.380
1.231
Air
—
28.97
0.287
1.169
1.004
0.717
1.400
Ammonia :
NH 3
17.031
0.4882
0.694
2.130
1.642
1.297
Argon
■ Ar
39.948
0.2081
1.613
0.520
0.312 -
1.667
Butane
C 4 H ]0
58.124
0.1430
2.407
1.716
1.573
1.091
Carbon dioxide
CO,
44.01
0.1889
1.775
0.842
0.653
1.289
Carbon monoxide
CO
28.01
0.2968
1.13
1.041
0.744
1.399
Ethane
30.07
0.2765
1.222
1.766
1.490
1.186
Ethanol
C ; H 5 OM
46.069 ■;.
0.1805
1.883
1.427
1.246
1.145
Ethylene
c 2 o 4
29.054
: 0.2964
1.138
1.548 V
1.252
1.237
Helium :
He
V 4.003
: 2.0771
; 0.1615
5.193
3.11b
Hydrogen
H 2
2.016
4.1243
0.0813
14.209
10.085
1.409
Methane
CH 4
16.043
0.5183
0.648
2.254
1.736
1.299
Methanol
CH 3 OH
32.042
0.2595
1.31
1.405
1.146
1.227
Neon
Ne :
: 20.183
0.4120
0.814
1.03
0.618
1.667
Nitric oxide
NO
30.006
0.2771
1,21
0.993
0.716
} 1.387
Nitrogen
28.013
0.2968
1.13
: . 1.042
0.745
1.400
Nitrous oxide
N 2
44.013
0.1889
1.775
0.879
0.690
1.274
H-octane
C^Hig
1 14.23
0.07279
0.092
1.711
1.638
1.044
Oxygen
o 2
31.999
0.2598
1.292
0.922
0.662
1.393
Propane
; 44.094
0.1886
1.808
1.679
1.490
1.126
R-12
CC1 2 F 2
120.914
0.06876
4.98
0.616
0.547
1.126
R-22 ...
CIlClFj
86.469
0.09616
3.54
0.658
0.562
. 1.171
R-32
CF 2 H 2
52.024
0.1598
2.125
0.822
0.662
1.242
R-125
CHF 2 CF 3
120.022
0.06927
4.918
0.791
0.721
1.097
R-134a
CF 3 CH 2 F
102.03
0.08149
4.20
0.852
0.771
1.106
Sulfur dioxide
so 2
64.059
0.1298
2.618
0.624
0.494
1.263
Sulfur trioxide
■■. so 3 -.-v.
■ 80.053
: 0.10386
3.272
0.635
■ 0.531
1.196
*Or saturation pressure if it is less than 100 kPa.
APPENDIX A SI UNITS; SINGLE- STATE PROPERTIES H 659
Table A.6
Constant-Pressure Specific Heats of Various Ideal Gases^
C P0
= C + C$ + C 2 6
+ c 3 e 3
(kj/kg K)
8 = 7XKelvin)/1000
Gas
Formula
c
c,
Steam
H 2
1.79
0.107
0.586
-0.20
Acetylene
1.03
2.91
-1.92
0.54
Air
1.05
-0.365
0.85
-0.39
Ammonia
NH 3
1.60 ;
'" 1.4
•0.7
Argon " ■ : ■
. : At .
0.52 : ;
■■
: - ;
Butane
C^H]^
0.163
5.70
-1.906
-0.049
Carbon dioxide
co 2
0.45
1.67
-1.27
0.39
Carbon monoxide
CO
1.10
-0.46
1.0
-0.454
Ethane
C 2 Hg
0.1S
5.92
-2.31
0.29
Ethanol
Q,H 5 OH
0.2
-4.65
: -1.82
0.03
Ethylene
C2H4
1.36
5.58
-3.0
0.63
Helium
He
5.193
Hydrogen
1 1 AC
4.6
-6.85
3.79
Methane
CH 4
1.2
3.25
0.75
-0.71
Methanol
CH3OH
0.66
2.21
0.81
-0.89
Neon
Ne
1.03 '■,
Nitric oxide :
NO
0.98
-0.031
0.325
-0.14
Nitrogen
N 2
1.11
-0.48
0.96
-0.42
Nitrous oxide
N 3
0.49
1.65
-1.31
0.42
»-octane
CgH ]3
-0.053
6.75
-3.67
0.775
Oxygen
o 2
0.88
-0.0001
0.54
-0.33
Propane
-0.096
6.95
-3.6
0.73
R-12*
CC1 2 F 2
0.26
1.47
-1.25
0.36
R-22*
CHC1F 2
0.2
1.87
-1.35
0.35
R-32*
0.227
2.27
-0.93
0.041
R-125*
CHF 2 CF 3
0.305
1.68
-0.284
R-134a*
CF 3 CH 2 F
0.165
2.81
-2.23
1.11
Sulfur dioxide
S0 2
0.37
1.05
-0.77
0.21
Sulfur trioxide
S0 3
0.24
1.7
-1.5
0.46
Approximate forms valid from 250 K to 1200 K.
♦Formula limited to maximum 500 K.
660 ^ APPENDIX A SI UNITS: SINGLE- STATE PROPERTIES
Table A7.1
JdeaUGas Properties of Air, Standard Entropy at 0.1-MPa (1-bar) Pressure
T
u
h
4
T
It
s r
(K)
(W/kg)
<kJ/kg)
(kJ/kg-K)
(K)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
200
142.77
t 7
ZwA I
O-4OZ0U
1 1 Of)
1 IUU
845.45
1161.18
g ,24449
220
157.07
ZZU.ZZ
U.JJSIZ
t 1 5n
1 1 JU
889.21
1219.30
8.29616
240
171.3a
1Af\ 77
Z4U.Z /
0.04j j J
17flft
I ZUU
n-t-i -17
jj j>. j> /
1277.81
8.34596
260
185.70
zoU.iz
O. /ZjOZ
19S0.
1336 68
8.39402
2S0
*}f\{\ A7
/.UU.U2
7©a 10
/It 7QQQfi
nnn
1 _J uu
1395.89
8,44046
290
7 (V7 1 Ct
O.OJjZl
1 i5n
1 067 Q4
1455.43
8,48539
298.15
213.04
2ye.o2
O.ouJUj
1/iriA
iHUU
1 1 1 1 «
1515.27
8.52891
300
214.36
iaa An
300.4/
D.oOyZD
140U
1 1 90
i 1 Jj.iU
1575.40
8.57111
320
22b. ii
320.58
A CJ1/11 1
o.yj4i j
1 3UU
1 705 ?5
8.61208
340
243.11
J4U. /U
£ QCK1 s
i 55n
1JJU
1251.55
1696.45
8.65185
360
257.53
7 £A Q/C
/.UjZ/O
lOUU
1 ?QS OS
1757.33
8.69051
380
27i.yy
oQI AC
T 1 ATI <;
i OJU
8.72811
400
286.49
4U1.3U
7 1 ^09^
/.ijyzo
I /uu
not sn
1 jy l >ou
1879.76
8.76472
420
301.04
4zl.5y
7 TACT 1 ;
/.Zuo / J
175n
1 4.-10 Q7
1941.28
8.80039
440
315.64
A A 1 G7
44 1 ,y3
/.ZjOU /
1 Rfin
8.83516
460
330.31
4DZ.34
7 1A1/I7
/.3UI4Z
IojU
9064 Sfi
480
345.04
45Z.K1
7 1MQQ
1581.59
2126 95
8.90219
500 '
359.84
jUj.30
/.jooyz
i Q5n
iyju
1629.47
2189.19
8,93452
520
1HA 71
3 /4. IS
^71 AO
/ ,4Z /JO
zuuu
1677.52
2251.58
8.966U
540
7QA Aft
389. W
j44.oy
/ .4004Z
?n5n
ZUJU
17?5 71
2314.13
8,99699
560
404.74
/T7
565.4/
7 <fl,177
/.JU-4ZZ
7 inn
ZiUU
1 774. Of!
9.02721
580
419.87
586.35
7 ^Af\QA
/.34U84
1S7? Id.
Q 05678 *
600
435.10
607.32
7.57638
2200
1871.16
2502.63
9.08573
620
450.42
628.38
7.61090
2250
1919.91
2565.73
9.11409
640
465.83
649.53
7.64448
2300
1968.79
2628.96
9.14189
660
481.34
670.78
7.67717
2350
2017.79
2692.31
9.16913
680
496.94
692.12
7.70903
2400
2066.91
2755.78
9.19586
700
512.64
713.56
7.74010
2450
2116.14
2819.37
9.22208
720
528.44
735.10
7.77044
2500
2165.48
2883.06
9.24781
740
544.33
756.73
7.80008
2550
2214.93
2946.86
9.27308
760
560.32
778.46
7.82905
2600
2264.48
3010.76
9.29790
780
576.40
800.28-
7.85740
2650
2314.13
3074.77
9.32228
800
592.58
822.20
7.88514
2700
2363.88
3138.87
9.34625
850
633.42
877.40
7.95207
2750
2413.73
3203.06
9.36980
900
674.82
933.15
8.01581
2800
2463.66
3267.35
9.39297
950
716.76
989.44
8.07667
2850
2513.69
3331.73
9.41576
1000
759.19
1046.22
8.13493
2900
2563.80
3396.19
9.43818
1050
802.10
1103.48
8.19081
2950
2613.99
3460.73
9.46025
1100
845.45
1161.18
8.24449
3000
2664.27
3525.36
9.48198
APPENDIX A SI UNITS: SINGLE-STATE PROPERTIES ffl 661
TABLE A7.2
The hentropic Relative Pressure and Relative Volume Functions
T[K]
P,
v r
T[K]
P r
200
0.2703
493.47
700
23.160
20.155
1900
1327.5
0.95445
220
0.3770
389.15
720
25.742
18.652
1950
1485.8
0.87521
240
0.5109
313.27
740
28.542
17.289
2000
1658.6
0.80410
260
0.6757
256.58
760
31.573
16.052
2050
1847.1
0.74012
280
0.8756
213.26 :
780
34.851
14.925
2100
2052.1
0.68242
290
0.9899
195.36
800
38.388
13.897
2150
2274.8
0.63027
298.15
1.0907
182.29
850
48.468
1 1.695
2200
2516.2
0.58305
300
1.1146
179.49
900
60.520
9.9169
2250
2777.5
0.54020
320
1.3972
152.73
950
74.815
8.4677
2300
3059.9
0.50124
340
1.7281
131.20
1000
91.651
7.2760
2350
3364.6
0.46576
360
2.1 123
113.65
1050
111.35
6.2885
2400
3693.0
0.43338
380
2.5548
99.188
1100
134.25
5.4641
2450
4046.2
0.40378
400
3.0612
87.137
1150
160.73
4.7714
2500
4425.8
0.37669
420
3.6373
77.003
1200
191.17
4.1859
2550
4833.0
0.35185
440
4.2892
68.409
1250
226.02
3.6880
2600
5269.5
0.32903
460
5.0233
61.066
1300
265.72.
3.2626
2650
5736.7
0.30805
480
5.8466
54.748
1350
310.74
2.8971
2700
6236.2
0.28872
500
6.7663
49.278
1400
361.62
2.5817
2750
6769.7
0.27089
520
7.7900
44.514
1450
418.89
2.3083
2800
7338.7
0.25443
540
8.9257
40.344
1500
483.16
2.0703
2850
7945.1
0.23921
560
10.182
36.676
1550
554.96
1.8625
2900
8590.7
0.22511
580
11.568
33.436
1600
634.97
1.6804
2950
9277.2
0.21205
600
13.092
30.561
1650
723.86
1.52007
3000
10007.
0.19992
620
14.766
28.001
1700
822.33
1.37858
640
16.598
25.713
1750
931.14
1.25330
660
18.600
23.662
1800
1051.05
1.14204
680
20.784
21.818
1850
1182.9
1.04294
700
23.160
20.155
1900
1327.5
0.95445
The relative pressure and relative volume are temperature functions calculated with two
scaling constants A u A 2 .
P r = exp[s° T /R - Ad; u r = A 2 TlP r
such that for an isentropic process (s l — s 2 )
where the near equalities are for the constant heat capacity approximation.
662 B APPENDIX A SI UNITS: SINGLE-STATE PROPERTIES
TABLE A.8
Ideal-Gas Properties of Various Substances, Entropies at OA-MPa (1-bar) Pressure, Mass Basis
NITROGEN, DIATOMIC (N 2 ) OXYGON, DIATOMIC (Oj)
R = 0.2968 kJ/kg-K R = 0.2598 kJ/kg-K
M= 28.013 M= 31.999
T
it
h
ft
a
ft
s r
( K )
(kJ/kg)
(KJ/Kg-JVJ
(kJ/kg-K)
200
148.39
207.75
6.4250
129.84
181.81
6.0466
250
185.50
259.70
6.6568
162.41
227.37
6.2499
300
222.63
311.67
6.8463
195.20
273.15
6.4168
350
259.80
363.68
7.0067 ... ■;
228.37
319.31
6.5590
400
297.09
415.81
7.1459 :
262.10
366.03
6.6838
450
334.57 ;
468.13
7.2692 ■
296.52
; 413.45
6.7954
500
372.35
520.75
7.3800
331.72
461.63
6.8969
550
410.52
573.76
7.481 1
367.70
510.61
6.9903
600
449.16
627.24
7.5741
404.46
560.36
7.0768
650
488.34
681.26
7.6606
441.97
610.86
7.1577
700
528.09
735.86
7.7415
480.18
662.06
7.2336
750
568.45
791.05
7.8176
519.02
713.90
7.3051
800
609.41
846.85
7.8897
558.46
766.33
7.3728
850
650.98
903.26
7.9581
598.44
819.30
7.4370
900
693.13
960.25
8.0232
638.90
872.75
7.4981
950
735.85
1017.81
8.0855
679.80
926.65
7.5564
1000
779.11
1075.91
8.1451
721.11
980.95
7.6121
1100
867.14
1193.62
8.2572
804.80
1090.62
7.7166
1200
957.00
1313.16
8.3612
889.72
1201.53
7.8131
1300
1048.46
1434.31
8.4582
975.72
1313.51
7.9027
1400
1141.35
1556.87
8.5490
1062.67
1426.44
7.9864
1500
1235.50
1680.70
8.6345
1150.48
1540.23
8.0649
1600
1330.72
1805.60
8.7151
1239.10
1654.83
8.1389
1700
1426.89
1931.45
8.7914
1328.49
1770.21
8.2088
1800
1523.90
2058.15
8.8638
1418.63
1886.33
8.2752
1900
1621.66
2185.58
8.9327
1509.50
2003.19
8.3384
2000
1720.07
2313,68
8.9984
1601.10
2120.77
8.3987
2100
1819.08
2442.36
9.0612
1693.41
2239.07
8.4564
2200
1918.62
2571.58
9.1213
1786.44
2358.08
8.5117
2300
2018.63
2701.28'
9.1789
; 1880.17
2477.79
8.5650
2400
2119.08
2831.41
9.2343
1974.60
2598.20
8.6162
2500
2219.93
2961.93
9.2876
2069.71
2719.30
8.6656
2600
2321.13
3092.81
9.3389
2165.50
2841.07
8.7134
2700
2422.66
3224.03
9.3884
2261.94
2963.49
8.7596
2800
2524.50
3355.54
9.4363
2359.01
• 3086.55
* 8.8044
2900
2626.62
3487.34
9.4825
2546.70
3210.22
8.8478
3000
2729.00
3619.41
9.5273
2554.97
3334.48
8.8899
Appendix A SI Units: Single-State Properties H 663
TABLE A.8 (continued)
Ideal-Gas Properties of Various Substances, Entropies at 0.1-M Pa (1-bar) Pressure, Mass Basis
CARBON DIOXIDE (COJ WATER (H 2 0)
if = 0.1889 kJ/kg-K R = 0.4615 kJ/kg-K
M = 44.010
Af= 18.015
T
it
A
«
h
(K)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
200
97.49
135.28
4.5439
276.38
368.69
9.7412
250
126.21
173.44
4.7139
345.98
461.36
10.1547
300
157.70
214.38
4.8631
415.87
554.32
10.4936
350.
.; . 191.78 :';
257.90
4.9972
486.37
647.90
10.7821
400
228.19
303.76
5.1196
557.79
742.40
11.0345
450
266.69
351.70 :.;
5.2325
630.40
838.09
11.2600
500
307.06
401.52
5.3375
704.36
935.12
11.4644
550
349.12
453.03
5.4356
779.79
1033.63
11.6522
600
392.72
506.07
5.5279
856.75
1133.67
11.8263
650
437.71 ■
; 560.51
5.6151
935.31
1235.30
11.9890
700
483.97
,\ 616.22
5.6976
: 1015.49
1338.56
12.1421
750
531.40
673.09
5.7761 \
1097.35 :
1443.49
12.2868
800
579.89
731.02
5.8508
1180.90
1550.13
12.4244
850
629.35
789.93
5.9223
1266.19
1658.49
12.5558
900
676.69
849.72
5.9906
1353.23
1768.60
12.6817
950
730.85
910.33
6.0561
■ 1442.03
1880.48
12.8026
1000
782.75
971.67 /
6.1190
1532.61
1994.13
12.9192
noo
888.55
1096.36
6.2379
1719.05
2226.73
13.1408 -
1200
996.64
1223.34
6.3483
1912.42
2466.25
13.3492
1300
1106.68
1352.28
6.4515
2112.47
2712.46
13.5462
1400
1218.38
1482.87
6.5483
2318.89
2965.03
13.7334
1500
1331.50
1614.88
6.6394
2531.28
3223.57
13.9117
1600
1445.85
1748.12
6.7254
2749.24
3487.69
14.0822
.1700
1561.26
1882.43
6.8068
2972.35
3756.95
14.2454
1800
1677.61
2017.67
6.8841
3200.17
4030.92
14.4020
1900
1794.78
2153.73
6.9577
3432.28
4309.18
14.5524
2000
1912.67
2290.51
7.0278
3668.24
4591.30
14.6971
2100
2031.21
2427.95
7.0949 :
3908.08
4877.29
14.8366
2200
2150.34 ;
2565.97 ' :
7.1591 ;
4151.28
5166.64
14.9712
2300
2270.00
2704.52
7.2206
4397.56
5459.08
15.1012
2400
2390.14
2843.55
7.2798
4646.71
5754.37
15.2269
2500
2510.74
2983.04
7.3368
4898.49
6052.31
15.3485
2600
2631.73
3122.93
7.3917
5152.73
6352.70
15.4663
2700
2753.10 :
3263.19
7.4446
5409.24
6655.36
15.5805
2800
2874.81
3403.79
7.4957 ■ : : :
5667.86
6960.13
15.6914
2900
2996.84
3544.71 "
7.5452 V
5928.44
7266.87
15.7990
3000
3119.18
3685.95
7.5931
6190.86
7575.44
15.9036
664 H appendix A Si units: Single-state properties
TABLE A.9
Ideal-Gas Properties of Various Substances (SI Units), Entropies at 0,1-MPa (1-bar) Pressure,
Mole Basis
Nitrogen, diatomic (Nj) nitrogen, monatomic (N)
Jjfen = o kJ/kmol h/,m = 472 680 kJ/kmol
M= 28.013 M= 14.007
7
(It - h° m )
~4
* T
K
kJ/kmol
kJ/kmol K
kJ/kmol
kj/kmol
—8670
U
— £1 Q7
O ly l
o
100
— 5/06
I jy.olZ
— A\ 1 Q
1 10 ^93
i ju. -jy j
200
—2857
145.001
298
lyi.ouy
u
300
■ ■' XA
54
lyL.ioy
10
JO
1 <;3 479
400
2y/I
Zw.lol
.. ^ii /
159.409
500
5911
2U0. /4U
41 Q£
164.047
600
0094
2 12.1 / /
700
1 1937
21 0.605
171.041
800
15046
071 A1 £
221.U10
173.816
900
18223
llH.loi
1000
1 1 A CI
21463
22S,I/I
1 J. 'IRQ
178 455
I / O.H.J.J
1 100
j!j l.j 14
1 ooo /
180.436
1200
2m uy
lR74fi
1 6 /HO
182.244
1300
31503
TJiT 0/11
230. y4j
ZUOZ;>
1R1 QftR
1400
- 34yjo
zzyuj
1SS 44R
1500
38405
1/1 1 001
241.8ol
24yoz
1 9f, SSI
1600
41904
244.139
27060
: 188.224
1700
45430
246.276
29139
189.484
1800
48979
248.304
31218
190.672
1900
52549
250.234
33296
191.796
2000
56137
252.075
35375
192.863
2200
63362
255,518
39534
194.845
.2400
70640
258.684
43695
196.655
2600
77963
261.615
47860
198.322
2800
85323
264.342
52033
199.868
3000
92715
266.892
56218
201.311
3200
100134
269.286
60420
202.667
3400
107577 i
■ 271.542
64646
203.948
3600
115042
273.675
68902
: 205.164
3800
122526
275.698
73194
206.325
4000
130027
277.622
77532
207.437
4400
145078
281.209
86367
209.542
4800
160188
284.495
95457 ■
211.519
5200
175352
287.530
104843
213.397
5600
190572
290.349
114550
215.195
6000
205848
292.984
124590
216.926
APPENDIX A SI UNITS: S INGLE-STATE PROPERTIES ffl 665
TABLE A.9 (continued)
Ideal-Gas Properties of Various Substances (SI Units), Entropies at OJ-MPa (1-bar) Pressure,
Mole Basis
Oxygen, diatomic (Oj) oxygen, monatomic (O)
hj m = kJ/kmol h}^ = 249 1 70 kJ/kmol
M = 31.999 M= 16.00
T
A
(h - h% s )
s r
K
kJ/kmol
kJ/kmol K
kJ/kmoi
kJ/kmol
-8683
-6725
100
-5777
173.308
-4518
135.947
200
-2868
193.483
-2186
152.153
298
205.148
161.059
300
V- . 54 ■
205.329
41
161.194
400
3027
213.873
2207
167.431
500
6086
220.693
4343
172.198
600
9245
226.450
6462
176.060
700
12499
231.465
8570
179.310
800
15836
235.920
10671
182.116
900
19241
239.931
12767
184.585
1000
22703
243.579
14860
186.790
1100
26212
246.923
16950
188.783
1200
29761
250.011
19039
190.600
1300
33345
252.878
21126
192.270
1400
36958
255,556
23212
193.816
1500
40600
258.068
25296
195.254
1600
44267
260.434
27381
196.599
1700
47959
262.673
29464
197.862
1800
51674
264.797
31547
199.053
1900
55414
266.819
33630
200.179
2000
59176
268.748
35713
201.247
2200
66770
272.366
39878
203.232
2400
74453
275.708
44045
205.045
2600
82225
278.818
48216
206.714
2800
90080
281.729
52391
208,262
3000
98013
284.466
56574.
209.705
3200
106022
" : 287.050
.60767
211.058
3400
114101
289.499
64971
212.332
3600
122245
291.826
69190
213.538
3800
130447
294.043
73424
214.682
4000
138705
296.161
77675
215.773
4400
155374
300.133
86234
217.812
4800
172240 ;
303.801
94873
: 219.691
5200
189312
307.217
103592
221.435
5600
206618
310.423
112391
223.066
6000
224210
313.457
121264
224.597
666 M APPENDIX A SI UNITS: SINGLE- STATE PROPERTIES
TABLE A.9 (continued)
Ideal-Gas Properties of Various Substances (SI Units), Entropies at O.I-MPa (1-bar) Pressure,
Mole Basis
Carbon Dioxide (COJ
CARBON MONOXIDE (CO)
'7,298 —
-393 522 kJ/kmol
110 527 kJ/kmol
M= 44.01
M
= 28.01
T
(J, _ To
V" "29S
s r
K
kJ/kmoi
kJ/kmol K
kJ/kmol
kJ/kmol K
u
-9364
-8671
1 nn
-6457
179.010
-5772
165.852
inn
zuu
-3413
199.976
-2860
186.024
213.794
197.651
inn
69
214.024
54
197.831
Ann
4UU
4003
: 225,314
2977
206.240
8305
234.902
5932
212.833
OW
12906
243.284
8942
218.321
i fin
17754
250,752
12021
223.067
Bflfl
ouu
22806
257.496
15174
227,277
yuu
28030
263.646
18397
231.074
i nfin
1UUU
33397
269.299
21686
234.538
1 1 nn
1 1UU
38885
274.528
25031
237.726
nrifi
44473
279.390
28427
240.679
ijW
50148
283.931
31867
243.431
55895
288.190
35343 ■
246.006
1 snn
61705
292.199
38852
248.426
iouu
67569
295.984
42388
250.707
1 /uu
73480
299.567
45948
252.866 '
1800
79432
302.969
49529
254.913
1900
85420
306.207
53128
256.860
2000
91439
309.294
56743
258.716
2200
103562
315.070
64012
262.182
2400
115779
320.384
71326
265.361
2600
128074
325.307
78679
268.302
2800
140435
329.887
86070
271.044
3000
152853
334.170
93504
273.607
3200
165321
338.194
100962
276.012
3400
177836
341.988
108440
278.279
3600
190394
345.576
115938
280.422
3800
202990
348.981
123454
282.454
4000
215624
352.221
130989
284.387
4400
240992
358.266
146108
287.989
4800
266488
363.812
161285
■291.290
5200
292112
368.939
176510
294,337
5600
317870
373.711
191782
297.167
6000
343782
378.180
207105
299.809
Appendix A SI Units: Single-State Properties H 667
TABLE A.9 (continued)
Ideal-Gas Properties of Various Substances (SI Units), Entropies at 0.1-MPa (1-bar) Pressure,
Mole Basis
WATER (HjO)
HYDROXYL (OH)
1,0 — _
241 826 kj/kmol
Tt% gs = 38 987 kJ/kmol
M
= 18.015
M= 17.007
T
(11 fl 198 )
S T
(T, — TA\
\' 1 "19$}
si
K
kJ/kmoi
kJ/kmol K
kJ/kmol
kJ/kmol
a
u
-9904
-9172
i f\n
1U0
-6617
152.386
-6140
149.591
2i)U
-3282
175.488
-2975
171.592
zya
188.835
183.709
300
62
189.043
55
183.894
Hill)
3450
198.787
3034
192.466
DUU
6922
206.532
5991
199.066
ouu
10499
213.051
8943
204.448
Iw
14190
218.739
11902
209.008
18002
223.826
. 14881
212.984
AAA
21937
228.460
17889
216.526
1 AAA
26000
232.739
20935
219.735
1 i aa
1 1UU
30190
236.732
24024
222.680
1 1AA
34506
240.485
27159
225.408
1300
38941
244.035
30340
227.955
1 ^flA
43491
247.406
33567
230.347
1 <AA
48149
250.620
36838
232.604
1 /TAA
loOU
52907
253.690
40151
234.741
1 *7AA
57757
256.631
43502
236.772
1800
62693
259.452
46890
238.707
1900
67706
262.162
50311
240.556
2000
72788
264.769
53763
242.328
2200
83153
269.706
60751
245.659
2400
93741
274.312
67840
248.743
2600
104520
278.625
75018
251.614
2800
115463
282.680
82268
254,301
3000
126548
286.504
89585
256.825
3200
137756
290.120
96960
259.205
3400
149073
293.550
104388
261.456
3600
160484
296.812
111864
263.592
3800
171981
299.919
119382
265.625
4000
183552
302.887
126940
267.563
4400
206892
308.448
142165
271.191
4800
230456
313.573
157522
274.531
5200
. 254216
318.328
173002
277.629
5600
278161
322.764
188598
280.518
6000
302295
326.926
204309
283.227
668 H Appendix A SI Units: Single-State Properties
TABLE A.9 (continued)
f deal-Gas Properties of Various Substances (SI Units), Entropies at 0.1-MPa (1-bar) Pressure,
Mole Basis
HYDROGEN (Hj)
Hydrogen, monatomic (H)
1,0 —
U KJ/KlllOl
ft/aw ~
= 217 999 kJ/kmol
iVJ —
M= 1.008
T
(A - Ti% s )
s° T
(ft - O s° T
K
kJ/kmoi
KJ/kmol Jv
kJ/kniol
kJ/kmoi K
-8467
-6197
100
-5467
100.727
-4119
92.009
200
-2774
119.410
-2040
106.417
298
130.678
114.716
300
53
130.856
38
114.845
400
2961
139.219
2117
120.825
500
5883
145.738
4196
125.463
600
8799
151.078
6274
129.253
700
11730
155.609
8353
132.457
800
14681
159.554
10431
135.233
900
17657 -
163.060
12510
137.681
1000
20663
166.225
14589
139.871
1100
23704
169.121
16667
141.852
1200
26785
171.798
18746
143.661
1300
29907
174.294
20825
145.324
1400
33073
176.637
22903
146.865
1500
36281
178.849
24982
148.299
1600
39533
180.946
27060
149.640
1700
42826
182.941
29139
150.900
1800
46160
184.846
31218
152.089
1900
49532
186.670
33296
153.212
2000
52942
188.419
35375
154.279
2200
59865
191.719
39532
156.260
2400
66915
194.789.
43689
158.069
2600
74082
197.659
47847
159.732
2800
81355
200.355
52004
161.273
3000
88725
202.898
56161
162.707
3200
96187
205.306
60318
164.048
3400
103736
'207.593
64475
165.308
3600
111367
209.773
68633
166.497
3800
119077
211.856
72790
167.620
4000
126864
213.851
76947
168.687
4400
142658
217.612
85261
170.668
4800
158730
221.109
93576
172.476
5200
175057
224.379
101890
174.140
5600
191607
227.447
110205
175.681
6000
208332
230.322
118519
177.114
Appendix A SI Units: Single-State Properties M 669
TABLE A.9 (continued)
Ideal-Gas Properties of Various Substances (SI Units), Entropies at 0.1-MPa (1-bar) Pressure,
Mole Basis
Nitric Oxide (NO)
fjza = 90 291 kJ/kmol
M = 30.006
Nitrogen dioxide (N0 2 )
Ajc»g = 33 lOOkJ/kmol
M = 46.005
T
K
kJ/kmol
S T
kJ/kmol K
kJ/kmol
kJ/kmol K
o
—9192
— 1(11
V
100
-6073
177.031
200
-2951
198.747
— 14Q*i
99S 8<\9
298
o
210,759
n
V
94ft 014
ZtU.UJH
300
55
210.943
68
94ft 9^1
400
3040
219.529
1Q97
911 149
500
6059
226 263
OU77
96ft 61fi
600
9144
211 886
i 9 < i'> < ;
i £ J J J
9£a 7<;s
700
12308
236.762
177-SO
Z / J.700
800
15548
241 088
22138
9S9 S1 1
900
18858
244.985
9718(1
988 4^(1
1000
22229
248 516
OZ jH*f
9Q1 HSQ
zyj.o&j'
1100
25653
251.799
J / DUO
9Q8 QfJ4
1200
29120
254.816
42946
Iftl 1
JUJ.JJ 1
1300
32626
257 621
481S1
107 276
1400
36164
260 243
53808
1500
39729
262.703
59309
315.715
1600
A1 fl
zoxuiy
64846
319.289
1700
46929
267.208
70414
322.664
1800
50557
269.282
76008
325.861
i Qfin
54201
271.252
81624
328.898
2000
57859
273.128
87259
331.788
2200
65212
276.632
98578
337.182
2400
72606
279.849
109948
342.128
2600
80034
282.822
121358
346.695
2800
87491
285.585
132800
350.934
3000
94973
288.165
144267
354.890
3200
102477
290.587
155756
358.597
3400
11 0000
292.867
167262
362.085
3600
117541
295.022
178783
365.378
3800
125099
297.065
190316
368.495
4000
132671
299.007
201860
371.456
4400
147857
302.626
224973
376.963
4800
163094
305.940
248114
381.997
5200
178377
308.998
271276
386.632
5600
193703
311.838
294455
390.926
6000
209070
314.488
317648
394.926
670 H appendix A SI Units: Single-State Properties
TABLE A. 10
Enthalpy of Formation and Absolute Entropy of Various Substances at 25°C, 100 kPa Pressure
Substance
Formula
Xf
Sri
State
1
h f
kJ/kmol
£0
s f
kJ/kmolK
Water
H 2
18.015
gas
-241 826
188.834
Water
H 2
18.015
liq
-285 830
69.950
Hydrogen peroxide
HA
34.015
gas
-136 106
232.991
Ozone
3
47.998
gas
+ 142 674
238.932
Carbon (graphite)
■ .. c
12.011
solid
5.740
Carbon monoxide
CO
28.011
gas
-110 527
197.653
Carbon dioxide
co 2
44.010
gas
-393 522
213.795
Methane
CH 4
16.043
gas
-74 873
186.251
Acetylene
C2H 2
26.038
gas
+226 731
200.958
Etbene
CjH^
28.054
gas
+52 467
219.330
Ethane
C 2 H 6
30.070
gas
-84 740
229.597
Propene
C 3 H,
42.081
gas
+20 430
267.066
Propane
C 3 H S
44.094
gas
-103 900
269.917
Butane
QH l0
58.124
gas
-126 200
306.647
Pentane
C 3 H 12
72.151
gas
-146 500
348.945
Benzene
78.114
gas
+82 980
269.562
Hexane
86.178
gas
-167 300
387.979
Heptane
100.205
gas
-187 900
427.805
>!-Octane
C g H 13
114.232
gas
-208 600
466.514
H-Octane
C 3 H lg
1 14.232
liq
-250 105
360.575
Methanol
CHjOH
32.042
gas
-201 300
239.709
Methanol
CH 3 OH
32.042
liq
-239 220
126.809
Ethanol
CiHjOH
46.069
gas
-235 000
282.444
Ethanol
CjHjOH
46.069
liq
-277 380
160.554
Ammonia
NH 3
17.031
gas
-45 720
192.572
T-r-Diesel
£-14.4^24.9
198.06
liq
-174 000
525.90
Sulfur
S
32.06
solid
32.056
Sulfur dioxide
so 2
64.059
gas
-296 842
248.212
Sulfur trioxide
so 3
80.058
gas
-395 765
256.769
Nitrogen oxide
N 2
44.013
gas
+82 050
219.957
Nitromethane
CH 3 N0 2
61.04
liq
-113 100
171.80
Appendix A SI Units: Single-State Properties H 671
3
1
+
ft
+
I
O
u
d
+
O
u
o
+
+
o
u
ft
d
u
o
CN
+
ft
o
a
w
ft
K
a.
S
ro r-; o c-;
.— > O CO r~«
I (
Ol N N N
[III
t I I
I 1 I i 1 I I i
1 ] I
oo oi h h
O 00 On Tf
w ^_ h q v)
o! d od vi ci o
VO r-t w
1111,1
Oi M « 0\ V3
N tn in m
VI OJ O t~-;
r-^ to r-°
vi in io n w
co (si co i-c
O OO GO O CO
■I I
CO M CO est est
I I I i I
1 I I I
t— (si tn i-H t~-
I f
in i-i vo o
« M N ^ rn M
[ I I I I I
. . m ^ m «
C\ M " V) ^ * "
vi h; n m m
S vl m d d
1 I I I I
m r- est oo
o M 0\ ^ oo
M N 0\ M w ^
oi m -t \6 oo
t-i ^-1
co >n *o
ro On i*
O -h" O
cn tn *t
T 1 1
5
hoo^oninwvovirt
votnoinfsiooNvtf'O
M O N
(N \0 1 — I
CO CST CN
i I t
t~- ■<+ On t~- in
I I
00 *0 -^J" Tf
fN On 00 r-
—l On On O
CO f-I O O
III!
t— CO On
^ CO oo
oo rq *o vo
*- * co ^d" in
I f
1 ' CO cs)
N vo M
CO co fN
vo vd ON
CO CN
I I
NO On
in est
oo vq
CO* ON
(S — < w _t
co co ^
O CO 60
CO VQ
^ On f-~ ON
<-< CN (N
O O t—
^ vo oi
I I I I I
On VO
I I I
CO cN
1 I
m o t~i co
I I
vo in in
tn^f-— <Ocooovo
t"^ CO* ON o" vo* vo" 06 ^
CO CN
I I
CO CO
VO Ol
On vo
On 00 *0
I I I
I I
3 CO
O (Sj
VO CM
CO CO
I I
too\t~-t~~ooNinw
^ cn vo Tt ^ o o
CO rt CO Ov 00 ON (SJ
oo" tn est o*
CO VO fx) NO
est
I I I I I I I I
M « O0
(fl ft >d N
I I
^ O t"-; CN O VO
I
I I i
0O \f — <
vo tn
CN On
CO Tf
— ' On*
I
^, i9> D f?;y <Nr ^ ,;iOOC < | t~-ON
fN N IO iO M N - ' Oi N <3\ N *0
■ — "o\ONOr~-inoooo\
<n co co
v> co
fill
I
o o cn m >n in
I I
co o o oo
CO O0 CC
CSt On O
ON CO CO
I I
*~i ^o o on tn o
3
to tj- in oo i-i o
cc oo ro cn in o
On in CN! o
CN T 1 1 < T — ( f — i
I I I I I I
oo no in co
I I I I
^ Tf CO ^ On
CO 00 est CO OO
on -^r t-; V~i in
csi— i o b o n ci in
I I I
00 CN
ON CO i — I _
\o to <n cn on
cNCNcstcNcNcococococo^r
tn o in O
■rh in tn io
Appendix B
SI Units: Thermodynamic
Tables
i.
673
TABLE B.l
Thermodynamic Properties of Water
TABLE B. 1.1
Saturated Water „_^_
SPEcrFic Volume, mVkg Internal energy, kJ/kg
Temp,
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
CP)
(kPa)
v f
v fs
v g
"/
u *
0.01
0.6113
0.001000
206.131
206.132
2375.33
2375.33
5
0.8721
0.001000
147.117
147.118
20.97
2361.27
2382.24
10
1.2276
0.001000
106.376
106.377
41.99
2347.16
2389.15
15
1.705
0.001001
77.924
77.925
62.98
2333.06 ,
2396.04
20
2.339 .
0.001002
57.7887
57.7897 ;
83.94
2318.98
2402.91
25
3.169 v
0.001003 ^
43.3583
43.3593 : : :
■ 104.86 :
2304.90
2409.76
30
4.246
0.001004
32.8922
32.8932
125.77
2290.81
2416.58
35
5.628
0.001006
25.2148
25.2158
146.65
2276.71
2423.36
.40
7.384
0.001008
19.5219
19.5229
167.53
2262.57
2430.11
45
9.593 '
0.001010
15.2571
15.2581
188.41
2248.40
2436.81
50
12.350
0.001012
12.0308 .
12.0318
209.30 ;
2234.17
2443.47
55
15.758
0.001015
9.56734
: 9.56835
230.19
2219.89
2450.08
60
19.941
0.001017
7.66969
7.67071
251.09
2205.54
2456.63
65
25.03
0.001020
6.19554
6.19656
272.00
2191.12
2463.12
70
31.19
0.001023
5.04114
5.04217
292.93
2176.62
2469.55
75
38.58
0.001026
4.13021
4.13123
313.87
2162.03
2475.91
80
47.39
0.001029
3.40612
3.40715
334.84
2147.36
2482.19
85
57.83
0.001032
2.82654
2.82757
355.82
2132.58
2488.40
90
70.14
0.001036
2.35953
2.36056
376.82
2117.70
2494.52
95
84.55
0.001040
1.98082
1.98186
397.86
2102.70
2500.56
100
101.3
0.001044
1.67185
1.67290
418.91
2087.58
2506.50
105
120.8
0.001047
1.41831
1.41936
440.00
2072.34
2512.34
no
143.3
0.001052
1.20909
1.21014
461.12
2056.96
2518.09
115
169.1
0.001056
1.03552
1.03658
482.28
2041.44
2523.72
120
.198.5
O001060
OjfStWO
0.89186
503.48
2025.76
2529.24
125
232 ~1
0.001065
0.76953
0.77059
524.72
2009.91
2534.63
130
270.1
0.001070
0.66744
0.66850
546.00
1993.90
2539.90
135
313.0
0.001075
0.58110
0.58217
567.34
1977.69
2545.03
140
361.3
0.001080
0.50777
0.50885
588.72
1961.30
2550.02
145
415.4
0.001085
0.44524
0.44632
610.16
1944.69
2554.86
150
475.9
0.001090
0.39169
0.39278
631.66
1927.87
2559.54
155
543.1
0.001096
0.34566
0.34676
653.23
1910.82
2564.04
160
617.8
0.001102
0.30596
0.30706
674.85
1893.52
2568.37
165
700.5
0.001108
0.27158
0.27269
696.55
: 1875.97
2572.51
170
791.7
0.001114
0.24171
0.24283
718.31
1858.14
2576.46 :
175
S92.0
0.001121
0.21568
0.21680
740.16
1840.03
2580.19
180
1002.2
0.001127
0.19292
0.19405
762.08
1821.62
2583.70
185
1122.7
0.001134
0.17295
0.17409
784.08
1802.90
2586.98
190
1254.4
0.001141
0.15539
0.15654
806.17
1783.84
2590.01
674
Appendix B SI Unsts: thermodynamic Tables H 675
TABLE B.l.l {continued)
Saturated Water
Enthalpy, kJ/kg Entropy, kJ/kg-K
Temp.
Press.
Sat. Liquid
h f
Evap.
Sat. Vapor
K
Sat. Liquid
s f
Evap.
Sat. Vapor
s t
0.01
0.6113
0.00
2501.35
2501.35
9.1562
9.1562
5
0.8721
20.98
2489.57
2510.54
0.0761
8.9496
9.0257 .
10
1.2276
41.99
2477.75
2519.74
0.1510
8.7498
8.9007
15
1.705
62.98
2465.93
2528.91
0.2245 \
8.5569
8.7813
20
2.339
83.94 ■■
2454.12 ■-
2538.06
\ 0.2966
8.3706
8.6671
25 : K\
3.169
104.87 ; V :
2442.30 V
2547.17
0.3673 ■
8.1905 ■
8.5579
30
4.246
125.77
2430.48
2556.25
0.4369
8.0164
8.4533
35
5.62S
146.66
2418.62
2565.28
0.5052
7.8478
8.3530
40
7.384
167.54
2406.72
2574.26
0.5724
7.6845
8.2569
45 .
9.593
188.42
2394.77
2583.19 :
0.6386 v
7.5261
8.1647
50
12.350
209.31 \
2382.75
2592.06
0.7037 : :
7.3725 -
8.0762 ;
55
15.758
230.20
2370.66
2600.86 ;
0.7679
7.2234
7.9912
60
19.941
251.11
2358.48
2609.59
0.8311
7.0784
7.9095
65
25.03
272.03
2346.21
2618.24
0.8934
6.9375
7.8309
70
31.19
292.96
2333.85
2626.80
0.9548
6.8004
7.7552
75
38.58
313.91
2323.37
2635.28
1.0154
6.6670
7.6824
80 :
47.39
334.88
2308.77
2643.66
1.0752
6.5369
7.6121
85
57.83
355.88
■ 2296.05
2651.93
1.1342
6.4102
7.5444
90
70.14
376.90
2283.19
2660.09
1.1924
6.2866
7.4790
95
84.55
397.94
2270.19
2668.13
1.2500
6.1659
7.4158
100
101.3
419.02
2257.03
2676.05
1.3068
6.0480
7.3548
105
120.8
440.13
2243.70
2683.83
1.3629
5.9328
7.2958
110
143.3
461.27
. 2230.20
2691.47
1.4184
5.8202
7.2386
115
169.1
482.46
2216.50
2698.96
.1.4733
5.7100
7.1832
120
198.5
503.69
2202.61
2706.30
1.5275
5.6020
7.1295
125
232.1
524.96
2188.50
2713.46
1.5812
5.4962
7.0774
130
270.1
546.29
2174.16
2720.46
1.6343
5.3925
7.0269
135
313.0
567.67
2159.59
2727.26
1.6869
5.2907
6.9777
140
361.3
589. U
2144.75 .
2733.87 ; ;
1.7390
5.1908
6.9298
145
: 415.4
610.61
2129.65
2740.26
1.7906
5.0926
6.8832
150
475.9
632.18
2114.26
2746.44
1.8417
4.9960
6.8378
155
543.1
653.82
2098.56
2752.39
1.8924
4.9010
6.7934
160
617.8
675.53
2082.55
2758.09
1.9426
4.8075
6.7501
165 V
700.5
697.32 . :
2066.20
2763.53
1.9924
4.7153
6.7078 :
170
791.7 :
719.20
2049.50
2768.70
2.0418
4.6244
6.6663
175
892.0
741.16
2032.42
2773.58
2.0909
4.5347
6.6256
180
1002.2
763.21
2014.96
2778.16
2.1395
4.4461
6.5857
185
1 122.7
785.36
1997.07
2782.43
2.187S
4.3586
6.5464
190
1254.4
807.61
1978.76
2786.37
2.2358
4.2720
6.5078
676 Appendix B SI Units: thermodynamic Tables
TABLE B.1.1 (continued)
Saturated Water
Specific Volume, m 3 /kg Internal energy, kJ/kg
Temp,
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
CO
(kPa)
v f
v ,
»/
li fs
" s
195
1397.8
0.001149
0.13990
0.14105
828.36
1764.43
2592.79
200_
1553.8
0.001156
0.12620
0.12736
850.64
1744.66
2595.29
205
1723.0
0.001 164
0.11405
0.11521
873.02
1724.49
2597.52
210
1906.3
0.001173
: 0.10324 V
0.10441
895.51
1703.93
2599.44 -
215 ;-
2104.2
0.001181
0.09361 ■;.:'/:
0.09479
918.12
1682.94
2601.06 ■
220
2317.8
0.001190 ; :
0.08500
0.08619
940.85 -V.-
1661.49
2602.35
225
2547.7
0.001199
0.07729
0.07849
963.72
1639.58
2603.30
230
2794.9
0.001209
0.07037
0.07158
986.72
1617.17
2603.89
235
3060.1
0.001219
0.06415
0.06536
1009.88
1594.24
2604.11
240
3344.2
0.001229
■ 0.05853
0.05976
1033.19
1570.75
2603.95
245
3648.2
0.001240 ■
■ : . 0.05346
0.05470
1056.69
; 1546.68
2603.37
250
3973.0
0.001251
■■; 0.04887
0.05013
1080.37
; 1522.00
2602.37
255
4319.5
0.001263
0.04471
0.04598
1104.26
1496.66
2600.93
260
4688.6
0.001276
0.04093 '
0.04220
1128.37
1470.64
2599.01
265
5081.3
0.001289
0.03748
0.03877
1152.72
1443.87
2596.60
270
5498.7
0.001302
0.03434
0.03564
1177.33
1416.33
2593.66
275
5941.8
0.001317
0.03147
0.03279
1202.23
1387.94
2590.17
280
6411.7
0.001332
0.02884
0.03017
1227.43
1358.66
2586.09
285
6909.4
0.001348
0.02642
0.02777
1252.98
1328.41
2581.38
290
7436.0
0.001366
0.02420
0.02557
1278.89
1297.11
2575,99
295
7992.8
0.001384
0.02216
0.02354
1305.21
1264.67
2569.87
300
8581.0
0.001404
0.02027
0.02167
1331.97
1230.99
2562.96
305
9201.8
0.001425
0.01852 •
0.01995
1359.22
,1195.94
2555.16
310
- 9856.6
0.001447
0.01690
0.01835
1387.03
1159.37
2546.40
315
10547
0.001472
0.01539
0.01687
1415.44
1121.11
2536.55
320*
11274
0.001499
0.01399
0.01549
1444.55
1080.93
2525.48
325
12040
0.001528
0.01267
0.01420
1474.44
1038.57
2513.01
330
12845
0.001561
0.01144
0.01300
1505.24
993.66
2498.91
335
13694
0.001597
0.01027
0.01186
1537.11
945.77
: 2482.88
340
14586
0.001638
/ 0.00916
0.01080
1570.26
894.26
2464.53
345
15525
0.001685
0.00810
0.00978
1605.01
838.29
2443.30
350
16514
0.001740
0.00707
0.00881
1641.81
776.58
2418.39
355
17554
0.001807
0.00607
0.00787
1681.41
707.11
2388.52
360
18651
0.001892
; 0.00505
0.00694
1725.19
626.29
2351.47
365
19807
0.002011 :
0.00398
■ 0.00599
1776.13
526.54
2302.67
370
21028 ;
0.002213
: 0.00271
0.00493
1843.84
384.69
2228.53
374.1
22089
0.003155
0.00315
2029.58
2029.58
Appendix B SI Units: Thermodynamic Tables B 677
TABLE B.1,1 {continued)
Saturated Water
Enthalpy, k j/kg entropy, kj/kg-K
Temp.
(°C)
Press.
(kPa)
lJ(Ui IjIU uiu
J
Evap.
fs
Sat, Vapor
ft.
Sat. Liquid
s f
Evap.
s fs
Sat. Vapor
s g
195
1397.8
829.96
1959.99
2789.96
2.2835
4.1863
6.4697
200
1553.8
852,43
1940.75
2793.18
2.3308
4.1014
6.4322
205
1723.0
875.03
1921.00
2796.03
2.3779
4.0172
6.3951
flirt
210
: 1906.3
897.75 ;
1900.73
: ; : ; 2798.48
2.4247 .; ■■'
3.9337
6.3584
215
2104.2
920.61
1879.91
■;■ 2800.51
2.4713
3.8507 ;
6.3221 :
220
2317.8
943.61
: 1858.51
2802.12
2.5177
3.7683
6.2860
225
2547.7
966.77
1836.50
2803.27
2.5639
3.6863
6.2502
230
2794.9
990.10
1813.85
2803.95
2.6099
3.6047
6.2146
235
3060.1
1013.61
1790.53
2804.13
2.6557
3.5233
6.1791
240
3344.2
1037.31
: 1766.50
2803.81
2.7015
3.4422
6.1436
245
3648.2
1061.21
1741.73
2802.95
2.7471
3.3612
6.1083
250
3973.0
1085.34
1716.18
2801.52
2.7927
3.2802
6.0729
255
4319.5
1 109.72
1689.80
2799.51
2.8382
3.1992
6.0374
260
4688.6
1134.35
1662.54
2796.89
2.8837
3.1181
6.0018
265
50S1.3
1159.27
1634.34
2793.61
2.9293
3.0368
5.9661
2. /O
5498.7
1184.49
1605.16
2789.65
2.9750
2.9551
5.9301
275
5941.8
1210.05
1574.92
2784.97
3.0208
2.8730
5.8937
280
6411.7
1235.97
1543.55
2779.53
3.0667
2.7903
5.8570
6909.4
1262.29
1510.97
2773.27
3.1129
2.7069
5.8198
290
7436.0
1289.04
1477.08
2766.13
3.1593
2.6227
5.7821
295
7992.8
1316.27
1441.78
2758.05
3.2061
2.5375
5.7436
300
■8581.0
1344.01
1404.93
2748.94
3.2533
2.4511
5.7044
305
9201.8
1372.33
1366.38
2738.72
3.3009
2.3633
5.6642
310
9856.6
1401.29
1325.97
2727.27
,3.3492
2.2737
5.6229
315
10547
1430.97
1283.48
2714.44
3.3981
2.1821
5.5803
320
11274
1461.45
1238.64
2700.08
3.4479
2.0882
5.5361
325
12040
1492.84
1191.13
2683.97
3.4987
1.9913
5.4900
330
: 12845
1525.29
1140.56
2665.85
3.5506
1.8909
5.4416
335
13694
1558.98
1086.37
2645.35
3.6040
1.7863
5.3903
340
14586
1594.15
1027.86
2622.01
3.6593
1.6763
5.3356
345
15525
1631.17
964.02
2595.19
3.7169
1.5594
5.2763
350
16514
1670.54
893.38
2563.92
3.7776
1.4336
5.2111
355
17554
1713.13
813.59
2526.72
3.8427
1.2951
5.1378
360
18651
1760.48
720.52
2481.00
3.9146
1.1379
5.0525
365
19807
1815.96
605.44
2421.40
3.9983
0.9487
4.9470
370
21028
1890.37
441.75
2332.12
4.1104
0.6868
4.7972
374.1
22089
2099.26
2099.26
4.4297
4.4297
678 B APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.1.2
Saturated Water Pressure
Entry
Press.
(kPa)
Temp.
CC)
Specific volume, m 3 /kg
INTERNAL ENERGY, kJ/kg
Sat. Liquid Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
a a 1 1 i
U.OI I J
u.u 1
0.001000
206.131
206.132
2375.3
2375.3
1
£ OS
129.20702
129.20802
29.29
2355.69
2384.98
1.5
1 J.UJ
87.97913
87.98013
54.70
2338.63
2393.32
2 ■■
1-7 CA
001001
67.00285
.67.00385
73.47
2326.02
2399.48
2.5 ; ,'•
54.25285
54.25385
88.47
2315.93
2404.40
24.U8
nnfiinfit
45.66402
45.66502
101.03
2307.48
2408.51
4
n no 1004
34.79915
34.80015
121.44
2293.73
2415.17
5
10 sis
no 1005
28.19150
28.19251
137.79
2282.70
2420.49
7.5
/tA 1Q
fi on 1 nns
19.23674
19.23775
168.76
2261.74
2430.50
10
/t< si
4j.o 1
n nmoio
14.67254
14.67355 : :
191.79
.:'], 2246.10
2437.89 ;
15
AT
jj.y /
fi fimfti4
10.02117
■ 10.02218 "
225.90
2222.83
2448.73
20
*ca n*:
7.64835
■/ ■'. 7.64937
251.35 ;.■
v 2205.36 : 'V
2456.71
25
64. y /
n fimfi90
6.20322
6.20424
271.88
2191.21
2463.08
30
£A 1 A
fi nntn??
5.22816
5.22918
289.18
2179.22
2468.40
40
/ J.O /
fi fifilfi26
3.99243
3.99345
317.51
2159.49
2477.00
50
51. JJ
nnnift3n
ViOyiyJU
3.23931
V 3.24034 ■
-■■ 340.42 /
. 2143.43 .
2483.85
91.77
0.001037
V 2.21607 ;
2.21711
394.29 :
2112.39
2496.67
100
99.62
0.001043
• 1.69296
1.69400 ' :-.
417.33 ' ■
2088.72 .
2506.06
125
105.99
0.001048
1.37385
1.37490
444.16
2069.32
2513.48
150
111.37
0.001053
1.15828
1.15933
466.92
2052.72
2519.64
175
116.06
0.001057
1.00257
1.00363
486.78
2038.12
2524.90
200
120.23
; 0.001061
0.88467
0.88573
504.47
2025.02
2529.49
225 ■
124.00
0.001064 :
0.79219
0.79325 :
520.45 v
2013.10
; 2533.56
250
127.43
i 0.001067
0.71765
; ; : 0.71871 .
535.08
2002.14 -
2537.21
275
130.60
0.001070
0.65624
0.65731
548.57
1991.95
2540.53
300
133.55
0.001073
0.60475
0.60582
561.13
1982.43
2543.55
325
13630
0.001076
0.56093
0.56201
572.88
1973.46
2546.34
350
C 138:88 / :
0.001079
0.52317
0.52425
583.93 "
1964.98
2548.92
"375
V .■ 141.32
0.001081
0.49029-
0.49137
594.38
1956.93
2551.31
400
143.63
0.001084
0.46138
0.46246
604.29
1949.26
2553.55
450
147.93
0.001088
0.41289
0.41398
622.75
1934.87
2557.62
500
151.86
0.001093
0.37380
0.37489
639.66
1921.57
2561.23
550
155.48
0.001097
0.34159
0.34268
655.30
1909.17
2564.47
600
158.85 ; ;
0.001 101 /
v ■ - 0.31457
0.31567
> ■ 669.88 ; ;
1897.52
2567.40
650
162.01 ;
0.001104
0.29158
0.29268
683.55
1886.51
2570.06
700
,. 164.97 : -
0.001108 ■ :
0.27176
0.27286
696.43 :
1876.07
2572:49
750
167.77
0.001111
0.25449
0.25560
708.62
1866.11
2574.73
800
170.43
0.001115
0.23931
0.24043
720.20
1856.58
2576.79
Appendix B SI Units: Thermodynamic Tables H 679
TABLE B.1.2 (Continued)
Saturated Water Pressure Entry
Enthalpy, kJ/kg Entropy, kJ/kg-K
Press,
(kPa)
Temp.
(°C)
Sat. Liquid
h f
Eva p.
urn* Ljjuuiu
s r
j
JS
Sat. Vapor
s„
s
U.Ol U
nnt
(J.U 1
0.00
2501.3
2501.3
9.1562
9.1562
o.ys
29.29
1 A O A on.
2484.89
2514.18
0.1059
8.8697
8.9756
1 t
I J.UJ
ca Trt
54.70
2470.59
2525.30
0.1956
8.6322
8.8278
IT Cfl
1 /.jU
73.47 ;
2460.02
2533.49
0.2607
8.4629
8.7236
Z.J
01 no
88.47 - :
2451.56
2540.03
0.3120
8.3311
8.6431
in
24.05
101.03
2444.47
2545.50
0.3545
8.2231
8.5775
121.44
2432.93
2554.37
0.4226
8.0520
8.4746
-J.U
32. bis
137.79
2423.66
2561.45
0.4763
7.9187
8.3950
1 K
/.J
Aft on
168.77
2406.02
2574.79
0.5763
7.6751
8.2514
45.5 1
191.81
2392,82
2584.63
0.6492
7.5010
8.1501
225.91
2373.14
2599.06
0.7548
7.2536
8.0084
oU.Oo
251.38
2358.33
2609.70
0.8319
7.0766
7.9085
z J
o4.y /
271.90
2346.29
2618.19
0.8930
6.9383
7.8313
1A
^Crt 1 n
oy.iu
289.21
2336.07
2625.28
0.9439
6.8247
7.7686
4U
/j.b7
317.55
2319.19
2636.74
1.0258
6.6441
7.6700
■\fi
jU
01.55
340.47
2305.40
2645.87
1.0910
6.5029
7.5939
ID
yl.lt
OA 1C
384.36
2278.59
2662.96
1.2129
6.2434
7.4563
1 fin
A in A A
: 417.44
2258.02
2675.46
1.3025
6.0568
7.3593
iz j
1 Ac on
i Lfj.yy
444.30
2241.05
2685.35
1.3739
5.9104
7.2843
111.37
467.08
2226.46
2693.54
1.4335
5.7897
7.2232
175
116.06
486.97
2213.57
2700.53
1.4848
5.6868
7.1717
200
120.23
504.68
2201.96
2706.63
1.5300
5.5970
7.1271
225
124.00
520.69
2191.35
2712.04
1.5705
5.5173
7.0878
250
127.43
535.34
2181.55
2716.89
1.6072
5.4455
7.0526
275
130.60
548.87
2172.42
2721.29
1.6407
5.3801
7.0208
300
133.55
561.45
2163.85
2725.30
1.6717
5.3201
6.9918
325
136.30
573.23
2155.76
2728.99
1.7005
5.2646
6.9651
350 :
138.88
584.31 ..
2148.10
2732.40
1.7274
5.2130
6.9404
375
141.32
594.79
2140.79
2735.58
1.7527
5.1647
6.9174
400
143.63
" :'' 604.73
2133.81
2738.53
1.7766
5.1193
6.8958
450
147.93
623.24
2120.67
2743.91
1.8206
5.0359
6.8565
500
151.86
640.21
2108.47
2748.67
1.8606
4.9606
6.8212
550
155.48
655.91
2097.04
2752.94
1.8972
4.8920
6.7892
600
158.85
670.54
2086.26
;': 2756.80
1.9311
4.8289
6.7600
650
162.01;
684.26
2076.04
2760.30
1.9627
4.7704
6.7330
700
164.97
697.20 ;■
2066.30 ,
2763.50
1.9922
4.7158
6.7080
750
167.77
709.45
2056.98
2766.43
2.0199
4.6647
6.6846
800
170.43
721.10
2048.04
2769.13
2.0461
4.6166
6.6627
680 M APPENDIX B SI UNITS: THERMODYNAMIC TABLES
TABLE B.1.2 (continued)
Saturated Water Pressure En try
Specific Volume, m 3 /kg internal energy, kJ/kg
Press.
Temp.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
(kPa)
(°C)
v f
v f s
"A
"s
850
172.96
0.001118
0.22586
0.22698
731.25
1847.45
2578.69
900
175.38
0.001121
0.21385
0.21497
741.81
1838.65
2580.46
950
177.69
0.001124
0.20306
0.20419
751.94
1830.17
2582.11
1000
179.91
0.001127
0.19332
0.19444
761.67
1821.97
2583.64
1100
184.09
0.001133
0.17639 '
0.17753
780.08
1806.32
2586.40
1200
187.99
0.001139
0.16220
0.16333
797.27
1791.55
2588.82
1300
191.64
0.001144
0.15011
0.15125
813.42
1777.53
2590.95
1400
195.07
0.001149
0.13969
0.14084
828.68
1764.15
2592.83
1500
198.32
0.001154
0.13062
0.13177
843.14
1751.3
2594.5
1750
205.76
0.001166
0.11232
0.11349
876.44
1721.39
2597.83
2000
212.42
0.001177
0.09845
0.09963
906.42
1693.84
2600.26
2250
218.45
0.001187
0.08756
0.08875 •
933.81
1668.18
2601.98
2500
223.99
0.001197
0.07878 ■
0.07998
959.09
1644.04
2603.13
2750
229.12
0.001207
0.07154
0.07275
982.65
1621.16
' 2603.81
3000
233.90
0.001216
0.06546
0.06668
1004.76
1599.34
2604.10
3250
238.38
0.001226
0.06029
0.06152
1025.62
1578.43
2604.04
3500
242.60
0.001235
0.05583
0.05707
1045.41
1558.29
2603.70
4000
250.40
0.001252
0.04853
0.04978
1082.28
1519.99
2602.27
5000
263.99
0.001286
0.03815
0.03944
1147.78
1449.34
2597.12
6000
275.64
0.001319
0.03112
0.03244
1205.41
1384.27
2589.69
7000
285.88
0.001351
0.02602
0.02737
1257.51
1322.97
2580.48
8000
295.06
0.001384
0.02213
0.02352
1305.54
1264.25
2569.79
9000.
303.40
0.001418
0.01907
0.02048
1350.47
1207.28
2557.75
10000
311.06
0.001452
0.01657
0.01803
1393.00
1151.40
2544.41
1J00O
318.15
0.001489
0.01450
0.01599
1433.68
1096.06
2529.74
12000
324.75
0.001527
'0.01274
0.01426
1472.92
1040.76
2513.67
13000
330.93
0.001567
0.01121
0.01278
1511.09
984.99
2496.08
14000
336.75
0.001611
0.00987
' 0.01149
1548.53
928.23
2476.76
-15000
342.24
0.001658
0.00868
0.01034
1585.58
869.85
2455.43
16000
347.43
0.001711
0.00760
0.00931
1622.63
809.07
2431.70
17000
352.37
0.001770
0.00659
0.00836
1660.16
744.80
2404.96
18000
357.06
0.001840
0.00565
0.00749
1698.86
675.42
2374.28
19000
361.54
0.001924
0.00473
0.00666
1739.87
598.18
2338.05
20000
365.81
0.002035
0.00380
0.00583
1785.47
507.58
2293.05
21000
369.89
0.002206
0.00275
0.00495
: 1841.97
: 388.74
2230.71
22000
373.80
0.002808
0.00072
0.00353
1973.16
108.24
■ 2081.39
22089
374.14
0.003155
0.00315
2029.58
2029.58
Appendix b si units: thermodynamic tables B 681
TABLE B.1.2 (Continued)
Saturated Water Pressure Entry
Enthalpy, kJ/kg entropy, kj/kg-K
Press.
(kPa)
Temp.
(°C)
Sat. Liquid
h f
Evap.
fiat Vnnnr
s
Sat. Liquid
J
Evap.
°/s
Sat. Vapor
O ^A
GDV
172.96
732.20
2039.43
2771.63,
2.0709
4.5711
6.6421
Qflfl
yuu
I / J.JiS
742.82
2031.12
2773.94
2.0946
4.5280
6.6225
l /7.t>y
753.00
2023.08
2776.08
2.1171
4.4869
6.6040
1 AAA
1 TO A 1
i /y.yi
762.79
2015.29
2778.08
2.1386.
4.4478
6.5864
i JUU
i o A AA
184.09
781.32
2000.36
2781.68
2.1791
4.3744
6.5535
lOAA
187.99
798.64
1986,19
2784.82
2.2165
4.3067 -
6.5233
1 JUU
i y 1.64
814.91
1972.67
2787.58
2.2514
4.2438
6.4953
1 A AA
14UU
1 AC An
830.29
1959.72
2790.00
2.2842
4.1850
6.4692
1 1 AA
I jUU
1 AC TO
O A A O i"T
844.87
1947,28
2792.15
2.3150
4.1298
6.4448
205.76
878.48
1917.95
2796.43
2.3851
4.0044
6.3895
1AAA
908.7/
1890.74
2799.51
■ 2.4473
3.8935
6.3408
218.45
936.48
1865.19
2801.67
2.5034
3.7938
6.2971
nc AA
ail AA
22 3,yy
962.09
1840.98
2803.07
2.5546
3.7028
6.2574
TJ^A
229.12
985,97
1817.89
2803.86
2.6018
3.6190
6.2208
JUUU
OH AA
1008.41
1795.73
2804.14
2.6456
3.5412
6.1869
n^A
1029,60
1774.37
2803.97
2.6866
3.4685
6.1551
I^AA
Ovll £TA
242. ou
1 F\At\ Ti
1049.73
1753.70
2803.43
2.7252
3.4000
6.1252
.f AAA
250.40
1087.29
1714.09
2801.38
2.7963
3.2737
6.0700
0/^1 AA
1 154.21
1640.12
2794,33
2.9201
3.0532
5.9733
6000
275.64
1213.32
1571.00
2784.33
3.0266
2.8625
5.8891
7000
285.88
1266.97
1505.10
2772.07
3.1210
2.6922
5.8132
8000
295.06
1316.61
1441.33
2757.94
3.2067
2.5365
5.7431
9000
303.40
1363.23
1378.88
2742.11
3.2857
2.3915
5.6771
10000
311.06
1407.53
1317.14
2724.67
3.3595
2.2545
5.6140
1 1000
318,15
1450.05
1255.55
2705.60
3.4294
2.1233
5.5527
12000
324.75
1491.24
1193.59
2684.83
3.4961
1.9962
5.4923
13000
330.93
1531.46
1130.76 .
2662.22
3.5604
1.8718
5.4323
14000
336.75
1571.08
1066.47
2637.55
3.6231
1.7485
5.3716
15000
342.24
1610.45
1000.04
2610.49
3.6847
1.6250
5.3097
16000
347.43
1650.00
930.59
2580.59
3.7460
1.4995
5.2454
17000
352.37
1690.25
856.90
2547.15
3.8078
1.3698
5.1776
18000
357.06
1731.97
777.13
2509.09
3.8713
1.2330
5.1044
19000
361.54
1776.43
688.11
2464.54
3.9387
1.0841
5.0227
20000
365.81
1826.18
583.56
2409.74
4.0137
0.9132
4.9269
21000
369.89
1888.30
446.42
2334.72
4.1073
0.6942
4.8015
22000
373.80
2034.92
124.04
2158.97
4.3307
0.1917
4.5224
22089
374.14
2099.26
2099.26
4.4297
4.4297
682 ■ APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.1.3
Superheated Vapor Water
Sat.
150
200
<°C)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
P = 10 kPa (45.81)
Sat.
14.67355
2437.89
2584.63
8.1501
50
14.86920
2443.87
2592.56
8.1749
100
17.19561
2515.50
2687.46
8.4479
150
19.51251
2587.86
2782.99
8.6881
200
21.82507
:2661.27
2879.52
8.9037
250
24.13559
2735.95
2977,31
9.1002
300
26.44508
2812.06
3076.51
9.2812
400
31.06252
2968.89
3279.51
9.6076
500
35.67896
3132.26
3489.05
9.8977
600
40.29488
3302.45
3705.40
10.1608
700
44.91052
3479.63
3928.73
10.4028
800
49.52599
3663.84
4159.10
10.6281
900
54.14137
3855.03
4396.44
10.8395
1000
58.75669
4053.01
4640.58
11.0392
1100
63.37198
4257.47
4891.19
11.2287
1200
67.98724
4467.91
5147.78
11.4090
1300
72.60250
4683.68
5409.70
14.5810
100 kPa (99.62)
300 kPa (133.55)
0.60582
0.63388
0J1629
2543.55 2725.30 6.9918
2570.79 2760.95 7.0778
2650.65 2865.54 7.3115
v
(m 3 /kg)
it ft s
(kJ/kg) (kJ/kg) (kJ/kg-K)
i> = 50kPa (81.33)
3.24034
2483.85
2645.87
7.5939
3.41833
2511.61
2682.52
7.6947
; 3.88937
2585.61
: 2780.08
7.9400
4.35595
2659.85 :
2877.64 ,
8.1579
4.82045
2734.97
2975.99
8.3555
5.28391
2811.33
3075.52
8.5372
6.20929
2968.43
3278.89
8.8641
7.13364
3131.94
3488.62
9.1545
8.05748
3302.22
3705.10
9.4177
8.98104
3479.45
3928.51
9.6599
9.90444
3663.70
4158.92
9.8852
10.82773
3854.91
4396.30
10.0967
11,75097
4052.91
4640.46
10.2964
12.67418
4257.37
4891.08
10.4858
13.59737
4467.82
5147.69
10.6662
14.52054
4683.58
5409.61
10.8382
200 kPa (120.23)
Sat.
1.69400
2506.06
2675.46
7.3593
0.88573
2529.49
150
1.93636
2582.75
2776.38
7.6133
0.95964
2576.87
200
2.17226
2658.05
2875.27
7.8342
1.08034
2654.39
250 '
2.40604
2733.73
2974.33
8.0332
1.19880
2731.22
300
2.63876
2810.41
3074.28
8.2157
1.31616
2808.55
MOO
3.10263
2967.85
3278.11
8.5434
1.54930
2966.69
500
3.56547
3131.54
3488.09
8.8341
1.78139
3130.75
600
4.02781
3301.94
3704.72
9.0975
2.01297
3301.36
700
4.48986
3479.24
3928.23
9.3398
2.24426
3478.81
800
4.95174
3663.53
4158.71
9.5652
2.47539
3663.19
900
5.41353
3854.77
4396.12
9.7767
2.70643
3854.49
1000
5.87526
4052.78
4640.31
9.9764
2.93740
4052.53
1100
6.33696
4257.25
4890.95
10.1658
3.16834
4257.01
1200
6.79863
4467.70
5147.56
10.3462
3.39927
4467.46
1300
7.26030
4683.47
5409.49
10.5182
3.63018
4683.23
2706.63
2768.80
2870.46
2970.98
3071.79
3276.55
3487.03
3703.96
3927.66
4158.27
4395.77
4640.01
4890.68
5147.32
5409.26
400 kPa (143.63)
7.1271
7.2795
7.5066
7.7085
7.8926
8.2217
8.5132
8.7769
9.0194
9.2450
9.4565
9.6563
9.8458
10.0262
10.1982
0.46246 2553.55 2738.53 6.8958
0.47084 2564.48 2752.82 6.9299
0.53422 2646.83 2860.51 7.1706
APPENDIX B SI UNITS: THERMODYNAMIC Tables H 683
TABLE B.1.3 (con tinned)
Superheated Vapor Water
Temp.
a
h
s
V
u
h
s
<°C)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
: 300 kPa (133.55)
400 kPa (143.63)
250
0.79636
2728.69
2967.59
7.5165
0.59512
2726.11
2964.16
7.3788
300
0.87529
2806.69
3069.28
7.7022
0.65484
2804.81
3066.75
7.5661
400
1.03151
2965.53
3274.98
8.0329
0.77262
2964.36
3273.41
7.8984
500
1.18669
3129.95
3485.96
8.3250
0.88934
3129.15
3484.89
8.1912
600
1.34136
3300.79
3703.20
8.5892
1.00555
3300.22
3702.44
8.4557
700
1.49573
3478.38
3927.10
8.8319
1.12147
3477.95
3926.53
8.6987
800
1.64994
3662.85
4157.83
9.0575
1.23722
3662.51
4157.40
8.9244
900
1.80406
3854.20
4395.42
9.2691
1.35288
3853.91
4395.06
9.1361
1000
1.95812
4052.27
4639.71
9.4689
1.46847
4052.02
4639.41
9.3360
1 100
2.11214
4256.77
4890.41
9.6585
1.58404
4256.53
4890.15
9.5255
1200
2.26614
4467.23
5147.07
9.8389
1.69958
4466.99
5146.83
9.7059
1300
2.42013
4682.99
5409.03
10.0109
1.81511
4682.75
5408.80
9.8780
500 kPa (151.86)
600 kPa (158.85)
Sat.
0.37489
2561.23
2748.67
6.8212
0.31567
2567.40
2756.80
6.7600
200
0.42492
2642.91
2855.37
7.0592
0.35202
2638.91
2850.12
6.9665
250
0.47436
2723.50
2960.68
7.2708
0.39383
2720.86
2957.16
7.1816
300
0.52256
2802.91
3064.20
7.4598
0.43437
2801.00
3061.63
7.3723
350
0.57012
2882.59
3167.65
7.6328
0.47424
2881.12
3165.66
7.5463
400
0.61728
2963.19
3271.83
7.7937
0.51372
2962.02
3270.25
7.7078
500
0.71093
3128.35
3483.82
8.0872
0.59199
3127.55
3482.75
8.0020
600
0.80406
3299.64
3701.67
8.3521
0.66974
3299.07
3700.91
8,2673
700
0.89691
3477.52
3925.97
8.5952
0.74720
3477.08
3925.41
8.5107
800
0.98959
3662.17
4156.96
8.8211
0.82450
3661.83
4156.52
8.7367 .
900
1.08217
3853.63
4394.71
9.0329
0.90169
3853.34
4394.36
8.9485
1000
- 1.17469
4051.76
4639.11
9.2328
0.97883
4051.51 •
4638.81
9.1484
1100
1.26718
4256.29
4889.88
9.4224
1.05594
4256.05
4889.61
9.3381
1200
1.35964
4466.76
5146.58
9.6028
1.13302
4466.52
5146.34
9.5185
1300
1.45210
4682.52 -
5408.57
9.7749
1.21009
4682.28
5408.34
9.6906
800 kPa (170.43)
1000 kPa (179.91)
Sat.
0.24043
2576.79
2769.13
6.6627
0.19444
2583.64
2778.08
6.5864
200
0.26080
2630.61
2839.25
6.8158
0.20596
2621.90
2827.86
6.6939
250
0.29314
2715.46
2949.97
7.0384
0.23268
2709.91
2942.59
6.9246
300
0.32411
2797.14
3056.43
7.2327
0.25794
2793.21
3051.15
7.1228
350
0.35439
2878.16
3161.68
7.4088
; 0.28247
2875.18
3157.65
7.3010
400
0.38426
2959.66
3267.07
7.5715
0.30659
2957.29
3263.88
7.4650
500
0.44331
3125.95
3480.60
7.8672
0.35411
3124.34
3478.44
7.7621
600
0.50184
3297.91
3699.38
8.1332
0.40109
3296.76
3697.85
8.0289
684 tl APPENDIX B SI UNITS: THERMODYNAMIC TABLES
TABLE B.1.3 (continued)
Superheated Vapor W ater ___________
Temp. v u h s
(°C) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K)
800 kPa (170.43)
/UU
U.JDUU /
3476.22
3924.27
8.3770
nrtn
sUu
U.OIolJ
3661.14
4155.65
8.6033
yuu
ft KIM ft
3852.77
4393.65
8.8153
1000
0.73401
4051.00
4638.20
9.0153
1100
0.79188
4255.57
4889.08
9.2049
1200
0.84974
4466.05
5145.85
9.3854
1300
0.90758
4681.81
5407.87
9.5575
1200 kPa (187.99)
Sat.
2588.82
2784.82
6.5233
200
f\ i /win
2612.74
2815.90
6.5898
250
2704.20
2935.01
6.8293
300
n o i is?
2789.22
3045.80
7.0316
350
2872.16
3153.59
7.2120
4UU
2954.90
3260.66
7.3773
5UU
3122.72
3476.28
7.6758
600
3295.60
3696.32
7.9434
700
U.J 1 Lyn
3474.48
3922.01
8.1881
800
U.f 1 J. / /
3659.77
4153.90
8.4149
yuo
3851.62
4392.23
8.6272
1000
0.48919
4049.98
4637.00
8.8274
1100
0.52783
4254.61
4888.02
9.0171
1200
0.56646
4465.12
5144.87
9.1977
1300
• 0.60507
4680.86
5406.95
9.3698
1600 kPa (201.40)
Sat.
0.12380
2595.95
2794.02
6.4217
250
0.14184
2692.26
2919.20 '
6.6732
300
0.15862
2781.03
3034.83
6.8844
350
0.17456
2866.05
3145.35
7.0693
400
0.19005
2950.09
3254.17
7.2373
500
0.22029
3119.47
3471.93
7.5389
600
0.24998
3293.27
3693.23
7.8080
700
0.27937
3472.74
3919.73
8.0535
800
0.30859
3658.40
4152.15
8.2808
900
0.33772
3850.47
4390.82
8.4934
1000
0.36678
4048.96
4635.81
8.6938
1100
0.39581
4253.66
4886.95
8.8837
1200
0.42482
4464.18
5143.89
9.0642
1300
0.45382
4679.92
5406.02
9.2364
v u h s
(nrVkg) (kJ/kg) (kJ/kg) (kJ/kg-K)
1000 kPa (179.91)
0.44779
3475.35
3923.14
8.2731
0.49432
3660.46
4154.78
8.4996
0.54075
3852.19
4392.94
8.7118
0.58712
4050.49
4637.60
8.9119
0.63345
4255.09
4888.55
9.1016
0.67977
4465.58
5145.36
9.2821
0.72608
4681.33
5407.41
9.4542
1400 kPa (195.07)
U.14US4
2592.83
2790.00
6.4692
U.14JUZ
2603.09
2803.32
6.4975
0.16350
2698.32
2927.22
6.7467
0.18228
2785.16
3040.35
6.9533
0.20026
2869.12
3149.49
7.1359
0.21780
2952.50
3257.42
7.3025
0.25215
3121.10
3474.11
7.6026
0.28596
3294.44
3694.78
7.8710
0.31947
3473.61
3920.87
8.1160
0.35281
3659.09
4153.03
8.3431
0.38606
3851.05
4391.53
8.5555
0.41924
4049.47
4636.41
8.7558
0.45239
4254.14
4887.49
8.9456
0.48552
4464.65
5144.38
9.1262
0.51864
4680.39
5406.49
9.2983
1800 kPa (207.15)
0.11042
2598.38
2797.13
6.3793
0.12497
2686.02
2910.96
6.6066
0.14021
2776.83
3029.21
6.8226
0.15457
2862.95
3141.18
7.0099
0.16847
2947.66
3250.90
7.1793
0.19550
3117.84
3469.75
7.4824
0.22199
3292.10
3691.69
7.7523
0.24818
3471.87
3918.59
7.9983
0.27420
3657.71
4151.27
8.2258
0.30012
3849.90
4390.11
8.4386
0.32598
4048,45
4635.21
8.6390
0.35180
4253.18
4886.42
8.8290
0.37761
4463.71
5143.40
9.0096
0.40340
4679.44
5405.56
9.1817
APPENDIX B SI UNITS: THERMODYNAMIC TABLES M 685
TABLE B.1.3 (continued)
Superheated Vapor Water
Temp.
u
it
h
s
V
ii
It
(°C)
(m 3 /kg)
(kJ/kg)
(kj/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
2000 kPa (212.42)
2500 kPa (223.99)
Sat.
0.09963
2600.26
2799.51
6.3408
0.07998
2603.13
2803.07
6.2574
250
0.11144
2679.58
2902.46
6.5452
0.08700
2662.55
2880.06
6.4084
300
0.12547
2772.56
3023.50
6.7663
0.09890
2761.56
3008.81
6.6437
350 ;
, 0.13857
2859.81
; 3136.96
6.9562
0.10976
2851.84
3126.24
V 6.8402
400 ;
0.15120 : .
2045.21
3247.60
7.1270 .. ■;
0.12010 :
2939.03
V 3239.28 : -
7.0147 V ;
450
0.16353
3030.41
3357.48
7.2844
0.13014
3025.43
3350.77
7.1745
500
0.17568
3116.20
3467.55
7.4316
0.13998
3112.08
3462.04
7.3233
600
0.19960
3290.93
3690.14
7.7023
0.15930
3287.99
3686.25
7.5960
700
0.22323
3470.99
3917.45
7.9487
0.17832
3468.80
3914.59
7.8435
800
0.24668
3657.03
4150.40
8.1766
0.19716
3655.30
4148.20
8.0720 : V
900
■.: 0.27004
3849.33
4389.40
8.3895
0.21590
3847.89
4387.64
8.2853
1000
0.29333
4047.94
4634.61
8.5900
0.23458
4046.67
4633.12
8.4860
1 100
0.31659
4252.71
4885.89
8.780O
0.25322
4251.52
4884.57
8.6761
1200
0.33984
4463.25
5142.92
8.9606
0.27185
4462.08
5141.70
8.8569
1300
0.36306
4678.97
5405.10
9.1328
0.29046
4677.80
5403.95
9.0291
3000 kPa (233.90)
4000 kPa (250.40)
Sat.
0.06668
2604.10
2804.14
6.1869
0.04978
2602.27
2801.38
6.0700
250
0.07058
2644.00
2855.75
6.2871
300
0.08114
2750.05
2993.48
6.5389
0.05884
2725.33
2960.68
6.3614
350
0.09053
2843.66
3115.25
6.7427
0.06645
2826.65
3092.43
6.5820
400
0.09936
2932.75
3230.82
6.9211
0.07341
2919.88
, 3213.51
6.7689
450
0.10787
3020.38
3344.00
7.0833
0.08003
3010.13
; 3330.23
6.9362
500
0.11619
3107.92
3456.48
7.2337
0.08643
3099.49
,3445.21
7.0900
600
0.13243
3285.03
3682.34
7.5084
0.09885
3279.06
3674.44
7.3688
700
0.I483S
3466.59
3911.72
7.7571
0.11095
3462.15
3905.94
7.6198
800
0.16414
3653.58
:4146.00
7.9862
0.12287
3650.11
4141.59
7.8502
900
0.17980
3846.46
4385.87
8.1999
0.13469
3843.59
4382.34
8.0647
1000
0.19541
4045.40
4631.63 ;
8.4009
0.14645
4042.87
4628.65
8.2661
1100
0.21098
4250.33
4883.26
8.5911
0.15817
4247.96
4880.63
8.4566
1200
022652
4460.92
5140.49
8.7719
0.16987
4458.60
5138.07
8.6376
1300
0.24206
4676.63
5402.81
8.9442
0.18156
4674.29
5400.52
8.8099
686 M APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.1.3 (continued)
Superheated Vapor Water „ -
Temp. v u h s v u h s
(°C) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K)
5000 kPa (263.99) 6000 kPa (275.64)
Sat.
0.03944
2597.12
2794.33
J.7/JJ
0.03244
2589.69
2784.33
5.8891
300
0.04532
2697.94
2924.53
n nifii6
2667.22
2884.19
6.0673
350
0.05194
2808.67
3068.39
ft AACtO
O f]422^
2789.61
3042.97
6.3334
400
0.05781
2906.58
3195.64
O.OHjo
04719
2892.81
3177.17
6.5407
450
0.06330
2999.64
3316.15
A 81 ;
2988.90
3301.76
6.7192
500 .
0.06857
3090.92
3433.76
: o.y / JO
05665
3082.20
3422.12 :
6.8802
550
0.07368
3181.82
3550.23
/.lil /
06101
3174.57
3540.62
7.0287
600
0.07869
3273.01
3666,47
06525
3266.89
3658.40
7.1676
700
0.08849
3457.67
3900.13
7
fl (V7^S?
3453.15
3894.28
7.4234
800
0.09811
3646.62 v
4137.17 :
7 7/i /in
. U,UO lv\J . .
3643.12
: 4132.74
7.6566
900 ;
0.10762 ;
i; 3840.71
: 4378.82
3837.84
4375.29 .
7.8727
1000
. U.l I l<J)
; 4040.35
4625.69
8.1612
0.09749
4037.83
4622.74 :
8.0751
1100
0.12648
4245.61
4878.02
8.3519
0.10536
4243.26
4875.42
8.2661
1200
0.13587
4456.30
5135.67
8.5330
0.11321
4454.00
5133.28
8.4473
1300
0.14526
4671.96
5398.24
8.7055
0.12106
4669.64
5395.97
8.6199
8000 kPa (295.06) ■ /.
10000 kPa(311.06)
Sat
0.02352
2569.79
2757.94
; 5.7431
0.01803
2544.41
2724.67
; 5.6140
300
0.02426
2590.93
2784.98
5.7905
350
0.02995
2747.67
2987.30
6.1300
0.02242
2699.16
2923.39
5.9442
400
0.03432
2863.75
3138.28
6.3633
0.02641
2832.38
3096.46
6.2119
450
0.03817
2966.66
3271.99
6.5550
: 0.02975
2943.32
3240,83
6.4189
500
0.04175
3064.30
3398.27
. 6.7239
0.03279
3045.77
3373.63
6.5965
550
0.04516
3159.76
3521.01 :
6.8778
: 0.03564
3144.54
3500.92
6.7561
600
0.04845
3254.43
3642.03
7.0205
0.03837
3241.68
3625.34
6.9028
700
0.05481
3444.00
3882.47
7.2812
0.04358
3434.72
3870.52
7.1687
800
0.06097
3636.08
4123.84
7.5173
0.04859
3628.97
4114.91
7.4077
900
0.06702
■ 3832.08
4368.26
7.7350
0.05349 :
3826.32
4361.24
7.6272
1000
0.07301
4032.81
J 4616.87 ;
7.9384 ;
0.05832
4027.81
4611.04
7.8315
1 100
0.07896
4238.60
4870.25
8.1299
0.06312
4233.97
: 4865.14
8.0236
1200
0.08489
4449.45
5128.54
8.3115
0.06789
4444.93
5123.84
8.2054
1300
0.09080
4665.02
5391.46
8.4842
0.07265
4660.44
5386.99
8.3783
APPENDIX B SI UNITS: THERMODYNAMIC TABLES H 687
TABLE B.1.3 {continued)
Superheated Vapor Water
Temp.
V
it
h
s
. V
u
h
s
ro
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
fkJ/kg-K)
15000 kPa (342.24)
20000 kPa (365.81)
Sat.
0.01034
2455.43
2610.49
5.3097
0.00583
2293.05
2409.74
4.9269
350
0.01 147
2520.36
2692.41
5.4420
— .
400
0.01565
2740.70
2975.44
5.8810
0.00994
2619.22
2818.07
5.5539
450
0.0)845
2879.47
3156.15
6.1403
0.01270
2806.16
3060.06
5.9016
500
0.02080
2996.52
3308.53
6.3442 ■■ ;
0.01477
2942.82
3238.18 ;
6.1400
550 ■■■
0.02293
3104.71
3448.61
6.5198
0.01656
3062.34
3393.45
6.3347
600
0.02491
3208.64
3582.30
6.6775
0.01818
3174.00
3537.57
6.5048
650
0.02680
3310.37
3712.32
6.8223
0.01969
3281.46
3675.32
6.6582
700
0.02861
3410.94
3840.12
6.9572
0.02113
3386.46
3809.09
6.7993
800
0.03210
; 3610.99
4092.43
7.2040
0.02385
3592.73
4069.80
7.0544
900
0.03546
3811.89
4343.75
7.4279
0.02645
3797.44
4326.37
7.2830
1000
0.03875
4015.41
4596.63
7.6347
0.02897
4003.12
; 4582.45
7.4925
1100
0.04200
4222.55
4852.56
7.8282
0.03145
4211.30
4840.24
7.6874
1200
0.04523
4433.78
5112.27
8.0108
0.03391
4422.81
5100.96
7.8706
1300
0.04845
4649.12
5375.94
8.1839
0.03636
4637.95
5365.10
8.0441
30000 kPa
40000 kPa
375
0.001789
1737.75
1791.43
3.9303
0.001641
1677.09
1742.71
3.8289
400
0.002790
2067.34
2151.04
4.4728
0.001908
1854.52
1930.83
4.1134
425
0.005304
2455.06
2614.17
5.1503
0.002532
2096.83
2198.11
4.5028
450
0.006735
2619.30
2821.35
5.4423
0.003693
2365.07
2512.79
4.9459
500
0.008679
2820.67
3081.03
5.7904
0.005623
2678.36
2903.26
5.4699
550
0.010168
2970.31
3275.36
6.0342
0.006984
2869.69
3149.05
5.7784
600
?. 0.011446
3100.53
3443.91
6.2330
0.008094
3022.61
3346.38
6.0113
650
0.012596
3221.04
3598.93
6.4057
0.009064
3158.04
3520.58
6.2054
700
0.013661
3335.84
3745.67
6.5606
0.009942
3283.63
3681.29
6.3750
800
0.015623
3555.60
4024.31
6.8332
0.011523
3517.89
3978.80
6.6662
900
0.017448
3768.48
4291.93
7.0717
0.012963
3739.42
4257.93
6.9150
1000
0.019196
3978.79 '
4554.68
7.2867
0.014324
3954.64
4527.59
7.1356
1100
0.020903
4189.18
4816.28
7.4845
0.015643
4167.38
4793.08
7.3364
1200
0.022589
4401.29
5078.97
7.6691
0.016940
4380.11
5057.72
7.5224
1300
0.024266
4615.96
5343.95
7.8432
0.018229
4594.28
5323.45
7.6969
688 m APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.i,4
Compressed Liquid Water
Temp.
V
it
h
s
V
it
h
s
(°C)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
fir T/1j-f¥ Tf\
500 kPa (151.86)
2000 kPa (212.42)
Sat.
A AH 1 AA1
639.66
640.21
1.8606
0.001177
906.42
908.77
2.4473
0.01
A AAAAfiO
0.0 1
0.51
0000
0.000999
0.03
2.03
0.0001
20
A AA1 f\f\1
83.91
84.41
0.2965
0.001001
83.82
85.82
.2962
; 40 ■;■
V 167.47
167.98
0.5722
0.001007
167.29 : ;
169.30
.5716
60
A AA1 ATT
O.uOlOl /
251.00 -
251.51
0.8308
0.001016
250.73
252.77
.8300
80
A AA 1 AIO
334.73 .■:
335.24
1.0749
0.001028
334.38 :
336.44
1.0739
100
A AAt f\A1
418.80
419.32
1.3065
0.001043
418.36
420.45
1.3053
120
A AA1 A£A
503.37
503.90
1.5273
0.001059
502.84
504.96
1.5259
140
A AA 1 C\Q(\
U.UU1U6U
588.66
589.20
1.7389
0.001079
588.02
590.18
1.7373
160
0.001101
674.14
676.34
1.9410
180
0.001 127
761.46
763.71
: 2.1382
200
001156
850.30
852.61
2.3301
5000 kPa (263.99)
10000 kPa (311.06)
Sat
0.001286
1 147.78
1154.21
2.9201
0.001452
1393.00
1407.53
3.3595
0.000998
0.03
; 5.02
0.0001
0.000995
0.10
10.05
0.0003
20
0.00 i 000
■■: ■ 83.64
88.64
0.2955
0.000997
83.35
93.32
0.2945
40
■ 0.001006
166.93
171.95
0.5705 ;
0.001003
166.33
176.36
0.5685
60
0.001015
250.21
255.28
0.8284
0.001013
249.34
259.47
0.8258
80
0.001027
333.69
338.83
1.0719
0.001025
332.56
342.81
1.0687
100
0.001041
417.50
422.71
1.3030
0.001039
416.09
426.48
1.2992
120
0.001058
501.79
507.07
1.5232
0.001055
500.07
510.61
1.5188
140
0.001077
586.74
: 592.13 ■
1.7342
.. 0.001074
584.67
595.40
1.7291
160
0.001099
672.61
678.10
1.9374
0.001195
670.11
681.07
1.9316
* 180
0.001124
759.62
765.24
2.1341
0.001120
756.63
767.83
2.1274
200
0.001153
848.08
853.85
2.3254
0.001148
844.49
855.97
2.3178
220
0.001187
938.43
944.36
2.5128
0.001181
934.07
945.88
2.5038
240
0.001226
1031.34
1037.47
2.6978
0.001219
1025.94
1038.13
2.6872
260
; : 0.001275 : :
1127.92 .
1134.30
2.8829
0.001265
1121.03
1133.68
2.869S
280
0.001322
1220.90
1234.11
3.0547
300
0.001397
1328.34
1342.31
3.2468
TABLE B.1,4 (continued)
Compressed Liquid Water
Temp.
V
u
h
s
V
it
ft
s
CC)
(m 3 /kg)
(kJ/kg)
(kJ/fcg)
(kJ/kg-K)
(mVkg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
15000 kPa (342.24)
20000 kPa (365.81)
Sat,
0.001658
1585.58
1610.45
3.6847
0.002035
1785.47
1826.18
4.0137
0.000993
0.15
15.04
0.0004
0.000990
0.20
20.00
0.0004
20
0.000995
83.05
97.97
0.2934
O.00O993
82.75
102.61
0.2922
40
0.001001
165.73
,"/ 180.75
; 0.5665 ■'
y 0.000999 ..
165.15
185.14
; 0.5646 :
60
0.001011
248.49
263.65
0.8231
0.001008
247.66
267.82
0.8205
80
0.001022
331.46
346.79
1.0655
0.001020
330.38
350.78
1.0623 ■
100
0.001036
414.72
430.26
1.2954
001034
413.37
434.04
1.2917
120
0.001052
498.39
514.17
1.5144
0.001050
496.75
517.74
1.5101
140
O.0OKI71
582.64
598.70
1.7241
0.001068
580.67
602.03
1.7192
ioU
A AA1 AAO
u.uuiuyz
667.69
684.07
t A*} CO
f\ flA 1 AOrt
665.34
687.11
1.9203
1 OA
loU
A AA 1 1 1 iC
U.uUli io
753.74
770.48
O 1 ")AA
z. Jzuy
a riA t 1 11
750.94 .
773.18
2.1146
zuu
fl AA1 1 A1
841.04
858.18
tit A1
A AA1 1 1 A
837.70
860.47
2.3031
Z2.U
A AA f 1 *1Z
U.UUI 1 /5
929.89
947.52
A AA1 1 £C\
925.89
949.27
2.4869
1A A
z4U
A AA1 O 1 1
U.UUJ2J 1
1020.82
1038.99
2,6770
A AA1 OAC
U.UU12U5
1015.94
1040.04
2.6673
1£A
ZOU
A AA lO^C
U.UU1235
1114.59
1133.41
Z.BO ID
A AA1 £.
1108.53
1133.45
2.8459
1QA
A AA1 *3 AO
1212.47
1232.09
i Aim
■ A A A 1 1 A*7
0.001297
1204.69
1230.62
3.0248
A AA1 T7*7
U.UUI3//
1316.58
1337.23
3.2259
A A A 1 'i cr\
1306.10
1333.29
3.2071
jzU
A AA 1 ^
0.0U1472
1431. Q5
1453.13
3.4z4o
n tif\ t A A A
0.001444
1415.66
1444.53
3.3978
A AA1 £1 1
U.UU10.J1
1567.42
1591.88
A AA1
U.UUiOOO
1539.64
1571.01
3.6074
A AAt O^l
U.UUlozi
1702.78
1739.23
3.8770
30000 kPa
50000 kPa
0.000986
0.25
29.82
0.0001
0.000977
0.20
49.03 ■
-0.0014
20
0.000989
82.16
111.82
0.2898
0.000980
80.98
130.00
0.2847
40
0.000995
164.01
193.87
0.5606
0.000987
161.84
211.20
0.5526
60
0.001004
246.03
276.16
0.8153
0.000996
242.96
292.77
0.8051
80
0.001016
328.28
358.75
1.0561
0.001007
324.32
374.68
1.0439
100
0.001029
410.76
441.63
1.2844
0.001020
405.86
456.87
1.2703
120
0.001044
493.58
524.91
1.5017
0.001035
487.63
539.37
1.4857
140
0.001062
576.86
608.73
1.7097
0.001052
569.76
622.33
1.6915
160
0.001082
660.81
693.27
1.9095
0.001070
652.39
705.91
1.8890
180
0.001105
745.57
778.71
2.1024
0.001091
735.68
790.24
2.0793
200
0.001130
831.34
865.24
2.2892
0.001115
819.73
875.46
2.2634
220
0.001159
918.32
953.09
2.4710
0.001141
904.67
961.71
2.4419
240
0.001192
1006.84
1042.60
2.6489
0.001170
990.69
1049.20
2.6158
260
0.001230
1097.38
1134.29
2.8242
0.001203
1078.06
1138.23
2.7860
280
0.001275
1190.69
1228.96
2.9985
0.001242
1167.19
1229.26
2.9536
300
0.001330
1287.89
1327.80
3.1740
0.001286
1258.66
1322.95
3.1200
320
0.001400
1390.64
1432.63
3.3538
0.001339
1353.23
1420.17
3.2867
340
0.001492
1501.71
1546.47
3.5425
0.001403
1451.91
1522.07
3.4556
360
■ 0.001627
1626.57
1675.36
3.7492
0.001484
1555.97
1630.16
3.6290
380
0.001869
1781.35
1837.43
4.0010
0.001588
1667.13
1746.54
3.8100 "
689
690 M APPENDIX B SIUNFTS: THERMODYNAMIC TABLES
TABLE B.1.5
Saturated Solid-Saturated Vap or, Water
Specific Volume, m 3 /kg Internal energy, kJ/kg
Temp.
Press.
Sat. Solid
Evap.
Sat. Vapor
Sat. Solid
Evap.
Sat. Vapor
(°C)
(kPa)
v t
"g
U.Ol I J
0010908
206.152
206.153
-333.40
2708.7
2375.3
u
U.D1UO
0010908
206.314
206.315
-333.42
2708.7
2375.3
— z
U.j 1 / /
00 1 0905
241.662
241.663
-337.61
2710.2
2372.5
„/(
0,4376
0.0010901
283.798
283.799
- 341.78
27U.5
2369.8
-6
0.3689 ;
0.0010898
334.138
334.139
-345.91 : v :
2712.9
2367.0 V.:
-8 :
0.3102
0.0010894
394.413
394.414 ■■■?
-350.02
2714.2
2364.2
-10
0.2601
0.0010891
466.756
466.757
-354.09
2715.5
2361.4
-12
0.2176
0.0010888
553.802
553.803
-358.14
2716.8
2358.7
-14
0.1815
0.0010884
658.824
658.824
-362.16
2718.0
2355.9
-16
0.1510
0.0010881
785.906
785.907
-366.14
: 2719.2
2353.1
— IS
0.1252
0.0010878
940.182
940.183
-370.10
2720.4
2350.3
-20
0.10355
0.0010874
1128.112
1128.113
-374.03
2721.6
2347.5
-22
0.08535
0.0010871
1357.863
1357.864
-377.93
2722.7
2344.7
-24
0.07012
0.0010868
1639.752
1639.753
-381.80
2723.7
2342.0
-26
0.05741
0.0010864
1986.775
1986.776
-385.64
2724.8
2339.2
-28
0.04684
0.0010861
2415.200
2415.201
-389.45
2725.8
2336.4
-30
0.03810
0.0010858
2945.227
2945.228
-393.23
2726.8
2333.6
-32
0.03090
0.0010854
3601.822
3601.823
-396.98
2727.8
2330.8
-34
0.02499
0.0010851
4416.252
4416.253
-400.71
2728.7
2328.0
-36
0.02016
0.0010848
5430.115
5430.116
-404.40
2729.6
2325.2
-38
0.01618
0.0010844
6707.021
6707.022
-408.06
2730.5
2322.4
-40
0.01286
0.0010841
_ . 8366.395
8366.396
-411.70
2731.3
2319.6
Appendix B SI Units: Thermodynamic Tables H 691
TABLE B.1.5 {continued)
Saturated Solid-Saturated Vapor, Water
ENTHALPY, kJ/kg ENTROPY, kJ/kg-K
Temp-
CQ
Press.
(kPa)
Sat. Solid
Evap.
K
Sat. Vapor
K
Sat. Solid
Evap.
Sat. Vapor
0.01
0.6113
-333.40
2834.7
2501.3
-1.2210
10.3772
9.1562
0.6108
-333.42
2834.8
2501.3
-1.2211
10.3776
9.1565
-2
0.5177
-337.61
2835.3
2497.6
-1.2369
10.4562
9.2193
-4
0.4376
-341.78 :
"2835.7
2494.0
-1.2526
10.5358
9.2832
0.3689
-345.91
2836.2 ;.
2490.3
-1.2683
10.6165
9.3482
-8 ^v.
0.3102
-350.02
2836.6
2486.6
-1.2839
10.6982
9.4143 "■
~10
0.2601
-354.09
2837.0
2482.9
-1.2995
10.7809
9.4815
™12
0.2176
-358.14
2837.3
2479.2
-1.3150
10.8648
9.5498
-14
0.1815
-362.16
2837.6
2475.5
-1.3306
10.9498
9.6192
-16
; 0.1510 .-.
-366.14
; : 2837.9
2471.8 .. ; ^
• 1.3461
11.0359
9.6898
-18
■ 0.1252
-370.10
■ 2838.2
2468.1 .:
-1.3617
li.1233
9.7616
-20
0.10355
-374.03
2838.4
2464.3
-1.3772
11.2120
9.8348
-22
0.08535
-377.93
2838.6
2460.6
-1.3928
11.3020
9.9093
-24
0.07012
-381.80
2838.7
2456.9
-1.4083
11.3935
9.9852
-26
0.05741
-385.64
2838.9
2453.2
-1.4239
11.4864
10.0625
-28 :
0.04684
-389.45
2839.0
'.. 2449.5
1.4394 : -
1 1.5808
10.1413
-30
0.03810
-393.23
2839.0
2445.8
-1.4550
; 11.6765
10.2215
-32
0.03090
-396.98
2839.1
2442.1 V
-1.4705
11.7733
10.3028
-34
0.02499
-400.71
2839.1
2438.4
-1.4860
11.8713
10.3853
-36
0.02016
-404.40
2839.1
2434.7
-1.5014
11.9704
10.4690
-38
0.01618
-408.06
2839.0
2431.0
-1.5168 .
12.0714
10.5546
-40
0.01286
-411.70
2838.9
2427.2
-1.5321
12.1768
10.6447
692 APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.2
Thermodynamic Properties of Ammonia
Table B.2.1
Saturated Ammonia
Specific Volume, m 3 /kg internal energy, kJ/kg
Temp.
Press,
Sat. Liquid
Evap.
Sat. Vapor
Sat. XjiQuid
,Sat Van
v f
v fs
11 r
f
JS
S
-50
40.9
0.001424
2.62557
2.62700
-43.82
1309.1
1265.2
-45
54.5
0.001437
2.00489
2.00632
-22.01
1293.5
1271.4
-40
71.7
0.001450-
1.55111
1.55256
-0.10
1277.6
1277.4
-35 ; .
93.2 ; ..
0.001463
: ; 1.21466 .,
1.21613
21.93
1261.3 V ;
; 1283.3
-30
119.5 '
0.001476 :
0.96192
0.96339
44.08
; 1244.8
1288.9
-25 ;
151.6 . ■;"
0.001490 ;
■ 0.76970
■ 0.77119 ; ;
66.36
1227.9
1294.3
-20
190.2
0.001504
0.62184
0.62334
88.76
1210.7
1299.5
-15
236.3
0.001519
0.50686
0.50838
111.30
1193.2
1304.5
-10
290.9
0.001534
0.41655
0.41808
133.96
1175.2
1309.2
-5
354.9
0.001550
0.34493
0.34648
156.76
1157.0
1313.7
429.6
0.001566
0.28763
: 0.28920
179.69
: 1138.3
; 1318.0
5
515.9
0.001583
0.24140
0.24299
202.77
1119.2
1322.0
10
615.2
0.001600
0.20381
0.20541
225.99
1099.7
1325.7
15
728.6
0.001619
0.17300
0.17462
249.36
1079.7
1329.1
20
857.5
0.001638
0.14758
0.14922
272.89
1059.3
1332.2
25
1003.2
0.001658
0.12647
0.12813
296.59
1038.4
1335.0
30
1167.0
0.001680
0.10881
0.11049
320.46
1016.9
1337.4
35
1350.4
0.001702
0.09397
0.09567
344.50
994.9
1339.4
40
1554.9
0.001725
0.08141
0.08313
368.74
972.2
1341.0
45
1782.0
0.001750
0.07073
0.07248
393.19
948.9
1342.1
50
2033.1
0.001777
0.06159
0.06337
417.87
924.8
1342.7
55
2310.1
0.001804
0.05375
0.05555
442.79
899.9
1342.7
60
2614.4
0.001834
0.04697
0.04880
467.99
874.2
1342.1
65
2947.8
0.001866
0.04109
0.04296
493.51
847.4
1340.9
70
3312.0
0.001900
0.03597
0.03787
519.39
819.5
1338.9
75
3709.0
0.001937
0.03148
0.03341
545,70
790.4
1336J
80
4140.5
0.001978
0.02753
0.02951
572.50
759.9
1332.4
85
4608.6
0.002022
0.02404
0.O2606
599.90
727.8
1327.7
90
5115.3
: 0.002071
0.02093
0.02300
627.99
693.7
1321.7
95
5662.9
0.002126
0.01815
0.02028
656.95
657.4
1314.4
100
6253.7
0.002188
0.01565
0.01784
686.96
618.4
1305.3
105
6890.4
0.002261
0.01337
0.01564
718.30
575.9
1294.2
110
7575.7
0.002347
0.01128
0.01363
751.37
529.1
1280.5
115
8313.3
0.002452
0.00933
0.01178
786.82
476.2
1263.1-
120
9107.2
0.002589
0.00744
0.01003
825.77
414.5
1240.3
125
9963.5
0.002783
0.00554
0.00833
870.69
337.7
1208.4
130
10891.6
0.003122
0.00337
0.00649
929.29
226.9
1156.2
132.3
11333.2
0.004255
0.00426
1037.62
1037.6
Appendix B si units: thermodynamic tables M 693
TABLE B.2.1 {continued)
Saturated Ammonia
Enthalpy, kJ/kg Entropy, kJ/kg-K
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
CQ
(kPa)
*/
h /$
*f
s f
s fi
-50
40.9
-43.76
1416.3
1372.6
-0.1916
6.3470
6.1554
-45
54.5
-21.94
1402.8
1380.8
-0.0950
6.1484
6.0534
-40
71.7
1388.8
1388.8
5.9567
5.9567
-35
93.2
22.06 \
■\ 1374.5
1396.5
v.-;,' 0.0935
5.7715
5.8650 \ :
-30
119.5
44.26 : ;:
1359.8
■ 1404.0, y::y\:
■ v; 0.1856
5.5922 .
5.7778
-25
151.6
66.58
1344.6 ; : V:
14H.2 ':; :■
0.2763 :.
5.4185 :
5.6947 ■:
-20
190.2
89.05
1329.0
1418.0
0.3657
5.2498
5.6155
-15
236.3
111.66
1312.9
1424.6
0.4538
5.0859
5.5397
-10
290.9
134.41
1296.4
1430.8
0.5408
4.9265
5.4673
■ -5 ■ V/
354.9
157.31
1279.4
1436.7 ■
0.6266
4.7711 : \
5.3977 ■
o
429.6
180.36
1261.8
1442.2 .
0.7114
4.6195 v
■ 5.3309
5
515.9 ;
203.58
1243.7
1447.3
: 0.7951
4.4715
5.2666
10
615.2
226.97
1225.1
1452.0
0.8779
4.3266
5.2045
15
728.6
250.54
1205.8
1456.3
0.9598
4.1846
5.1444
20
857.5
274.30
1185.9
1460.2
1.0408
4.0452
5.0860
25
1003.2
.298.25
; 1165.2
1463.5 ; ;
1.1210
3.9083
5.0293
30
1167.0
322.42
1143.9
1466.3
1.2005 V
3.7734
4.9738
35
1350.4
346.80
1121.8 :
1468.6 7
1.2792 ■
3.6403
4.9196
40
1554.9
371.43
1098.8
1470.2
1.3574
3.5088
4.Z662
45
1782.0
396.31
1074.9
1471.2
1.4350
3.3786
4.8136
50
2033.1
421.48
1050.0
1471.5
1.5121
3.2493
4.7614
55
2310.1
446.96
1024.1
1471.0
1.5888
3.1208
4.7095
60
2614.4
472.79
997.0
1469.7
1.6652
2.9925
4.6577
65
2947.8
499.01
968.5 :
1467.5
1.7415
2.8642
4.6057
70
3312.0
525.69
938.7
1464.4
1.8178
2.7354
4.3533
75
3709.0
552.88
907.2
1460.1
1.8943
2.6058
4.5001
80
4140.5
580.69
873.9
1454.6
1.9712
2.4746
4.4458
85 "
4608.6
609.21
838.6
1447.8
: 2.0488
2.3413
4.3901
90
5115.3
638.59
800.8
1439.4
2.1273
2.2051
■ 4.3325 ■;
95
5662.9
. 668.99
760.2
1429.2 ■
■ 2.2073
2.0650 :
■ 4.2723
100
6253.7
700.64
716.2
1416.9
2.2893
1.9195
4.2088
105
6890.4
733.87
668.1
1402.0
2.3740
1.7667
4.1407
110
7575.7
769.15
614.6
1383.7
2.4625
1.6040
4.0665
115
; 8313.3
807.21
: ■ 553.8 ..
1361.0 :
2.5566
1.4267 ■;:
3.9833
120 :
I 9107.2
\ 849.36 . V :
■ ■[ 482.3 .
1331.7 V :
2.6593
1.2268
3.8861
125 i- : :
9963.5
898.42
393.0
1291.4
/ 2.7775 : .
0.9870
3.7645
130
10892
963.29
263.7
1227.0
2.9326
0.6540
3.5866
132.3
11333
1085.85
1085.9
3.2316
3.2316
694 H APPENDIX B SI UNITS: THERMODYNAMIC TABLES
TABLE B.2.2
Superheated Ammonia
Temp.
V
it
h
s
V
u
ii
s
\ w
/ T11 3/l/ fI \
\m /Kg}
CkJAtel
50 kPa (-
-46.53)
100 kPa (-
-33.60)
Sat.
2.1752
1269.6
1378.3
6.0839
1.1381
1284.9
1398.7
5.8401
-30
2.3448
1296.2
1413.4
6.2333
1.1573
1291.0
1406.7
5.8734
-20
2.4463
1312.3
1434.6
6.3187
1.2101
1307.8
1428.8
5.9626
-10
2.5471
: 1328.4 :
1455.7
6.4006 : ;
1.2621
1324.6
1450.8 ■
6.0477
'i
2.6474 V
.': 1344.5
1476.9
6.4795
1.3136 : ' . ; . .
1341.3
1472.6
6.1291
10
2.7472 ■ -
1360.7 ;:
1498.1
6.5556 \ )
1.3647 A
1357.9
1494.4
6.2073 :
20
2.8466
1377.0
1519.3
6.6293
1.4153
1374.5
1516.1
6.2826
30
2.9458
1393.3
1540.6
6.7008
1.4657
1391.2
1537.7
6.3553
40
3.0447
1409.8
1562.0
6.7703
1.5158
1407.9
1559.5
6.4258
50
3.1435
v 1426.3 ;
1583.5
6.8379
1.5658
1424.7
1581.2
6.4943
60
3.2421
1443.0
1605.1 .
6.9038 .
1.6156
■ 1441.5
1603.1
6.5609
70
3.3406
1459.9
1626.9
6.9682
1.6653
1458.5
1625.1
\ 6.6258
SO
3.4390
1476.9
1648.8
7.0312
1.7148
1475.6
1647.1
6.6892
100
3.6355
1511.4
1693.2
7.1533
1.8137
1510.3
1691.7
6.8120
120
3.8318
1546.6
1738.2
7.2708
1.9124
1545.7
1736.9
6.9300
140
4.0280
1582.5
1783.9
7.3842
2.0109
1581.7
1782.8
7.0439
160
4.2240
1619.2
1830.4
7.4941 /
; 2.1093
■ 1618.5
1829.4
7.1540 :■
180
4.4199
1656.7
1877.7
/ 7.6008
2.2075
1656.0
1876.8
7.2609
200
4.6157
1694.9
1925.7
7.7045
2.3057
1694,3
1924.9
7.3648
150 kPa (-
25.22)
200 kPa (-
-18.86)
Sat.
0.7787
1294.1
1410.9
5.6983
0.5946
1300.6
1419.6
5.5979
-20
0.7977
1303.3
1422.9
5.7465
IU
n on £
U.OJJD
lion 7
0.6193
1316.7
1440.6
5.6791
0.8689
1337.9
1468.3
5.9189
0.6465
1334.5 :
1463.8
5.7659
10
0.9037
1355.0
1490.6
5.9992
0.6732
1352.1
3486.8
5.8484
20
0.9382
1372.0
1512.8
6.0761
0.6995
1369.5
1509.4
5.9270
30
0.9723
1389.0
1534.9
6.1502
0.7255
1386.8
1531.9
6.0025
40
1.0062
1406.0
1556.9
6.2217
0.7513
1404.0
1554.3
6.0751
50
1.0398
1423.0
1578.9
: 6.2910
0.7769
■ i42i.3 ■■
1576.6 :
6.1453 ;
60
1.0734
1440.0
4601.0
6.3583 : -
0.8023
1438.5
1598.9
6.2133
70
1.1068
1457.2
1623.2
6.4238
0.8275
1455.8
1621.3
6.2794
80
1.1401
1474.4
1645.4
6.4877
0.8527
1473.1
1643.7
6.3437
100
1.2065
1509.3
1690.2
6.6112
0.9028
1508.2
1688.8
6.4679
120
1.2726
1544.8
1735.6
6.7297
0.9527
1543.8
1734.4
6.5869
140
1.3386
1580.9
1781.7 :
6.8439 :
1.0024
1580.1
. 1780.6
6.7015
160
1.4044
1617.8
1828.4
. 6.9544 '■ ■" ■
1.0519
1617.0
1827.4
■ 6.8123 7
180
1.4701
1655.4
1875.9
7.0615
1.1014
1654.7
1875.0
6.9196
200
1.5357
1693.7
1924,1
7.1656
1.1507
1693.2
1923.3
7.0239
220
1.6013
1732.9
1973.1
7.2670
1.2000
1732.4
1972.4
7.1255
APPENDIX B SI UNITS: THERMODYNAMIC TABLES B 695
Table B.2.2 {continued)
Superheated Ammonia
Temp.
V
u
/i
s
V
u
h
s
(°C)
(m 3 /kg)
(kj/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
{kJ/kg-K)
300kPa(
-9.24)
400 kPa (-
-1.89)
Sat
0.40607
1309.9
1431.7
5.4565
0.30942
1316.4
1440.2
5.3559
0.42382
1327.5
1454.7
5.5420
0.31227
1320.2
1445.1
5.3741
10
0.44251
1346.1
1478.9
5.6290
0.32701
1339.9
1470.7
5.4663
20
0.46077
1364.4
1502.6
5.7113
0.34129
1359.1
1495.6
5.5525 ■.
30
0.47870
1382.3
1526.0
5.7896
0.35520
1377.7
1519.8
5.6338
40
0,49636
1400.1
1549.0 :
ft 1fi884
1396.1
1543.6
< 71 1 t
J.I 111
50
1417.8
1571.9
U.JSZZO
1414.2
1567.1
j./ojU
60
0.531 1 1
1435.4
1594.7
u.uuuu
U.J7JJU
1432.2
1590.4
j.ojDU
70
0.54827
1453.0
1617.5
6.0732
UnUOUU
1450.1
1613.6
D.yZ i ¥t
80
56532
1470.6
1640.2
6.1385
f\ d?ifin
1468.0
1636.7
c QQ/V7
100
0.59916
1506.1 ,".':■./■
1685.8
6.2642
0.44732
1503.9 V.-.
1682.8
& 1 1 70
120
0.63276
1542.0
1731.8
6.3842
ft 47979
1540.1
1729.2
f, 9ion
140
66618
1578.5
1778.3
6.4996
ft 4Q8DR
1576.8
1776.0
O.J JJZ
160
0.69946
1615.6
1825.4
O.Ol Kfy
ft 'iOIOI
1614.1
1823.4
0.40 / 1
180
0.73263
1653.4
1873.2
6.7188
ft *!4Si?7
1652.1
1871.4
f. <7S<
O.J / JJ
200
O 7657?
1692.0
1921.7
. U.OZ.J.J ■ ■ .
ft ^7^91
. 1690.8
1920.1
p.OotlO
220
: 0.79872
1731.3 ■
1970.9
6.9254
ft SQSflQ
1730.3
1969.5
O. / OZO
240
0.83167
1771.4 :'\):
2020.9
7 ft?47
: / .Wit /
ft 6?9S0
1770.5
2019.6
O.OOZJ
260
86455
1812.2
2071.6
7.1217
ft f&lfiA
1811.4
2070.5
c Q707
500 kPa(4.13)
600 kPa (9.28)
Sat.
0.25035
1321.3
1446.5
5.2776
0.21038
1325.2
1451.4
5.2133
10
0.25757
1333.5
1462.3
5.3340
0.21115
1326.7
1453.4
5.2205
20
0.26949
1353.6
1488.3
5.4244
0.22154
1347.9
1480.8
5.3156
30
0.28103
1373.0
1513.5
5.5090
0.23152
1368.2
1507.1
5.4037
40
0.29227
1392.0
1538.1
5.5889
0.24118
1387.8
1532.5
5.4862
50
0.30328
1410.6
1562.2
5.6647
0.25059
1406.9
1557.3
5.5641
60
0.31410
1429.0
1586.1
5.7373
0.25981
1425.7
1581.6
5.6383
70
0.32478
1447.3
1609.6
5.8070
0.26888
1444.3 . ..
1605.7
5.7094
80
0.33535
1465.4
1633.1
5.8744 *
0.27783
1462.8
1629.5
5.7778
100
0.35621
1501.7
1679.8
6.0031
0.29545
1499.5
1676.8
5.9081
120
0.37681
1538.2
1726.6
6.1253
0.31281
1536.3
1724.0
6.0314
140
0.39722
1575.2
1773.8
6.2422
0.32997
1573.5
1771.5
6.1491
160
0.41748
: ; 1612.7 :
1821.4 ;
6.3548
0.34699
1611.2
1819.4
6.2623 :
180
0.43764
1650.8 .
1869.6
6.4636
0.36389
1649.5
1867.8
6.3717
200
0.45771
1689.6
1918.5
6.5691
0.38071
1688.5
1916.9
6.4776
220
0.47770
1729.2
1968.1
6.6717
0.39745
1728.2
1966.6
6.5806
240
0.49763
1769.5
2018.3
6.7717
0.41412
1768.6
2017.0
6.6808
260
0.51749
1810.6
2069.3
6.8692
0.43073
1809.8
2068.2
6.7786
696 H Appendix B SI Units: Thermodynamic Tables
TABLE B.2.2 {continued)
Superheated Ammonia
Temp. v u h s v u h s
(°C) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K) (m 3 /kg> (kJ/kg) (kJ/kg) (kJ/kg-K)
800 kPa (17,85) 1000 kPa (24.90)
Sat.
0.15958
1330.9
1458.6
5.1110
0.12852
1334.9
1463.4
5.0304
20
0.16138
1335.8
1464.9
5.1328
30
0.16947
1358.0
1493.5
5.2287
0,13206
1347.1
1479.1
5.0826
40
0.17720
1379.0
1520.8
5.3171 ;
0.13868
1369.8
1508.5 ..
: 5.1778
50
0.18465
1399.3
1547.0 /;
5.3996
0.14499
/ 1391.3 \: :
1536.3
5.2654
60
0.19189
1419.0
1572.5 ;
5.4774
0.15106
1412.1
1563.1
5.3471
70
0.19896
1438.3
1597.5
5.5513
0.15695
1432.2
1589.1
5.4240
80
0.20590
1457.4
1622.1
5.6219
0.16270
1451.9
1614.6
5.4971
100
0,21949
1495.0
1670.6
5.7555
0.17389
1490.5
1664.3
5.6342
120
0.23280
1532.5
1718.7
5.8811
0.18477
1528.6
1713.4
5.7622
140
0.24590
1570.1
1766.9
6.0006
0.19545
1566.8 /
1762.2
5.8834
160
0.25886
1608.2
1815.3
6.1150
0.20597
1605.2
1811.2 ..
5,9992
180
0.27170
1646.8
1864.2
6.2254
0.21638
1644.2
1860.5
6.1105
200
0.28445
1686.1
1913.6
6.3322
0.22669
1683.7
1910.4
6.2182
220
0.29712
1726.0
1963.7
6.4358
0.23693
1723.9
1960.8
6.3226
240
0.30973
1766.7
2014.5
6.5367
0.24710
1764.8
2011.9
6.4241
260
0.32228
1808.1
2065.9
6.6350
0.25720
1806.4
2063.6
6.522y
280
0.33477
1850.2
2118.0
6.7310
0.26726
1848.8
2116.0
6.6194
300
0.34722
1893.1
2170.9
6.8248
0.27726
1891.8
2169.1
6.7137
1200 kPa (30.94)
1400 kPa (36.26)
Sat.
0.10751
1337.8
1466.8
4.9635
0.09231
1339.8
1469.0
4.9060
40
0.11287
1360.0
1495.4
5.0564
0.09432
1349,5
1481.6
4.9463
50
0.11846
1383.0
1525.1
5.1497
0.09942
1374.2
1513.4
5.0462
60
0.12378
1404.8
1553.3
5.2357
0.10423
1397.2
1543.1
5.1370
70
0.12890
1425.8
1580.5
5.3159
0.10882
1419.2
1571,5
5.2209
80
0.13387
1446.2
1606.8
5.3916
0.11324
1440.3
1598.8
5.2994
100
0.14347
1485.8 :
1658.0
5.5325
0.12172
1481.0
1651.4
5.4443
120
0.15275
1524.7
1708.0
5.6631
0.12986
1520.7
1702.5
5.5775
140
0.16181
1563.3
1757.5
5.7860
0.13777
1559.9
1752.8
5.7023
160
0.17071
1602.2
1807.1
5.9031
- 0.14552
1599.2
1802.9
5.8208
180
0.17950
1641.5
1856.9 :
6.0156
0.15315
1638.8
1853.2
5.9343
200
0.18819
1681.3
1907.1
6.1241
0.16068
1678.9
; 1903.8
6.0437
220
0.19680
: 1721.8
1957.9
6.2292
:' 0.16813
1719.6
1955.0 :
6.1495
240
0.20534
1762.9
2009,3
6.3313
0.17551
1761.0
2006.7
6.2523
260
0.21382
1804.7
2061.3
6.4308
0.18283
1803.0
2059.0
6.3523
280
0.22225
1847.3
2114.0
6.5278
0.19010
1845.8
2111.9
6.4498
300
0.23063
1890.6
2167.3
6.6225
0.19732
1889.3
2165.5
6.5450
320
0.23897
1934.6 :
2221.3
6.7151
0.20450
1933.5
2219.8
6.6380
APPENDIX B SI UNITS: THERMODYNAMIC TABLES 1 697
TABLE B.2.2 {continued)
Superheated Ammonia
Temp.
V
u
h
$
V
u
h
s
( m
(m /kg)
(kJ/kg)
(kj/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-
1600 fcPa (41.03)
2000 kPa (49.37)
Sat.
0.08079
1341.2
1470.5
4.8553
0.06444
1342.6
1471.5
H . /OoU
50
0.08506
1364.9
1501.0
4.9510
0.06471
1344.5
1473.9
4.7754
60
0.08951
1389.3
1532.5
5.0472
0.06875
1372.3
1509.8
4.8848
70
0.09372
1412.3 ■
1562.3
5,1351
fl (WOAfi
U.U/i'iO
1397.8
1542.7
4.9821
80 . ..
0.09774
1434.3
1590.6
5.2167
0.07595
1421.6
1573.5
5.0707
100
0.10539
1476.2
1644.8
5.3659
0.08248
1466.1 ■ ■"
1631.1
5.2294
200
0.11268
1516.6
1696.9
5.5018
0.08861
1508.3
1685,5
5.3714
140
0.11974
1556.4
1748.0
5.6286
0.09447
1549.3
1738.2
5.5022
160
0.12662
1596.1
1798.7
5,7485
0.10016
1589.9
1790.2
5.6251
180
0.13339
1636.1
1849.5
5.8631
0.10571
1630.6
1842.0
5.7420
200
. 0.14005
1676.5
1900.5
5.9734
0.11116
1671.6
1893.9
5.8540
220
0.14663
1717.4
1952.0
6.0800
0.11652
1713.1
1946.1
5.9621
240
0.15314
1759.0
2004.1
6.1834
0.12182
1755.2
1998.8
6.0668
260
0.15959
1801.3
2056.7
6.2839
0.12705
1797.9
2052.0
6.1685
280
0.16599
1844.3
2109.9
6.3819
0.13224
1841.3
2105.8
6.2675
300
0.17234
1888.0 ;
2163.7
6.4775
0.13737
1885.4
2160.1
6.3641
320
0.17865
1932.4
2218.2
6.5710
0.14246
1930.2
2215.1
6.4583
340
0.18492
1977.5
2273.4
6.6624
0.14751
1975.6
2270.7
6.5505
360
0.19115
2023.3
2329.1
6.7519
0.15253
2021.8
2326.8
6.6406
5000 kPa (88.90)
10000 kPa (125.20)
Sat.
0.02365
1323.2
1441.4
4.3454
0.00826
1206.8
1289.4
J. fDof
100
0.02636
1369.7
1501.5
4.5091
120
0.03024
1435.1
1586.3
4.7306
140
0.03350
1489.8
1657.3
4.9068
0.01195
1341.8
1461.3
4.1839
160
0.03643
1539.5
1721.7
5.0591
0.01461
1432.2
1578.3
4.4610
180
0.03916
1586.9
1782.7
5.1968
0.01666
1500.6
1667.2
4.6617
200
0.04174
1633.1
1841.8
5.3245
0,01842
1560.3
1744.5
4.8287
220
0.04422
1678.9
. 1900.0
5.4450 '
0.02001
1615.8
1816.0
4.9767
240
0.04662
1724.8
1957.9
5.5600
0.02150
1669.2
1884.2
5.1123
260
0.04895
1770.9
2015.6
5.6704
0.02290
1721.6
1950.6
5.2392
280
0.05123
1817.4
2073.6
5.7771
0.02424
1773.6
2015.9
5.3596
300
0.05346
1864.5
2131.8
5.8805
0.02552
1825.5
2080.7
5.4746
320
0.05565
1912.1
2190.3
5.9809
0.02676
1877.6
2145.2
5.5852
340
0.05779
1960.3
2249.2
6,0786
0.02796
1930.0
2209.6
5.6921
360
0.05990
2009.1
2308.6
6.1738
0.02913
1982.8
2274.1
5.7955
380
0.06198
2058.5
2368.4
6.266S
0.03026
2036.1
2338.7
5.8960
400
0.06403
2108.4
2428.6
6.3576
0.03137
2089.8
2403.5
5.9937
420
0.06606
2159.0
2489.3
6.4464
0.03245
2143.9
2468.5
6.0888
440 :
0.06806
2210.1
2550.4
6.5334
0.03351
2198.5
2533.7
6.1815
]
698 H APPENDIX B SI UNITS: THERMODYNAMIC TABLES |
Table B.3 \
Thermodynamic Properties ofR-12
Table B.3.1
Saturated R-12
Specific Volume, m 3 /kg Internal Energy, kJ/kg
Temp.
Press.
Sat. Liquid
Evap,
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
(°C)
(kPa)
v f
V A
"ft
" s
— yu
R
O 000608
4.41494
4.41555
-43.29
177.20
133.91
— isu
0006 1 7
2.13773
2.13835
-34.73
172.54
137.82
— 7n
IU
17 1
000627
1.12665
1.12728
-26.14
167.94
141.81
— Aft
ou
00 fx
0.000637
0.63727 :
0.63791
-17.50
163.36 V.
145.86
.Cf>
-1Q 1
.J J. I
000648
; : 0.38246
0.38310
-8.80
158.76
149.95
AG
f A A
0.000654
0.30203
0.30268
-4.43
156.44
152.01
— 4fl
fid 9
0.000659
0.24125
0.24191
-0.04
154.11
154.07
IS
000666
0.19473
0.19540
4.37
151.77
156.13
1 fin d
000672
0.15870
0.15937
8.79
149.40
158.19
— TO R
1A1 "J
000672
0.15736 \
0.15803 VV
8.98 ;. ;
149.30
158.28
123.7
000679
0.13049
0.13117 :
; 13.24
147.01
160.25
— zu
1 so
0.000685
0.10816
0.10885
17.71
144.59
162.31
K
— ID
ISO (\
000693
0.09033
0.09102
22.20
142.15
164.35
— 1 A
000700
0.07595
0.07665
26.72
139.67
166.39
om n
0.000708
0.06426
0.06496
31.26
137.16
168.42
. JUu.U
0.000716
0.05467 :
0.05539
35.83 ; : :
134.61
170.44
s
3fi? ft
000724
0.04676
0.04749
■ 40.43 ■
132.01
: . 172.44 :
1 n
423.3
0.000733 ;
0.04018
0.04091 V
. 45.06 . ■■' .;
129.36
174.42
15
491.4
0.000743
0.03467
0.03541
49.73
126.65
176.38
20
567.3
0.000752
0.03003
0.03078
54.45
123.87
178.32
25 ■
651.6
0.000763
0.02609
0.02685
59.21
121.03
180.23
30
... 744.9
0.000774 . .
; 0.02273
0.02351
64.02
118.09
182.11
35
847.7 -
0.000786
. : 0.01986
0.02064
68.88
115.06
183.95
40 .
960.7
0.000798 ;
■ : 0.01737 : ■
0.01817
73.82
111.92
185.74 \
45
1084.3
0.000811
0.01522
0.01603
78.83
108.66
187.49
50
1219.3
0.000826
0.01334
0.01417
83.93
105.24
189.17
55
1366.3
O.00O841
0.01170
0.01254
89.12
101.66
190.78
60
1525.9
0.000858
0.01025
0.01111
94.43
97.88 :
192.31
65
1698.8
0.000877
0.00897
0.00985
99.87
93.86
193.73 ...
70
1885.8 ■ ;
0.000897
0.00783
0.00873
105.46
89.56
195.03
75
2087.5
0.000920
O.O068O
0.00772
111.23
84.94
196.17
80
2304.6
0.000946
0.00588
0.00682
117.21
79.90
197.11
85
2538.0
0.000976
0.00503
0.00600
123.45
74.34
197.80
90
2788.5
0.001012
0.00425
0.00526
■ . 130.02
68.12
198.14
.95
/ 3056.9
0.001056
0.00351
0.00456
mm
60.98 -
\ 197.99 \,
100
3344.1 :
0.001113
0.00279
0.00390
144.59
V. 52.48
197.07
105
3650.9
0.001197
0.00205
0.00324
153.15
41.58
194.73
110
3978.5
0.001364
0.001 10
0.00246
164.12
24.08
188.20
112.0
4116.8
0.001792
0.00179
176.06
176.06
Appendix B SI Units: thermodynamic Tables El 699
TABLE B.3.1 {continued)
Saturated R-12
ENTHALPY, kJ/kg ENTROPY, kJ/kg-K
Temp.
CC)
Press.
(kPa)
Sat, Liquid
hf
Evap.
K
Sat Vapor
*.
Sat, Liquid
s f
Evap.
s /s
Sat. Vapor
s s
-90
2.8
-43.28
189.75
146.46
-0.2086
1.0359
0.8273
-80
6.2
-34.72
185.74
151.02
-0.1631
0.9616
0.7984
~70
12.3
-26.13
181.76
155.64
-0.1198
0.8947
0.7749
•60
V 22.6
.: -17.49
177.77
160.29
: -0.0783
0.8340
0.7557
-50
39.1
--8.78
173.73 : :
164.95
-0.0384
0.7785
0.7401
50.4 V
-4.40
171.68
167.28 :
-0.0190
0.7524
0.7334
-40
64.2
169.59
169.59
0.7274
0.7274
-35
80.7
4.42
167.48
171.90
0.0187
0.7032
0.7219
-30
100.4
8.86
165.34
174.20
0.0371
0.6799
0.7170
-29.8
101.3 .
9.05
165.24 /
,: 174.29 : :
0.0379
0.6790
0.7168
-25
123.7
13.33
163.15 ;
176.48 : .
0.0552
0.6574
0.7126
-20 :
150.9
17.82
160.92
/ 178.74
0.0731
0.6356
0.7087
-15
182.6
22.33
158.64
180.97
0.0906
0.6145
0.7051
-10
219.1
26.87
156.31
183.19
0.1080
0.5940
0.7019
-5
261.0
31.45
153.93
185.37
0.1251
0.5740
0.6991
308.6
36.05
151.48
187.53
0.1420
0.5545
0.6965
5 , ;
362.6
40.69
148.96
189.65 .
0.1587
0.5355
0.6942
10
423.3
45.37 ..
146.37
191.74
0.1752
0.5169
0.6921
15
491.4
50.10
143.68
193.78
0.1915
0.4986
0.6902
20
567.3
54.87
140.91
195.78
0.2078
0.4806
0.6884
25
651.6
59.70
138.03
197.73
0.2239
0.4629
0.6868
30
: 744.9
64.59
135.03
199.62
0.2399
0.4454
0.6853
35
847.7
. . 69.55 .
131.90
201.45
0.2559
0.4280
0.6839
40
960.7
74.59
128.61
203.20
0.2718
0.4107
0.6825
45
1084.3
79.71
125.16
204.87
0.2877
0.3934
0.6811
50
1219.3
84.94
121.51
206.45
0.3037
0.3760
0.6797
55
1366.3
90.27
117.65
207.92
0.3197
0.3585
0.6782
60
1525.9 . .
95.74
H3.52 .:
209.26
0.3358
0.3407
0.6765
65
1698.8
101.36
109.10
210.46
0.3521
0.3226
0.6747
70
1885.8
107.15 ..'
104.33
211.48
0.3686 .-
0.3040
0.6726
75
2087.5
113.15
99.14
212.29
0.3854
0.2847
0.6702
80
2304.6
119.39
93.44
212.83
0.4027
0.2646
0.6672
85
2538.0
125.93
87.11
213.04
0.4204
0.2432
0.6636
90
2788.5
■." 132.84 . .
79.96
212.80
0.4389
0.2202
0.6590
95 .
3056.9
140.23
71.71 ;
211.94
0.4583
0.1948
0.6531
100
3344.1 :.
148.31 .\. ;
61.81
210.12
0.4793
0.1656
0.6449
105
3650.9
157.52
49.05
206.57
0,5028
0.1297
0.6325
110
3978.5
169.55
28.44
197.99
0.5333
0.0742
0.6076
112.0
4116.8
183.43
183.43
0.5689
0.5689
700 B APPENDIX B SI UNITS: THERMODYNAMIC TABLES
TABLE B.3.2
Superheated R-12
Temp.
V
u
h
s
V
it
h
(°C)
(mVkg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
50 kPa (-
45.18)
100 kPa (-
-33.10)
Sat.
0.30515
151.94
167.19
0.7336
0.15999
158.15
174.15
0.7171
-30
0.32738
159.18
175.55
u./oyi
158.20
174.21
u. / 1 m
-20
0.341S6
164.08
181.17
0.7917
0.16770
163.22
179.99
0.7406
-io : : ;
0.35623
169.08
186.89
: 0.8139
0.17522
168.32
185.84
0.7633
0.37051
174.18
192.70
0.8356
0.18265
173.50
191.77
0.7854
io v'.
0.38472
179.38
198.61
0.8568
0.18999
178.77
197.77
0.8070
20
0.39886
184.67
204.62
0.8776
0.19728
184.13
203.85
0.8281
30
0.41296
190.06
210.71
0.8981
0.20451
189.57
210.02
0.8488
40
0.42701
195.54
216.89
0.9181
0.21169
195.09
216.26
0.8691
50
0.44103
201.11 ;
223.16
0.9378 7
0.21884
200.70 ; :
222.58 :
0.8889
60
0.45502
206.76 .
229.51
0.9572
0.22596 :
206.39 : . :
228.98
0.9084 [
70
0.46898
212.50 \
235.95
0.9762 : " :
0.23305 ;
212.15 ..■;.;/
235.46
0.9276 7
80
0.48292
218.31
242.46
. 0.9949
0.24011
218.00
242.01
0.9464
90
0.49684
224.21
249.05
1.0133
0.24716
223.91
248.63
0.9649
100
0.51074
230.18
255.71
1.0314
0.25419
229.90
255.32
0.9831
110
0.52463
: 236.22
262.45
1.0493 :;
0.26121
235.96
262.08
1.0009
120
0.53851 ;
242.33
269.26
1.0668 v;
0.26821
242.09 : :
268.91
1.0185 :
200 kPa (•
-12.53)
400 kPa (8.15)
Sat.
0.08354
165.36
182.07
0.7035
0.04321
173.69
190.97
0.6928
0.08861
172.08
189.80
0.7325
10
0.09255
177.51
196.02
0.7548
0.04363
174.76
192.21
0.6972
20 -
0.09642
183.00 :
202.28
0.7766 .
0.04584
: 180.57 ; ;
198.91
0.7204
30
0.10023
188.55
208.60
0.7978
0.04797
186.39
205.58
0.7428
* 40
0.10399
194.17
214.97
0.8184
0.05005
192.23
212.25
0.7645
50
0.10771
199.86
221.41
0.8387
0.05207
198.11
218.94
0.7855
60
0.11140
205.62
227.90
0.8585
0.05406
204.03
225.65
0.8060
70
0.11506
211.45
234.46
0.8779 :
0.05601
209.99
232.40
0.8259
80
0.11869
217.35
241.09
0.8969
0.05794
216.01
239.19
0.8454
90
0.12230
223.31
247.77
0.9156 : y.
0.05985 ;
: ; : 222.08
246.02
0.8645
100
0.12590
229.34
254.53
0.9339
0.06173
228.20
252.89
0.8831
110
0.12948
235.44
261.34
0.9519
0.06360
234.37
259.81
0.9015
120
0.13305
241.60
268.21
0.9696
0.06546
240.60
266.79
0.9194
130
0.13661
.. 247.82
275.15
..: 0.9870 ./■.■■/■
0.06730 :
246.89
273.81
0.9370
140 :
0.14016
254.11
282.14
1.0042 ■ •
0.06913 ;
253.22
280.88
0.9544
ISO
0.14370
260.45 ;
289.19 :
1.0210
: 0.07095 V-
: \ 259.61
287.99
} 0.9714
Appendix B SI Units: Thermodynamic Tables H 701
TABLE B.3.2 {continued)
Superheated R-12
Temp.
V
it
h
5
V
u
h
s
CO
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
500 kPa (15.60)
1000 kPa (41.64)
Sat.
0.03482
176.62
194.03
0.6899
0.01744
186.32
203.76
A £OTA
30
0,03746
185.23
203.96
0.7235
40
0.03921
191.20
210.8.1
0.7457
—
50
ft ft4flQ I
197.19
217.64
U. ib/Z
0.01837
191.95
210.32
0.7026
203.20
224.48
0.7881
0.01941
198.56
217.97
0.7259
70
209.24
231.33
0,02040
205.09
225.49
0.7481
SO
04S77
215.32
238.21
v.&Zal
0.02 1 34
211.57
232.91
0.7695
90
0.04734
221.44
245.11
U.54/3
0.02225
218.03
240.28
0.7900
100
227.61
252.05
V.oobZ
0.02313
224.48
247.61
0.8100
110
05041
233.83
259.03
U.iSo4/
0.02399
230.94
254.93
0.8293
120
0.05193
240.09
266.06
0.02483
237.41
262.25
0.8482
130
246.41
273.12
0.02566
243.91
269.57
0.8665
140
0549?
252.77
280.23
u.yj /y
0. 02647
250.43
276.90
0.8845
150
ft DSfidfl
259.19
287.39
A A<
0.02728
256.98
284.26
0.9021
160
265.65
294.59
u.y /is
0.02807
263.56
291.63
0.9193
170
272.16
301.83
0,02885
270.18
299.04
0.9362
180
ft 06nsn
278.72
309.12
0.02963
276.84
306.47
0.9528
1500 kPa (59.22)
2000 kPa (72.88)
Sat.
0.01132
192.08
209.06
0.6768
0.008 13
195,70
211.97
0.6713
70
0.01226
200.05
218.44
0.7046
80
0.01305
207.16
226.73
0.7284
0.00870
201.61
219.02
0.6914
90
0.01377
214.11
234.77
0.7508
0.00941
209.41
228.23
0.7171
100
0.01446
220.95
242.65
0.7722
0.01003
216.87
236.94
0.7408
110
0.01512
227.73
250.41
0.7928
0.01061
224.1 1
245.34
0.7630
120
0.01575
234.47
258.10
0.8126
0.01116
231.21
253.53
0.7841
130
0.01636
241.20
265.74
0.8318
0.01168
238.22
261.58
0.8043
140
0.01696
247.91
273.35
0.8504
0.01217
245.18
269.53
0.8238
150
0.01754
254.63
280.94
0.8686
0.01265
252.10
277.41
0.8426
160
0.01811
261.36
288.52
0.8863
0.01312
259.00
285.24
0.8609
170
0.01S67
268.10
296.11
0.9036
0.01357
265.90
293.04
0.8787
180
0.01922
274.87
303.70
0.9205
0.01401
272.79
300.82
0.8961
190
0.01977
281.65
311,31
0.9371
0.01445
279.69
308.59
0.9131
200
0.02031
288.47
318.93
0.9534
0.01488
286.61
316.36
0.9297
210
0.02084
295.31
326.58
0.9694
0.01530
293.54
324.14
0.9459
220
0.02137
302.19
334.24
0.9851
0.01572
300.49
331.92
0.9619
702 H Appendix b si units: Thermodynamic tables
Table b.4
Thermodynamic Properties ofR-22
Table b.4.1
Saturated R-22
• specific Volume, m 3 /kg internal energy, kJ/kg
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Sat. Vapor
v f
v fs
s
f
IS
it,
s
-70
20.5
0.000670
0.94027
0.94094
-30.62
230.13
199.51
-65
28.0
0.000676
0.70480
0.70547
-25.68
227.21
201.54
-60
37.5
0.000682
0.53647
0.53715
-20.68
224.25
203.57
-55
49.5
0.000689
0.41414
0.41483
-15.62
221.21
205.59
-50
64.4
0.000695
0.32386
0.32456
-10.50
218.11
207.61
-45
82.7
0.000702
0.25629
0.25699
-5.32
214.94
209.62
-40.8
101.3
0.000708
0.21191
0.21261
-0.87
212.18
211.31
-40
104.9
0.000709
0.20504
0.20575
-0.07
211.68
211.60
-35
131.7
0.000717
0.16568
0.16640
5.23
208.34
213.57
-30
163.5
0.000725
0.13512
0.13584
10.61
204.91
215.52
-25
201.0
0.000733
0.11113
0.11186
16.04
201.39
217.44
-20
244.8
0.000741
0.09210
0.09284
21.55
197.78
219.32
-15
295.7
0.000750
0.07688
0.07763
27.11
194.07
221.18
-10
354.3
0.000759
0.06458
0.06534
32.74
190.25
222.99
-5
421.3
0.000768
0.05457
0.05534
38.44
186.33
224.77
497.6
0.000778
.0.04636
0.04714
44.20
182.30
226,50
5
583.8
0.000789
0.03957
0.04036
50.03
178.15
228.17
10
680.7
0.000800
0.03391
0.03471
55.92
173.87
229.79
15
789.1
0.000812
0.02918
0.02999
61.88
169.47
231.35
20
909.9
0.000824
0.02518
0.02600
67.92
164.92
232.85
25
1043.9
0.000838
0.02179
0.02262
74.04
160.22
234.26
30
1191.9
0.000852
0.01889
0.01974
80.23
155.35
235.59
35
1354.8
0.000867
0.01640
0.01727
86.53
150.30
236.82
*40
1533.5 "
0.000884
0.01425
0.01514
92.92
145.02
237.94
45
1729.0
0.000902
0.01238
0.01328
99.42
139.50
238.93
50
1942.3
0.000922
0.01075
0.01167
106.06
133.70
239.76
55
2174.4
0.000944
0.00931
0.01025
112.85
127.56
240.41
60
2426.6
0.000969
0.00803
0.00900
119.83
121.01
240.84
65
2699.9
0.000997
0.00689
0.00789
127.04
113.94
240.98
70
2995.9
0.001030
0.00586
0.00689
134.54
106.22
240.76
75
3316.1
0.001069
0.00491
0.00598
142.44
97.61
240.05
80
3662.3
0.001118
0.00403
0.00515
150.92
87.71
238.63
85
4036.8
0.001 183
0.00317
0.00436
160.32
75.78
236.10
90
4442.5
0.001282
0.00228
0.00356
171.51
59.90
231.4.1
95
4883.5
0.001521
0.00103
. 0.00255
188.93
29.89
218.83
96.0
4969.0
0.001906
0.00191
203.07
203.07
Appendix B SI Units: thermodynamic Tables M 703
TABLE B.4.I {continued)
Saturated R-22
Enthalpy, kJ/kg entropy, kj/kg-K
Temp.
(°C)
Press.
(kPa)
Sat. Liquid
*/
h fg
L>di» v*tpur
Sat. Liquid
Sr
f
Eva p.
fs
i>at, Vapor
— IV
on <
—30.61
249.43
218.82
-0.1401
1.2277
1.0876
OJ
no n
—25.66
246.93
221.27
-0.1161
1.1862
1.0701
Aft
—20.65
244.35
223.70
-0.0924
1.1463
1.0540
ft
— JJ
AO S
— 15.59
241.70
226.12
-0.0689
1.1079
1.0390
04.4
— 10.46
238.96
228.51
-0.0457
1.0708
1.0251
— AS
SO *7
oz. /
—5.26
236.13
230.87
-0.0227
1.0349
1.0122
— 4U.O
1 A 1 1
—0,80
233.65
232.85
-0.0034
1.0053
1.0019
4 A
4U
iu4.y
233.20
233.20
1.0002
1.0002
is
jj
1 1 1 O"
5.33
230,16
235.48
0.0225
0.9664
0.9889
10,73
227.00
237.73
0.0449
0.9335
0.9784
OA 1 A
ZULU
16.19
223.73
239.92
0.0670
0.9015
0.9685
_-OA
— ZU
*)A A O
Z44.B
21.73
220.33
242.06
0.0890
0.8703
0.9593
2.yj. i
27.33
216.80
244.13
0.1107
0.8398
0.9505
1/1
— IU
J.D4.-J
33.01
213.13
246.14
0.1324
0.8099
0.9422
— <
j
/f O 1 1
38.76
209.32
248.09
0.1538
0.7806
0.9344
n
u
4y f.O
44.59
205.36
249.95
0.1751
0.7518
0.9269
D
50.49
201.25
251.73
0.1963
0.7235
0.9197
1 fl
OoU. /
5o.4o
196.96
253.42
0.2173
0.6956
0.9129
15
789.1
62.52
192.49
255.02
0.2382
0.6680
0.9062
20
909.9
68.67
187.84
256.51
0.2590
0.6407
0.8997
25
1043.9
74.91
182.97
257.88
0.2797
0.6137
0.8934
30
1191.9
81.25
177.87
259.12
0.3004
0.5867
0.8871
35
1354.8
87.70
172.52
260.22
0.3210
0.5598
0.8809
40
1533.5
94.27
166.88
261.15
0.3417
0.5329
0.8746
45
1729.0
100.98
160.91
261.90
0.3624
0.5058
0.8682
50
1942.3
107.85
154.58
262.43
0.3832
0.4783
0.8615
55
2174.4
114.91
147.80
262.71
0.4042
0.4504
0.8546
60
2426.6
122.18
140.50
262.68
0.4255
0.4217
0.8472
65
2699.9
129.73
132.55
262.28
0.4472
0.3920
0.8391
70
2995.9
137.63
123.77
261.40
0.4695 .
0.3607
0.8302
75
3316.1
145.99
113.90
259.89
0.4927
0.3272
0.8198
80
3662.3
155.01
102.47
257.49
0.5173
0.2902
0.8075
85
4036.8
165.09
88.60
253.69
0.5445
0.2474
0.7918
90
4442.5
177.20
70.04
247.24
0.5767
0.1929
0.7695
95
4883.5
196.36
34.93
231.28
0.6273
0.0949
0.7222
96.0
4969.0
212.54
212.54
0.6708
0.6708
704 B APPENDIX B SI UNITS: THERMODYNAMIC TABLES
/
Table b.4.2
Superheated R-22
Temp.
V
it
h
s
V
ii
n
s
{°C>
(m 3 /kg)
(KJ/kg)
(kJ/Kg;
\K J/ Kg-JV )
\m /KgJ
rig/
(kJ/kg-K)
50kPa (-
54.80)
100 kPa (-
-41.03)
Sat.
0.41077
205.67
226.21
1.0384
0.21525
211.19
232.72
1.0026
212.69
234.72
1.0762
0.21633
211.70
233.34
1.0052
-30
0.46064
217.57
240.60
1.1008
0.22675
216.68
239.36
1.0305
-20
0.48054
222.56
246.59
1.1250
0.23706
221.76
245.47
1.0551
-10
0.50036
227.66
252.68
1.1485
; 0.24728
226.94
251.67
1.0791
0.52010
232.87
258.87
1.1717
0.25742
232.21
257.96
1.1026
10
0.53977
238.19
265.18
1.1943
0.26748
237.60
264.35
1.1256
20
0.55939
243.62
271.59
1.2166
0.27750
243.08
270.83
1.1481
30
0.57897
249.17
278.12
1.2385
0.28747
248.67
277.42
1.1702
40
0.59851
254.82
284.74
1.2600
0.29739
254.36
284.10
1.1919
50
0.61801
260.58
291.48
1.2811
0.30729
260.16
290.89
1.2132
60
0.63749
266.44
298.32
1.3020
0.31715
266.06
297.77
1.2342
70
0.65694
272.42
305.26
1.3225
0.32699
272.06
304.76
1.2548
80
0.67636
278.50
312.31
1.3428
0.33680
278.16
311.84
1.2752
90
0.69577
284.68
319.47
1.3627
0.34660
284.37
319.03
1.2952
100
0.71516
290.96
326.72
1.3824
0.35637
290.67
326.31
1.3150
110
0.73454
297.34
-334.07
1.4019
0.36614
297.07
333.69
1.3345
150 kPa (
-32.02)
200 kPa (
-25.12)
bat.
ft 1 AT)1
1 \A 1A
236.83
0.9826
0.11237
217.39
239.87
0.9688
-20
0.15585
220.94
244.32
1,0129
0.11520
220.10
243.14
0.9818
-10
0.16288
226.20
250.63
1.0373
0.12065
225.44
249.57
1.0068
.
0.16982
231.55
257.02
1.0612
0. 12600
230.87
256.07
1.0310
10
0.17670
236.99
263.50
1.0844
0.13129
236.38
262.63
1.0546
v 20
0.18352
242.53
270.06
1.1072
0.13651
241.97
269.27
1.0776
30
0.19028
248.17
276.71
1.1295
0.14168
247.66
275.99
1.1002
40
0.19701
253.90
283.45
1.1514
0.14681
253.43
282.80
1.1222
50
0.20370
259.73
290.29
1.1729
0.15190
259.31
289.69
1.1439
60
0.21036
265.67
297.22
1.1940
0.15696
265.27
296.66
1.1652
70
0.21700
271.70
304.25
1.2148
0.16200
271.33
303.73
1.1861
80
0.22361
277.83
311.37
1.2353
' 0.16701
277.49
310.89
1.2066
90
0.23020
284.05
318.58
1.2554
0.17200
283.74
318.14
1.2269
100
0.23678
290.38
325.90
1.2753
0.17697
290.09
325.48
1.2468
110
0.24333
296.80
333.30
1.2948
0.18193
296.53
332.91
1.2665
120
0.24988
303.32
340.80
1.3142
0.18688
303.06
340.44
1.2858
130
0.25642
309.93
348.39
1.3332
0.19181
309.69
348.05
1.3050
APPENDIX B SI UNITS: THERMODYNAMIC TABLES M 705
TABLE B.4.2 {continued)
Superheated R-22
i em p.
V
u
h
s
V
h
s
(KJ/Kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
300 kPa (
-14.61)
400 kPa (
-6.52)
Sat.
0.07657
221.32
244.29
0.9499
0.05817
224.23
247.50
0.9367
-10
0.07834
223.88
247.38
0.9617
0.08213
229.47
254.10
0.9868
0.06013
228.00
252.05
0.9536
10
0.08583
235.11
260.86
1.0111
0.06306
233.80
259.02
0.9787
20
0.08947
240.83
267.67
1.0347
0.06591
239.64
266.01
1.0029
30
0.09305
246.62
274.53
1.0577
0.06871
245.55
273.03
1.0265
40
0.09659
252.48
281.46
1.0802
0.07146
251.51
280.09
1 .0494
50
0.10009
258.43
288.46
1.1022
0.07416
257.54
287.21
1.0717
60
0.10355
264.47
295.54
1.1238
0.07683
263.65
294.39
1.0936
70
0.10699
270.59
302.69
1.1449
0.07947
269.84
301.63
1.1150
80
0.11040
276.80
309.92
1.1657
0.08209
276.11
308.94
1.1361
90
0.11379
283.10
317.24
1.1861
0.08468
282.46
316.33
1.1567
100
0.11716
289.49
324.64
1.2062
0.08725
288.89
323.80
1.1770
no
0.12052
295.97
332.13
1.2260
0.08981
295.41
331.34
1.1969
120
0.12387
302.54
339.70
1.2455
0.09236
302.02
338.96
1.2165
130
0.12720
309.20
347.36
1.2648
0.09489
308.71
346.66
1.2359
140
0.13052
315.95
355.10
1.2837
0.09741
315.48
354.45
1.2550
500 kPa (0.15)
600 kPa (5.88)
Sat.
O.04692
226.55
250.00
0.9267
U.LO yz.y
228.46
252.04
0.9185
10
0.04936
232.43
257.11
0.9522
0.O4O18
231.00
255.11
0.9295
20
0.05175
238.42
264.30
0.9772
0.04228
237.15
262.52
0.9552
30
0.05408
244.44
271.48
1.0013
0.04431
243.30
269.89
0.9799
40
0.05636
250.51
278.69
1.0247
0.04628
249.48
277.25
1.0038
50
0.05859
256.63
285.93
1.0474
0.04820
255.70
284.62
1.0270
60
0.06079
262.82
293.22
1.0696
0.05008
261.97
292.02
1.0495
70
0.06295
269.08
300.55
1.0913
0.05193
268.30
299.46
1.0715
80
0.06509
275.40
307.95
1.1126
0.05375 '
274.69
306.94
1.0930
90
0.06721
281.81
315.41
1.1334 .
0^05555
281.14
314.48
1.1140
100
0.06930
288.29
322.94
1.1539
0.05733
287.67
322.07
1.1347
110
0.07138
294.85
330.54
1.1740
0.05909
294.27
329.73
1.1549
120
0.07345
301.49
338.21
1.1937
0.06084
300.95
337.46
1.1748
130
0.07550
308.21
345.96
1.2132
0.06258
307.71
345.26
LI 944
140
0.07755
315.02
353.79
1.2324
0.06430
314.54
353.12
1.2137
150
0.07958
321.90
361.69
1.2513
0.06601
321.46
361.07
1.2327
160
0.08160
328.87
369.67
1.2699
0.06772
328.45
369.08
1.2514
706 9 Appendix B SI Units: thermodynamic tables
TABLE B.4.2 (continued)
Superheated R-22
Temp.
V
u
h
5
V
u
h
s
(°C)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(KJ/Kg-K)
800 kPa (15.47)
1000 kPa (23.42)
Sat.
0.02958
231.50
255.16
0.9056
0.02364
233.82
257.46
0.8954
20
0.03037
234.44
258.74
30
0.03203
240.91
266.53
0.9446
0.02460
238.31
262.91
0.9136
40
0.03363
247.34
274.24
0.9690
0.02599
245.05
271.04
0.9400
50
0.03518
253.77
281.91
0.9931
0.02732
251.72
279.05
0.9651
60
0.03667
260.21
289.55
1.0164
0.02860
258.37
286.97
0.9893
70
0.03814
266.69
297.20
1.0391
0,02984
265.02
294.86
1.0126
80
0.03957
273.21
304.87
1.0611
0.03104
271.69
302.73
1.0352
90
0.04097
279.79
312.57
1.0826
0.03221
278.39
310.60
1.0572
100
0.04236
286.42
320.30
1.1036
0.03336
285.12
318.49
1.0786
no
0.04373
293.11
328.09
1.1242
0.03449
291.91
326.41
1.0996
120
0.04508
299.86
335.93
1.1444
0.03561
298.75
334.36
1.1200
130
0.04641
306.69
343.82
. 1.1642
0.03671
305.65
342.36
1.1401
140
0.04774
313.59
351.78
1.1837
0.03780
312.61
350.41
1.1599
150
0.04905
320.56
359.80
1.2029
0.03887
319.64
358.51
1.1792
160
0.05036
327.60
367.89
1.2218
0.03994
326.74
366.68
1.1983
170
0.05166
334.72
376.04
1.2404
0.04100
333.90
374.90
1.2171
1200 kPa (30.26)
1400 kPa (36.31)
Sat.
0.01960
235.66
259.18
0.8868
0.01668
237.12
260.48
0.8792
40
0.02085
242.58
267.60
0.9141
0.01712
239.89
263.86
0.8901
50
0.02205
249.55
276.01
0.9405
0.01825
247.22
272.77
0.9181
60 -
0.02319
256.43
284.26
0.9657
0.01930
254.38
281.40
0.9444
70
0.02428
263.28
292.42
0.9898
0.02029
261.45
289.86
0.9694
.80
0.02534
270.10
300.51
1.0131
0.02125
268.45
298.20
0.9934
90
0.02636
276.94
30S.57
1.0356
0.02217
275.44
306.47
1.0165
100
0.02736
283.79
316.62
1.0574
0.02306
282.42
314.70
1.0388
110
0.02833
290.68
324.68
1.0788
0.02393
289.42
322.92
1.0606
120
0.02929
297.61
332.76
1.0996
0.02477
296.44
331.13
1.0817
130
0.03024
304.59
340.87
1.1199
0.02561
303.50
339.35
1.1024
140
0.03117
311.62
349.02
1.1399
* 0.02643
310.61
347.60
1.1226
150
0.03208
318.71
357.21
1.1595
0.02723
317.76
355.89
1.1424
160
0.03299
325.86
365.45
1.1787
0.02803
324.97
364.21
1.1618
170
0.03389
333.07
373.74
1.1977
0.02882
332.23
372.57
1.1809
180
0.03479
340.35
382.09
1.2163
0.02960
339.55
380.99
1.1997
190
0.03567
347.69
390.50
1.2346
0.03037
346.94
389.45
1.2182
APPENDIX B SI UNITS: THERMODYNAMIC TABLES M 707
jj
TABLE B.4.2 (continued)
Superheated R-22
Temp.
it
It
5
V
it
/i
s
(°C)
(mVkg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(kJ/kg)
(kJ/kg)
fir Ifka-TCi
1600 kPa (41.75)
2000 kPa (51.28)
Sat.
0.01446
238.30
261.43
0.8724
0.01129
239.95
262.53
0.8598
50
0.01535
244.70
269.26
0.8969
60
0.01635
252,20
278.36
0.9246
0.01213
247.29
271.56
0.8873
70
0.01728
259.52
287.17
0.9507
0.01301
255.29
281.31
0.9161
80
0.01817
266.73
295.80
0.9755
0.01381
263.02
290.64
0.9429
90
0.01901
273.88
304.30
0.9992
0.01456
270.57
299.70
0.9682
100
0.01983
281.00
312.73
1.0221
0.01528
278.02
308.57
0.9923
110
0.02061
288.12
321.10
1.0442
0.01596
285,40
317.32
1.0155
120
0.02138
295.25
329.46
1.0658
0.01662
292.75
325.99
1.0378
130
0.02213
302.39
337.81
1.0867
0.01726
300.09
334.61
1.0594
140
0.02287
309.57
346.16
1.1072
0.01788
307.44
343.20
1.0805
150
0.02359
316.79
354.54
1.1272
0.01849
314.80
351.78
1.1010
160
0.02430
324.06
362.95
1.1469
0.01909
322.19
360,37
1.1211
170
0.02501
331.37
371.39
1.1661
0.01967
329.62
368.97
1.1407
180
0.02570
338.74
379.87
1.1851
0.02025
337.09
377.60
1.1600
190
0.02639
346.17
388.40
1.2037
0.02082
344.61
386.25
1.1788
200
0.02707
353.66
396.97
1.2220
0.02138
352.17
394.94
1.1974
3000 kPa (70.09)
4000 kPa (84.53)
Sat.
0.00688
240.75
261.38
0.8300
0.00443
236.40
254.13
0.7935
80
0.00775
251.29
274.53
0.8678
90
0.00847
260.65
286.04
0.9000
0.00504
245.48
265.63
0.8254
100
0.00910
269,37
296.66
0.9288
0.00580
257.78
281.00
0.8672
110
0.00967
277.72
306.74
0.9555
0.00641
268.13
293.75
0.9009
120
0.01021
285.84
316.47
0.9805
0.00692
277.58
305.27
0.9306
130
0.01072
293.80
325.96
1.0044
0.00739
286.52
316.08
0.9578
140
0.01120
301.67
335.27
1.0272
0.00782
295.14
326.42
0.9831
150
0.01166
309.47
344.47
1.0492
0.00823
303.54
336.45
1.0071
160
0.01211
317.24
353.58
1.0705
0.00861
311.81
346.25
1.0300
170
0.01255
325.00
362.65
1.0912
0.00898
319.97
355.89
1.0520
180
0.01298
332.75
371.68
1.1113
0.00933
328.08
365.41
1.0732
190
0.01339
340.52
380.70
1.1310
0.00968
336.15
374.85
1.0939
200
0.01380
348.31
389.71
1.1502
0.01001
344.20
384.24
1.1139
210
0.01420
356.12
398.73
L1691
0.01033
352.25
393.59
1.1335
220
0.01460
363.98
407.77
1.1876
0.01065
360.31
402.93
1.1526
230
0.01499
371.87
416.83
1.2058
0.01097
368.38
412.25
1.1713
708 M APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table B.5
Thermodynamic Properties ofR-134a
Table B.5.1
Saturated R-134a
Specific Volume, m 3 /kg internal energy, kJ/kg
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat, Liquid
Evap.
Sat. Vaj
(°c>
(kPa)
v f
v t
"f
"A
-70
8.3
0.000675
1.97207
1.97274
1 19.46
218.74
338.20
—65
11.7
0.000679
1.42915
1.42983
123.18
217.76
340.94
— 60
16.3
O.O00684
1.05199
1.05268
127.52
216.19
343.71
—55
22.2
0.000689
0.78609
0.78678
132.36
214.14
346.50
—50
29.9
0.000695
0.59587
0.59657
137.60
211.71
349.31
—45
39.6
0.000701
0.45783
0.45853
143.15
208.99
352.15
-40
51.8
0.000708
0.35625
0.35696
148.95
206.05
355.00
—35
66.8
0.000715
0.28051
0.28122
154.93
202.93
357.86
—30
85.1
0.000722
0.22330
1 0.22402
161.06
199.67
360.73
-26.3
101.3
0.000728
0.18947
0.19020
165.73
197.16
362.89
—25
107.2
0.000730
0.17957
0.18030
167.30
196.31
363.61
—20
133.7
0.000738
0.14576
0.14649
173.65
192.85
366.50
— 15
165.0
0.000746
0.11932
0.12007
180.07
189.32
369.39
-10
20 L7
0.000755
0.09845
0.09921
186.57
185.70
372.27
-5
244.5
0.000764
0.08181
0.08257
193.14
182.01
375.15
o
294.0
0.000773
0.06842
0.06919
199.77
178.24
378.01
5
350.9
0.000783
0.05755
0.05833
206.48
174.38
380.85
10
415.8
0.000794
0.04866
0.04945
213.25
170.42
383.67
15
489.5
0.000805
0.04133
0.04213
220.10
166.35
386.45
20
572.8
0.000817
0.03524
0.03606
227.03
162.16
389.19
25
666.3
0.000829
0.03015
0.03098
234.04
157.83
391.87
30
771.0
0.000843
0.02587
0.02671
241.14
153.34
394.48
k 35
887.6
0.000857
0.02224
0.02310
248.34
148.68
397.02
40
1017.0
0.000873
0.01915
0.02002
255.65
143.81
399.46
45
1160.2
0.000890
0.01650
0.01739
263.08
138.71
401.79
50
1318.1
0.000908
0.01422
0.01512
270.63
133.35
403.98
55
1491.6
0.000928
0.01224
0.01316
278.33
127.68
406.01
60
1681.8
0.000951
0,01051
0.01146
286.19
121.66
407.85
65
1889.9
0.000976
0.00899
0.00997
294.24
115.22
409.46
70
2117.0
0.001005
0.00765
0,00866
302.51
108.27
410.78
75
2364.4
0.001038
0.00645
0.00749
311.06
100.68
411.74
SO
2633.6
0.001078
0.00537
0.00645
319.96
92.26
412.22
85
2926.2
0.001128
0.00437
0.00550
329.35
82.67
412.01
90
3244.5
0.001195
0.00341
0.00461
339.51
71.-24
410.75
95
3591.5
O.001297
0.00243
0.00373
351.17
56.25
407.42
100
3973.2
0.001557
0.00108
0.00264
368.55
28,19
396.74
101.2
4064.0
0.001969
0.00197
382.97
382.97
Appendix B SI Units: Thermodynamic Tables B 709
TABLE B.5.1 {continued)
Saturated R-134a
Enthalpy, kJ/kg Entropy, kJ/kg-K
Temp.
CQ
Press.
(kPa)
Sat. Liquid
Evap.
Sat. Vapor
K
Sat. Liquid
s f
Evap,
s fs
Sat. Vapor
s s
-70
S.3
119.47
235.15
354.62
1 1 ^7^.
1.1 J /j
-65
11.7
123.18
234.55
357.73
U.OoZ J
1 1 1AQ
1.8094
-60
16.3
127.53
233.33
36(1 86
fl IfYX t
1 AQ,17
i iftno
I ./y/B
-55
22.2
132.37
231.63
364.00
V, /ZOO
l.Uoi 5
1.7874
-50
29.9
137.62
229.54
J\J /.ID
fl 1AQ1
1 A70£
1.7780
-45
39.6
143.18
227.14
370.32
(1 11 Aft
A ftft<£
V.yyjo
1 .7695
-40
51.8
148.98
224.50
373 48
ft 70Q1
ft QClCi
1. 7620
-35
66.8
154.98
221.67
376.64
A B7JS
(\ a^ao
u.yius
1.7553
-30
85.1
16L12
218.68
379.80
fl SdQQ
A aQQ/l
U.syy4
i ia m
I . /4yi
-26.3
101.3
165.80
216.36
382,16
u.aoyu
A Q7£"3
U.S /6i
1.7453
-25
107.2
167.38
215.57
382.95
U.O / J't
A OfTQ-)
1. /441
-20
133.7
173.74
212.34
U.ESJes
1.7395
-15
165.0
180.19
209.00
^80 ?ft
ft QIKQ
ft OAAjC
1.7354
-10
201.7
186.72
205.56
392.28
fi g*;a7
ft 70 T ^
1.7319
-5
244.5
193.32
202.02
3Q*i 3d
u.y / oo
ft 1C2A
1.7288
294.0
200.00
198.36
398.36
i aaaa
A 77£7
If, /ZOZ
I./ZOZ
5
350.9
206.75
194.57
401.32
i fl7i7
ri tone
I. /2jy
10
415.8
213.58
190.65
4(14 93
A £71*3
U.6/33
1 .72 1 8
15
489.5
220.49
186.58
4fl7 01
1 fl7T\
lf.64/0
1./200
20
572.8
227.49
182.35
409.84
1.0963
0.6220
1.7183
25
666.3
234.59
177.92
412.51
1.1201
0.5967
1.7168
30
771.0
241.79
173.29
415.08
1.1437
0.5716
1.7153
35
887.6
249.10
168.42
417.52
i.1673
0.5465
1.7139
40
1017.0
256.54
163.28
419.82
1.1909
0.5214
1.7123
45
1160.2
264.11
157.85
421.96
1.2145
0.4962
1.7106
50
1318.1
271.83
152.08
423.91
1.2381
0.4706
1.7088
55
1491.6
279.72
145.93
425.65
1.2619
0.4447
1.7066
60
1681.8
287.79
139.33
427.13
1.2857
0.4182
1.7040
65
1889.9
296.09
132.21
428.30
1.3099
0.3910
1.7008
70
2117.0
304.64
124.47
429.11
1.3343
0.3627
1.6970
75
2364.4
313.51
115.94
429.45
1.3592
0.3330
1.6923
80
2633.6
322.79
106.40
429.19
1.3849
0.3013
1.6862
85
2926.2
332.65
95.45
428.10
1.4117
0.2665
1.6782
90
3244.5
343.38
82.31
425.70
1.4404
0.2267
1.6671
95
3591.5
355.83
64.98
420.81
1.4733
0.1765
1.6498
100
3973.2
374.74
32.47
407.21
1.5228
0.0870
1.6098
101.2
4064.0
390.98
390.98
1.5658
1.5658
710 H APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table b.5.2
Superheated R-134a „____— — —
Temp v u h s v u h s
CC) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K)
50 kPa (-40.67) ____ 100 kPa (-26.54)
Sat.
354.61
373.06
1.7629
0.19257
362.73
381.98
1.7456
—20
A /tACA7
368.57
388.82
1.8279
0.19860
367.36
387.22
1.7665
1 f\
— 1U
A A')'}')'}
375.53
396.64
1.8582 '
0.20765
374.51
395.27
1.7978
u
A /fJO")!
u.'f jyzi
382.63
404.59
1.8878
0.21652
381.76
403.41
1.8281
1 A
10
IRQ Qfl
412.70
1,9170
0.22527
389.14
411.67
1.8578
OA
ZU
397.32
420.96
1.9456
0.23392
396.66
420.05
1.8869
'i A
U.407JO
404.90
429.38
1.9739
0.24250
404.31
428.56
1.9155
4U
412.64
437.96
2.0017
0.25101
412.12
437.22
1.9436
<a
446.70
2.0292
0.25948
420.08
446.03
1.9712
/^A
OU
428.63
455.60
2.0563
0.26791
428.20
454.99
1.9985
/U
436.86
464.66
2.0831
0.27631
436.47
464.10
2.0255
OA
445.26
473.88
2.1096
0.28468
444.89
473.36
2.0521
AA
yy)
483.26
2.1358
0.29302
453.47
482.78
2.0784
1 AA
1UU
462.53
492.81
2.1617
0.30135
462.21
492.35
2.1044
1 1 A
471.41
502.50
2.1874
0.30967
471.11
502.07
2.1301
120
0.63835
480.44
512.36
2.2128
0.31797
480.16
511.95
2.1555
130
0.65479
489.63
522.37
2.2379
0.32626
489.36
521.98
2.1807
150 kPa(
-17.29)
200 kPa (-
-10.22)
Sat.
0.13139
368.06
387.77
1.7372
0.10002
372.15
392.15
1.7320
-10
0.13602
373.44
393.84
1.7606
0.10013
372.31
392.34
1.7328
0.14222
380.85
402,19
1.7917
0.10501
379.91
400.91
1.7647
10 .
0.14828
388.36
410.60
1.8220
0.10974
387.55
409.50
1.7956
20
0.15424
395.98
419.11
1.8515
0.11436
395.27
418.15
1.8256
v30
0.16011
403.71
427.73
1.8804
0.11889
403.10
426.87
1.8549
40
0.16592
411.59
436.47
1.9088
0.12335
411.04
435.71
1.8836
50
0.17168
419.60
445.35
1.9367
0.12776
419.11
444.66
1.9117
60
0.17740
427.76
454.37
L9642
0.13213
427.31
453.74
1.9394
70
0.18308
436.06
463.53
1.9913
0.13646
435.65
462.95
1.9666
80
0.18874
444.52
472.83
2.0180
0.14076
444.14
472.30
1.9935
90
0.19437
453.13
482.28
2.0444
' 0.14504
452.78
481.79
2.0200
100
0.19999
461.89
491.89
2.0705
0.14930
461.56
491.42
2.0461
110
0.20559
470.80
501.64
2.0963
0.15355
470.50
501.21
2.0720
120
0.21117
479.87
511.54
2.1218
0.15777
479.58
511.13
2.0976
130
0.21675
489.08
521.60
2.1470
0.16199
488.81
521.21
2.1229
140
0.22231
498.45
531.80
2.1720
0.16620
498.19
531.43
2.1479
J
Appendix B si units: thermodynamic tables M lit
TABLE B.5.2 (continued)
Superheated R-134a
j. crop*
V
it
h
u
h
cc)
(m 3 /ke)
(kj/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
300 kPa (0.56)
400 kPa (8.84)
Sat.
0.06787
378.33
398.69
1.7259
0.05136
383.02
403.56
1.7223
10
0.07111
385.84
407.17
1.7564
0.05168
383.98
404.65
1.7261
20
0.07441
393.80
416.12
1.7874
0.05436
392.22
413.97
1.7584
30
0.07762
401.81
425.10
1.8175
0.05693
400.45
423.22
1.7895
40
0.08075
409.90
434.12
1.8468
0.05940
408.70
432.46
1.8195
50
0.08382
418.09
443.23
1.8755
0.06181
417.03
441.75
1.8487
60
0.08684
426.39
452.44
1.9035
0.06417
425.44
451.10
1.8772
70
0.08982
434.82
461.76
1.9311
0.06648
433.95
460.55
1.9051
80
0.09277
443.37
471.21
1.9582
0.06877
442.58
470.09
1.9325
90
0.09570
452.07
480.78
1.9850
0.07102
451.34
479.75
1.9595
100
0.09861
460.90
490.48
2.0113
0.07325
460.22
489.52
1.9860
110
0.10150
469.87
500.32
2.0373
0.07547
469.24
499.43
2.0122
120
0.10437
478.99
510.30
2.0631
0.07767
478.40
509.46
2.0381
130
0.10723
488.26
520.43
2.0885
0.07985
487.69
519.63
2.0636
140
0.11008
497.66
530.69
2.1136
0.08202
497.13
529.94
2.0889
150
0.11292
507.22
541.09
2.1385
0.08418
506.71
540.38
2.1139
160
0.11575
516.91
551.64
2.1631
0.08634
516.43
550.97
2.1386
500 kPa (15.66)
600 kPa (21.52)
Sat.
0.04126
386.82
407.45
1.7198
0.03442
390.01
410.66
1.7179
20
0.04226
390.52
411.65
1.7342
30
0.04446
398.99
421.22
1.7663
0.03609
397.44
419.09
1.7461
40
0.04656
407.44
430.72
1.7971
0.03796
406.11
428.88
1.7779
50
0.04858
415.91
440.20
1.8270
0.03974
414.75
438.59
1.8084
60
0.05055
424.44
449.72
1.8560
0.04145
423.41
448.28
1.8379
70
0.05247
433.06
459,29
1.8843
0.04311
432.13
457.99
1.8666
80
0.05435
441.77
468.94
1.9120
0.04473
440.93
467.76
1.8947
90
0.05620
450.59
478.69
1.9392
0.04632
449.82
477.61
1.9222
100
0.05804
459.53
488.55
1.9660
0.04788
458.82
487.55
1.9492
110
0.05985
468.60
498.52
1.9924
0.04943 y
467.94
497.59
1.9758
120
0.06164
477.79
508.61
2.0184
0.05095
477.18
507.75
2.0019
130
0.06342
487.13
518.83
2.0440
0.05246
486.55
518.03
2.0277
140
0.06518
496.59
529.19
2.0694
0.05396
496.05
528.43
2.0532
150
0.06694
506.20
539.67
2.0945
0.05544
505.69
538.95
2.0784
160
0.06869
515.95
550.29
2.1193
0.05692
515.46
549.61
2.1033
170
0.07043
525.83
561.04
2.1438
0.05839
525.36
560.40
2.1279
712 H APPENDIX B SI UNITS: THERMODYNAMtC TABLES
TABLE B.5.2 {continued)
Superheated R-134a
Temp.
V
u
It
s
V
i/
h
s
(°C)
(m 3 /kg)
(kJ/kg)
(kj/kg)
(kJ/kg-K)
(mVkg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
800 kPa (31.30)
1000 kPa (39.37)
Sat.
0.U2571
395.15
415.72
1 n 1 ^.(\
U.UiUJO
399.16
419.54
1 71
40
0.02711
403.17
■ 424.86
1.7446
0.02047
399.78
420.25
1.7148
50
0.02861
412.23
435.1 1
1.7768
0.02185
409.39
431.24
1.7494
60
0.03002
421.20
445.22
1.8076
0.02311
418.78
441.89
1.7818
70
0.03137
430.17
455.27
1.8373
0.02429
.428.05
452.34
1.8127
80
0.03268
439.17
465.31
1.8662
0.02542
437.29
462.70
1.8425
90
0.03394
448.22
475.38
1.8943
0.02650
446.53
473.03
1.8713
100
0.03518
457.35
485.50
1.9218
0.02754
455.82
483.36
1.8994
110
0.03639
466.58
495.70
1.9487
0.02856
465.18
493.74
1.9268
120
0.03758
475.92
505.99
1.9753
0.02956
474.62
504.17
1.9537
130
0.03876
485.37
516.38
2.0014
0.03053
484.16
514.69
. 1.9801
140
0.03992
494.94
526.88
2,0271
0.03150
493.81
525.30
2.0061
150
0.04107
504.64
537.50
2.0525
0.03244
503.57
536.02
2.0318
160
0.04221
514.46
548.23
2.0775
0.03338
513.46
546.84
2.0570
170
0.04334
524.42
559.09
2.1023
0.03431
523.46
557.77
2.0820
180
0.04446
534.51
570.08
2.1268
0.03523
533.60
568.83
2.1067
1200 kPa (46.31)
1400 kPa (52.42)
Sat.
0.01676
402.37
422.49
1.7102
0.01414
404.98
424,78
1.7077
50
0.01724
406.15
426.84
1.7237
60
0.01844
416.08
438.21
1.7584
O.015O3
413.03
434,08
1.7360
70
0.01953
425.74
449.18
1.7908
0.01608
423.20
445.72
1.7704
80,
0.02055
435.27
459.92
1.8217
0.01704
433.09
456.94
1.8026
90
0.02151
444.74
470.55
1.8514
0.01793
442.83
467.93
1.8333
v 100
0.02244
454.20
481.13
1.8801
0.01878
452.50
478.79
1.8628
110
0.02333
463.71
491.70
1.9081
0.01958
462.17
489.59
1.8914
120
0.02420
473.27
502.31
1.9354
0.02036
471.87
500.38
1.9192
130
0.02504
482.91
512.97
1.9621
0.02112
481.63
511.19
1.9463
140
0.02587
492.65
523.70
1.9884
0.02186
491.46
522.05
1.9730
150
0.02669
502.48
534.51
2.0143
0.02258
501.37
532.98
1.9991
160
0.02750
512.43
545.43
2.0398 '
0.02329
511.39
543.99
2.0248
170
0.02829
522.50
556.44
2.0649
0.02399
521.51
555.10
2.0502
180
0.02907
532.68
567.57
2.0898
0.02468
531.75
566.30
2.0752
APPENDIX B SI UNITS: THERMODYNAMIC TABLES M 713
TABLE B.5.2 {continued)
Superheated R-134a
Temp. v u h s v it h s
<°C) (m 3 /kg) (kJ/kg) (kJ/kg) (kJ/kg-K) <m 3 /kg) (kJ/kg) (kj/kg) (kJ/kg-K)
1600 kPa (57.90) 2000 kPa (67.48)
Sat.
0.01215
407.11
426.54
1.7051
0.00930
410.15
428.75
1.6991
60
0.01239
409.49 '
429.32
1.7135
70
0.01345
420.37
441.89
1.7507
0.00958
413.37
432.53
1.7101
80
0.01438
430.72
453.72
1.7847
0.01055
425.20
446.30
1.7497
90
0.01522
440.79
465.15
1.8166
0.01137
436.20
458.95
1.7850
100 '
0.01601
450.71
476.33
1.8469
0.01211
446.78
471.00
1.8177
110
0.01676
460.57
487.39
1.8762
0.01279
457.12
482.69
1.8487
120
0.01748
470.42
498.39
1.9045
0.01342
467.34
494.19
1.8783
130
0.01817
480.30
509.37
1.9321
0.01403
477.51
505.57
1.9069
140
0.018S4
490.23
520.38
1.9591
0.01461
487.68
516.90
1.9346
150
0.01949
500.24
531.43
1.9855
0.01517
497.89
528.22
1.9617
160
0.02013
510.33
542.54
2.0115
0.01571
508.15
539.57
1.9882
170
0.02076
520.52
553.73
2.0370
0.01624
518.48
550.96
2.0142
180
0.02138
530.81
565.02
2.0622
0.01676
528.89
562.42
2.0398
3000 kPa (86.20) 4000 kPa (100.33)
Sat.
0.00528
411.83
427.67
1.6759
0.00252
394.86
404.94
1.6036
90
0.00575
418.93
436.19
1.6995
100
0.00665
433.77
453.73
1.7472
no
0.00734
446.48
468.50
1,7862
0.00428
429.74
446.84
1.7148
120
0.00792
458.27
482.04
1.8211
0.00500
445.97
465.99
1.7642
130
O.00845
469.58
494.91
1.8535
0.00556
459.63
481.87
1.8040
140
0.00893
480.61
507.39
1.8840
0.00603
472.19
496.29
1.8394
150
0.00937
491.49
519.62
1.9133
0.0p644
484.15
509.92
1.8720
160
0.00980
502.30
531.70
1.9415
0.00683
495.77
523.07
1.9027
170
0.01021
513.09
543.71
1.9689
0.00718
507.19
535.92
1.9320
180
0.01060
523.89
555.69
1.9956
0.00752
518.51
548.57
1.9603
6000 kPa 10000 kPa
90 ;
0.001059
328.34
334.70
1.4081
0.000991
320.72
330.62
1.3856
100
0.001150
346.71
353.61
1.4595
0.001040
336.45
346.85
1.4297
110
0.001307
368.06
375.90
1.5184
0.001100
352.74
363.73
1.4744
120
0.001698
396.59
406.78
1.5979
0.001175
369.69
381.44
1.5200
130
0.002396
426,81
441.18
1.6843
0.001272
387.44
400.16
1.5670
140
0.002985
448.34
466.25
1.7458
0.001400
405.97
419.98
1.6155
150
O.0O3439
465.19
485.82
1.7926
0.001564
424.99
440.63
1.6649
160
0.003814
479.89
502.77
1.8322
0.001758
443.77
461.34
1.7133
170
0.004141
493.45
518.30
1.8676
0.001965
461.65
481.30
1.7589
180
0.004435
506.35
532.96
1.9004
0.002172
478.40
500.12
1.8009
714 H Appendix B SI Units: Thermodynamic Tables
Table B.6
Thermodynamic Properties of Nitrogen
Table b.6.1
Saturated Nitrogen
Specific volume, m 3 /kg
INTERNAL ENERGY, kJ/kg
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
(K)
(kPa)
v fi
"/
u fs
"i
63.1
12.5
0.001150
1.48074
1.48189
-150.92
196.86
45.94
65
17.4
0.001160
1.09231
1.09347
-147.19
194.37
47.17
70
38.6
0.001191
0.52513
0.52632
-137.13
187.54
50.40
75
76.1
0.001223
0.28052
0.28174
-127.04
180.47
53.43
77.3
101.3
0.001240
0.21515
0.21639
-122.27
177.04
54.76
80
137.0
0.001259
0.16249
0.16375
-116.86
173.06
56.20
85
229.1
0.001299
O.10018
0.10148
-106.55
165.20
58.65
90
360.8
0.001343
O.06477
0.06611
-96.06
156.76
60.70
95
541.1
0.001393
0.04337
0.04476
-85.35
147.60
62.25
100
779.2
0.001452
0.02975
0.03120
-74.33
137.50
63.17
105
1084.6
0.001522
0.02066
0.02218
-62.89
126.18
63.29
110
1467.6
0.001610
0.01434
0.01595
-50.81
113.11
62.31
115
1939.3
0.001729
0.0097 i
0.01144
-37.66
97.36
59.70
120
2513.0
0.001915
0.00608
0.00799
-22.42
76.63
54.21
125
3208.0
0.002355
0.00254
0.00490
-0.83
40.73
39.90
126.2
3397.8
0.003194
0.00319
18.94
18.94
Enthalpy, kl/kg
ENTROPY, kJ/kg-K
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
(K)
(kPa)
*/
h fi
K
s f
s s
63.1
12.5
-150.91
215.39
64.48
2.4234
3.4109
5.8343
65
17.4
-147.17
213.38
66.21
2.4816
3.2828
5.7645
*70
38.6
-137.09
207.79
70.70
2.6307
2.9684
5.5991
75
76.1
-126.95
201.82
74.87
2.7700
2.6909
5.4609
77.3
101.3
-122.15
198.84
76.69
2.8326
2.5707
5.4033
80
137.0
-116.69
195.32
78.63
2.9014
2.4415
5.3429
85
229.1
-106.25
188.15
81.90
3.0266
2.2135
5.2401
90
360.8
-95.58
180.13
84.55
3.1466
2.0015
5.1480
95
541.1
-84.59
171,07
86.47
3.2627
1.8007
5.0634
100
779.2
-73.20
160.68
87.48
3.3761
1.6068
4.9829
105
1084.6
-61.24
148.59
87.35
3.4883
1.4151
4.9034
110
1467.6
-48.45
134.15
85.71
3.6017
1.2196
4.8213
115
1939.3
-34.31
116.19
81.88
3.7204
1.0104
4.7307
120
2513.0
-17.61
91.91
74.30
3.8536
0.7659
4.6195 ,
125
3208.0
6.73
48.88
55.60
4.0399
0.3910
4.4309
126.2
3397.8
29.79
29.79
4.2193
4.2193
appendix B Si Units: thermodynamic tables
b 715
Table b.6.2
Superheated Nitrogen
Temp.
V
it
V
u
h
s
(K)
(m /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
100 fcPa (77.24 K)
200 kPa (83.62 K)
Sat.
0.21903
54.70
76.61
5.4059
0.11520
58.01
81.05
5.2673
100
0.29103
72.84
101.94
5.6944
0.14252
71.73
100.24
5 4775
120
0.35208
87.94
123,15
5.8878
0.17397
87.14
121.93
5.6753
140
0.41253
102.95
144.20
6.0501
0.20476
102.33
143.28
5.8399
160
117.91
165.17
o.iyui
U.23Diy
1 17.40
164.44
5.9812
180
0.53254
132.83
186.09
6.3132
0.26542
132.41
185.49
6.1052
200
0.59231
147.74
206.97
6.4232
0.29551
147.37
206.48
6.2157
220
0.65199
162.63
227.83
6.5227
0.32552
162.31
227.41
6.3155
240
0.71161
177.51
248.67
6.6133
0.35546
177.23
248.32
6.4064
260
0.77118
192.39
269.51
6.6967
0.38535
192.14
269.21
6.4900
280
0.83072
207.26
290.33
6.7739
0.41520
207.04
290.08
6.5674
300
0.89023
222.14
311.16
6.8457
0.44503
221.93
310.94
6.6393
350
1.03891
259.35
363.24
7.0063
0.51952
259.18
363.09
6.8001
400
1.18752
296.66
415.41
7.1456
0.59392
296.52
415.31
6.9396
450
1.33607
334.16
467.77
7.2690
0.66827
334.04
467.70
7.0630
500
1.48458
371.95
520.41
7.3799
0.74258
371.85
520.37
7.1740
600
1.78154
448.79
626.94
7.5741
0.891 14
448.71
626.94
7.3682
700
2.07845
527.74
735.58
7.7415
1.03965
527.68
735.61
7.5357
800
2.37532
609.07
846.60
7.8897
1.18812
609.02
846.64
7.6839
900
2.67217
692.79
960.01
8.0232
1.33657
692.75
960.07
7.8175
1000
2.96900
778.78
1075.68
8.1451
1.48501
778.74
1075.75
7.9393
TABLE B.6.2 (continued)
Superheated Nitrogen .
Temp.
(K)
V
(m 3 /kg) •
u
(kJ/kg)
it
(kJ/kg)
s
(kJ/kg-K)
V
(m 3 /kg)
u
(kJ/kg)
h
(kJ/kg)
s
(kJ/kg-K)
400 kPa (91.22 K)
600 kPa (96,37 K)
Sat
0.05992
61.13
85.10
5.1268
0.04046
62.57
86,85
C f\A 1 1
5,041 1
100
0,06806
69.30 •
96.52
5.24oo
0.042yy
66.41
92.20
5.095 /
120
0.08486
85.48
119.42
5.4556
A f\G£ 1 A
0. 055 10
83.73
116.79
5.3204
140
0.10085
i a i a<c
1 A 1 A(\
14 1.40
5.6250
0.06620
yy./5
1 10 An
< AV\<1
j,4y33
160
0.1 1647
116.38
162.96
5.7&yu
0.07689
115.34
161.47
5.6422
180
0.13186
131.55
184.30
5,894/
0,08/34
130.69
183,10
5./oy6
200
0.14712
146.64
205.49
6.0063
0.09766
145.91
204.50
5.8823
220
0.16228
161.68
226.59
6.1069
0.10788
161.04
225.76
5.9837
240
0.17738
176.67
247.62
6.1984
0.11803
176.11
246.92
6.0757
260
0.19243
191.64
268.61
6.2824
0.12813 '
191.13
268.01
6.1601
280
0.20745
206.58
289.56
6.3600
0.13820
206.13
289.05
6.2381
300
0.22244
221.52
310,50
6.4322
0.14824
221.11
310.06
6.3105
350
0.25982
258.85
362.78
6.5934
0.17326
258.52
362.48
6.4722
400
0.29712
296.25
415.10
6.7331
0.19819
295.97
414.89
6.6121
450
0.33437
333.81
467,56
6,8567
0.22308
333.57
467.42
6.7359
500
0,37159
371.65
520.28
. 6.9678
0.24792
371.45
520.20
6.8471
600
0.44595
448.55
626.93
7.1622
0.29755
448.40
626.93
7.0416
700
0.52025
527.55
735.65
7.3298
0.34712
527.43
735.70
7.2093
800
0.59453
608.92
846.73
7.4781
0,39666
608.82
846.82
7.3576
900
0.66878
692.67
960.19
7.6117
0.44618
692.59
960.30
7.4912
1000
0.74302
778.68
1075.89
7.7335
0.49568
778.61
1076.02
7.6131
800 kPa (100,38 K)
1000 kPa (103.73 K)
Sat.
0.03038
63.21
87.52
4.9 /OS
A f\**lA 1 £
0.02416
63.35
87.51
A AOT7
120
0.04017
81.88
114.02
^ ■*> 1 A T
j. ziyi
a ai t n
0.03 11 /
79.91
111.08
j. 133 /
140
0,04886
98.41
137.50
5.4U02
A rtl OA C
u. 03845
97.02
135.47
j,3zjy
160
0.05710
114.28
159.95
5.5501
A f\A
0.U4522
113.20
158,42
3.4/ 12.
.180
0.06509 .
129.82
181.89
5.6793
0.05173
128.94
180.67
5.6082
200
0.07293
145.17
203.51
5.7933
0.05809
144.43
202.52
5.7234
220
0.08067
160.40
224.94
5.8954
0.06436
159.76
224.11
5.8263
240
0.08S35
175.54
246.23
5.9880
0.07055
174.98
245.53
5.9194
260
0.09599
190.63
267.42
6.0728
0.07670
190.13
266.83
6.0047
280
0.10358
205.68
288.54
6.1511
0.08281
205.23
288.04
6.0833
300
0.11115
220.70
309.62
6.2238
0.08889
-■220.29
309.18
6.1562
350
0.12998
258.19
362.17
6.3858
0.10401 -v
257.86
361.87
6.3187
400
0.14873
295.69
414.68
6.5260
0.11905
295.42
414.47
6,4591
500
0.18609
371.25
520.12
6.7613
0.14899
371.04
520.04
6.6947
600
0.22335
448.24
626.93
6.9560
0.17883
448.09
626.92
6.8895
700
0.26056
527.31
735.76
7.1237
0.20862
527.19
735.81
7.0573
800
0.29773
608.73
846.91
7.2721
0.23837
608.63
847.00
7.2057
900
0.33488
692.52
960.42
7.4058
0.26810
692.44
960.54
7.3394
1000
0.37202
778.55
1076.16
7.5277
0.29782
778.49
1076.30
7.4614
Appendix B SI Units: Thermodynamic Tables
m 717
TABLE B.6.2 (continued)
Superheated Nitrogen
Tomn
1 efnp.
V
it
h
V
it
It
s
fnvVkel
(kJ/kg)
(kJ/kg)
fkJ/ke-Ki
(kJ/kg)
(kJ/kg)
1500 kPa (110.38 K)
2000 kPa (1 15.58 K)
Sat.
0.01555
62.17
85.51
4.8148
0.01100
59.25
81.25
4.7193
120
0.01899
74.26
102.75
4.9650
0.01260
66.90
92.10
4.8116
140
0.02452
93.36
130.15
5.1767
0.01752
89.37
124.40
5.0618
160
0.02937
110.44
154.50
5.3394
0.02144
107.55
150.43
5.2358
ISO
0.03393
126.71
177.60
5.4755
0.02503
124.42
174.48
5.3775
200
0.03832
142.56
200.03
5.5937
0.02844
140.66
197.53
' '5.4989
220
0.04260
158.14
222.05
5.6987
0.03174
156.52
219.99
5.6060
240
0.04682
173.57
243.80
5.7933
0.03496
172.15
242.08
5.7021
260
0.05099
188.87
265.36
5.8796
0.03814
187.62
263.90
5.7894
280
0.05512
204.10
286.78
5.9590
0.04128
202.97
285.53
5.8696
300
0.05922
219.27
308.10
6.0325
0.04440
218.24
307.03
5.9438
350
0.06940
257.03
361.13
6.1960
0.05209
256.21
360.39
6.1083
400
0.07949
294.73
413.96
6.3371
0.05971
294.05
413.47
6.2500
450
0.08953
332.53
466.82
6.4616
0.06727
331.95
466.49
6.3750
500
0.09953
370.54
519.84
6.5733
0.07480
370.05
519.65
6.4870
600
0.11948
447.71
626.92
6.7685
0.08980
447.33
626.93
6.6825
700
0.13937
526.89
735.94
6.9365
0.10474
526.59
736.07
6.8507
800
0.15923
608.39
847.22
7.0851
0.11965
608.14
847.45
6.9994
900
0.17906
692.24
960.83
7.2189
0.13454
692.04
961.13
7.1333
1000
0.19889
778.32
1076.65
7.3409
0.14942
778.16
1077.01
7.2553
3000 kPa (123.61 K)
10000 kPa
Sat.
0.00582
46.03
63.47
4.5032
140
01038
\Jm\J 1UJO
79.98
111.13
A RlC\f\
H.o /UU
U.UUZUU
0.84
20.87
A (YXTl
160
0.01350
101.35
141.85
5.0763
0.00291
47.44
76.52
4.4088
180
0.01614
119.68
168.09
5.2310
0.00402
82.44
122.65
4.6813
200
0.01857
136.78
192.49
5.3596
0.00501
108.21
158.35
4.8697
220
0.02088
153.24
215.88
5.4711
0.00590
129.86
188.88
5.0153
240
0.02312
169.30
238.66
5.5702 .
0.00672
149.42
216.64
5.1362
260
0.02531
185.10
261.02
5.6597
0.00749
167.77
242.72
5.2406
280
0.02746
200.72
283.09
5.7414
0.00824
185.34
267.69
5.3331
300
0.02958
216.21
^304.94
5.8168
0.00895
202.38
291.90
5.4167
350
0.03480
254.57
~358.96
5.9834
0.01067
243.57
350.26
5.5967
400
0.03993
292.70
412.50
6.1264
0.01232
283.59
406.79
5.7477
500
0.05008
369.06
519.29
6.3647
0.01551
362.42
517.48
5.9948
600
0.06013
446.57
626,95
6.5609
0.01861
441.47
627.58
6.1955
700
0.07012
525.9?
736.35
6.7295
0.02167
521.96
738.65
6.3667
800
0.08008
607.67
847.92
6.8785
0.02470
604.42
851.43
6.5172
900
0.09003
691.65
961.73
7.0125
0.02771
689.02
966.15
6.6523
1000
0.09996
777.85
1077.72
7.1347
0.03072
775.68
1082.84
6.7753
718 H Appendix B SI Units: Thermodynamic Tables
Table B.7
Thermodynamic Properties of Methane
Table b.7.1
Saturated Methane
Temp.
(K)
p
(kPa)
Specific Volume
INTERNAL ENERGY
v fs
tt f
ii/g
u g
90.7
11.7
0.002215
3.97941
3.98163
—358.10
496.59
138.49
95
19.8
0.002243
2.44845
2.45069
-343.79
488.62
144.83
100
34.4
0.002278
1.47657
1.47885
—326.90
478.96
152.06
105
56.4
0.002315
0.93780
0.94012
-309.79
468.89
159.1 1
no
88.2
0.002353
0.62208
0.62443
-292.50
458.41
165.91
111.7
101.3
0.002367
0.54760
0.54997
-286.74
454.85
168.10
115
132.3
0.002395
0.42800
0.43040
-275.05
447.48
172.42
120
191.6
0.002439
0.30367
0.30610
-257.45
436.02
178.57
125
269.0
0.002486
0.22108
0.22357
-239.66
423.97
184.32
130
367.6
0.002537
0.16448
0.16701
-221.65
411.25
189.60
135
490.7
0.002592
0.12458
0.12717
-203.40
397.77
194.37
140
641.6
0.002653
0.09575
0.09841
-184.86
383.42
198.56
145
823.7
0.002719
0.07445
0.07717
-165.97
368.06
202.09
150
1040.5
0.002794
0.05839
0.06118
-146.65
351,53
204.88
155
1295.6
0.002877
0.04605
0.04892
-126.82
333.61
206.79
160
1592.8
0.002974
0.03638
0.03936
-106.35
314.01
207.66
165
1935.9
0.003086
- 0.02868
0.03177
-85.06
292.30
207.24
170
2329.3
0.003222
0.02241
0.02563
-62.67
267.81
205.14
175
2777.6
0.003393
0.01718
0.02058
-38.75
239.47
200.72
180
3286.4
0.003623
0.01266
0.01629
-12.43
205.16
192.73
185
3863.2
0.003977
0.00846
0.01243
18.47
159.49
177.96
190
4520.5
0.004968
0.00300
0.00797
69.10
67.010
136.11
190.6
4599.2
0.006148
0.00615
101.46
101.46
Appendix B SI Units: Thermodynamic Tables
■ 719
TABLE B.7.1 {continued)
Saturated Methane
Temp.
( k )
P
(kPa)
Enthalpy
Entropy
h f
h h
h t
s f
*f
90.7
11.7
-358.07
543.12
185.05
4.2264
5.9891
10.2155
95
19.8
-343.75
537.18
193.43
4.3805
5.6545
10.035
100
34.4
-326.83
529.77
202.94
4.5538
5.2977
9.8514
105
56.4
-309.66
. 521.82
212.16
4.7208
4.9697
9.6905
110
88.2
-292.29
513.29
221.00
4.8817
4.6663
9.5480
111.7
101.3
-286:50
510.33
223.83
4.9336
4.5706
9.5042
115
132.3
-274.74
504.12
229.38
5.0368
4.3836
9.4205
120
191.6
-256.98
494.20
237.23
5.1867
4.1184
9.3051
125
269.0
-238.99
483.44
244.45
5.3321
3.8675
9.1996
130
367.6
-220.72
471.72
251.00
5.4734
3.6286
9.1020
135
490.7
-202.13
458.90
256.77
5.6113
3.3993
9.0106
140
641.6
-183.16
444.85
261.69
5.7464
3.1775
8.9239
145
823.7
-163.73
429.38
265.66
5.8794
2.9613
8.8406
150
1040.5
- 143.74
412.29
268.54
6.0108
2.7486
8.7594
155
1295.6
-123.09
393.27
270.18
6.1415
2.5372
8.6787
160
1592.8
-101.61
371.96
270.35
6.2724
2.3248
8.5971
165
1935.9
-79.08
347.82
268.74
6.4046
2.1080
8.5126
170
2329.3
-55.17
320.02
264.85
6.5399
1.8824
8.4224
175
2777.6
-29.33
287.20
257.87
6.6811
1.6411
8.3223
180
3286.4
-0.53
246.77
246.25
6.8333
1.3710
8.2043 .
185
3863.2
33.83
192.16
226.00
7.0095
1.0387
8.0483
190
4520.5
91.56
80.58
172.14
7.3015
0.4241
7.7256
190.6
4599.2
129.74
129.74
7.4999
7.4999
720 APPENDIX B SI UNITS: THERMODYNAMIC TABLES
Table b/7.2
Superheated Methane
Temp.
(K)
V
(mVkg)
u
(kJ/kg)
h
(kJ/kg)
s
(kJ/kg-K)
V
(m 3 /kg)
u
(kJ/kg)
h
(kJ/kg)
s
(kJ/kg-K)
100 kPa(111.50K)
200 kPa (120.61 K)
Sat.
0,55665
167.90
223.56
9.5084
0.29422
179.30
238.14
9.2918
125
0.63126
190.21
253.33
9.7606
0.30695
186.80
248.19
9.3736
150
0.76586
230.18
306.77
10.1504
0.37700
227.91
303.31
9.7759
175
0.89840
269.72
359.56
10.4759
0.44486
268.05
357.02
10.1071
200
1.02994
309.20
412.19
10.7570
: 0.51165
■ 307.88
410.21
10.3912
225
1.16092
348,90
464.99
11.0058
0.57786
347.81
: 463.38
10.6417
250
1.29154
389.12
518.27
11.2303
0.64370
388.19
516.93
10.8674
275
1.42193
430.17
572.36
11.4365
0.70931
429.36
571.22
11.0743
300
1.55215
472.36
627.58
11.6286
0.77475
471.65
626.60
11.2670
325
1.68225
516.00
684.23
11.8100
0.84008
515.37
683.38
11.4488
350
1.81226
561.34
742.57
1 1.9829
0.90530
560.77
741.83
11.6220
375
1.94220
608.58
802.80
12.1491
0.97046
608.07
802.16
11.7885
400
2.07209
657.89
865.10
12.3099
1.03557
657.41
864.53
11.9495
425
2.20193
709.36
929.55
12.4661
1.10062
708.92
929.05
12.1059
APPENDIX B SI UNITS: THERMODYNAMIC TABLES 68 721
TABLE B.7.2 (continued)
Superheated Methane
Temp,
V
h
5
V
h
s
(K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
400 kPa (131.42 K)
600 kPa (138.72 K)
191.01
252.72
197,54
260.51
H OA*; Si
150
0.18233
223.16
296.09
9.3843
0.11717
218.08
288.38
9.1390
175
0.21799
264.61
351,81
9.7280
0.14227
261.03
346.39
9.4970
200
0.25246
305.19
406.18
10,0185
0.16603
302.44
402.06
9.7944
225
0.28631
345.61
460.13
10.2726
0.18911
343.37
456.84
10.0525
250
0.3197a
386.32
514.23
1 A CAAT
10.5007
A A 1 t O A
0.21180
384.44
511.52
10.2830
275
A O A 1
0.35301
427.74
568.94
1 A TAAA
10.7092
0,23424
426.11
566.66
10.4931
300
0.38606
470.23
624.65
t A AA*> t
10.9031
0.25650
468.80
622.69
10.6882
325
0.41899
514.10
681.69
1 1.0857
0.27863
512.82
680.00
10.8716
350
0.45 1 83
559.63
740.36
1 1,2595
0.30067
558.48
738.88
11.0461
375
A iO^ ^ A
0.48460
607.03
800.87
1 1.4265
0,32264
605.99
799.57
1 1 .21 36
400
0.51731
656.47
863.39
1 1.5879
0.34456
655.52
862.25
11,3754
425
0.54997
708.05
928.04
11.7446
0.36643
707.18
927.04
11.5324
450
0.58260
761.85
994.89
1 1.8974
0.38826
761.05
994.00
11.6855
475
0.61520
817.89
1063.97
12.0468
0.41006
817.15
1063.18
11.8351
500
0,64778
876.18
1135.29
12.1931
0.43184
875.48
1134.59
11.9816
525
0,68033
936.67
1208.81
12.3366
0.45360
936.03
1208.18
12.1252
800 kPa (144.40 K)
1000 kPa (149.13 K)
Sat.
0.07941
201.70
265.23
8.8505
0.06367
204.45
268.12
8.7735
150
0.08434
212,53
280.00
8.9509
0.06434
206.28
270.62
8.7902
175
0.10433
257.30
340,76
9.3260
0.08149
253.38
334.87
9.1871
200
0.12278
299.62
397.85
9.6310
0.09681
296.73
393.53
9.5006
225
0.14050
341.10
453.50
9.8932
0.11132
338.79
450.11
9.7672
250
0.15781
382.53
508.78
10.1262
0.12541
380.61
506.01
10.0028
275
0.17485
424.47
564.35
10.3381
0.13922
422.82
562.04
10.2164
300
0.19172
467.36
620.73
10.5343
0.15285 ■
465.91.
618.76
10.4138
325
0.20845
511.55
678.31
10.7186
0.16635
510.26
676.61
10.5990
350
0.22510
557.33
737.41
10.8938
0.17976
556.18
735.94
10.7748
375
0.24167
604.95
798.28
11.0617 "
0.19309
603.91
797.00
10.9433
400
0.25818
654.57
861.12
11.2239
0.20636
653.62
859.98
11.1059
425
0.27465
706.31
926.03
11.3813
0.21959
705.44
925.03
11.2636
450
0.29109
760.24
993.11
11.5346
0.23279
759.44
992.23
11,4172
475
0.30749
816.40
1062.40
11.6845
0.24595
815.66
1061.61
11.5672
500
0.323S7
874.79
1133.89
11.8311
0.25909
874.10
1133.19
11.7141
525
0.34023
935.38
1207,56
11.9749
0.27221
934.73
1206.95
11.8580
550
0.35657
998.14
1283.45
12.1161
0,28531
997.53
1282.84
11.9992
722 ffl appendix b si Units: thermodynamic tables
TABLE B.7.2 (continued)
Superheated Methane
Temp.
V
u
h
s
V
h
s
(K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(kJ/kg-K)
(m 3 /kg)
(kJ/kg)
(kJ/kg)
(KJ/Kg-RJ
1500 kPa (158.52 K)
2000 kPa (165.86 K)
Sat.
0.04196
207.53
270.47
8.6215
0.03 062
207,01
268.25
8.4975
175
0.05078
242.64
318.81
c on i
U.UJJU*t
229.90
299.97
8.6839
200
0.06209
289.13
382.26
9.2514
0.04463
280.91
370.17
9.0596
225
0.07239
332.85
441.44
9.5303
0.05289
326.64
432 A3
9.3532
250
0.08220
375.70
499.00
9.7730
0.06059
370.67
491.84
9.6036
275
0.09171
418.65
556.21
9.9911
0.06796
414.40
550.31
9.8266
300
0.10103
462.27
613.82
10.1916
0.07513
458.59
608.85
10.0303
325
0.11022
507.04
672.37
10.3790
0.08216
503.80
668.12
10.2200
350
0.11931
553.30
732.26
10.5565
0.08909
550.40
728.58
10.3992
375
0.12832
601.30
793.78
10.7263
0.09594
598.69
790.57
10.5703
400
0.13728
651.24
857.16
10.8899
0.10274
648.87
854.34
10.7349
425
0.14619
703.26
922.54
11.0484
0.10949
701.08
920.06
10.8942
450
0.15506
757.43
990.02
11.2027
0.11620
755.43
987.84
11.0491
475
0.16391
813.80
1059.66
11.3532
0.12289
811.94
1057.72
11.2003
500
0.17273
872.37
1131.46
11.5005
0.12955
870.64
1129.74
11.3480
525
0.18152
933.12
1205.41
11.6448
0.13619
931.51
1203.88
11.4927
550
0.19031
996.02
1281.48
11.7864
0.14281
994.51
1280.13
11.6346
4000 kPa (186.10 K)
8000 kPa
Sat.
0.01160
172.96
219.34
o.UUJ J
200
0.01763
237.70
308.23
8.4675
0.00412
55.58
88.54
7.2069
225
0.02347
298.52
392.39
8.8653
0.00846
217.30
284.98
8.1344
250
0.02814
349.08
461.63
9.1574
0.01198
298.05
393.92
8.5954
275
0.03235
396.67
526.07
9.4031
0.01469
357.88
475.39
8.9064
^300
0.03631
443.48
588.73
9.6212
0.01705
411.71
548.15
9.1598
*325
0.04011
490.62
651.07
9.8208
0.01924
463.52
617.40
9.3815
350
0.04381
538.70
713.93
10.0071
0.02130
515.02
685.39
9.5831
375
0.04742
588.18
777.86
10.1835
0.02328
567.12
753.34
9.7706
400
0.05097
639.34
843.24
10.3523
0.02520
620.38
821.95
9.9477
425
0.05448
692.38
910.31
10.5149
0.02707
675.14
891.71
10.1169
450
0.05795
747.43
979.23
10.6725 •
0.02891
731.63
962.92
10.2796
475
0.06139
804.55
1050.12
10.8258
0.03072
789.99
1035.75
10.4372
500
0.06481
863.78
1123.01
10.9753
0.03251
850.28
1110.34
10.5902
525
0.06820
925.11
1197.93
11.1215
0.03428
912.54
1186.74
10.7393
550
0.07158
988.53
1274.86
11.2646
0.03603
976.77
1264.99
10.8849
575
0.07495
1053.98
1353.77
1 1.4049
0.03776
1042.96
1345.07
11.0272
Appendix C
ideal-Gas Specific heat
Three types of energy storage or possession were identified in Section 2.6, of which two,
translation and intramolecular energy, are associated with the individual molecules.
These comprise the ideal-gas model, with the third type, the system intermolecular poten-
tial energy, then accounting for the behavior of real (nontdeal-gas) substances. This ap-
pendix deals with the ideal-gas contributions. Since these contribute to the energy, and
therefore also the enthalpy, they also contribute to the specific heat of each gas. The dif-
ferent possibilities can be grouped according to the intramolecular energy contributions
as follows:
Monatomic Gases (inert gases Ah, He, ne,
XE, Kr, ALSO N, O, H 5 CL, F, . . .)
^ ~ ^translation ^~ ^electronic ~ ?h
dh_Mh_,dh r _ r , r _5 B , ,™
dT~ dT dT' ~ 1 ~ 2 ^ '
where the electronic contribution, f e (T), is usually small, except at very high T (common
exceptions are O, CI, F).
DIATOMIC AN© LINEAR POLYATOMIC GASES
(N 2 , O a , CO, OH, . . . , C0 2 , N 2 0, . . . )
In addition to translational and electronic contributions to specific heat, these also have
molecular rotation (about the center of mass of the molecule) and also (3a — 5) indepen-
dent modes of molecular vibration of the a atoms in the molecule relative to one another,
such that
Cm = C m + C nr +C POo + C POg = ^R + R +f v (T) +f e (T)
where the vibrational contribution is
/^^lW-l) J ], Xl = %
723
724 Appendix c ideal-gas specific heat
and the electronic contribution, f£T), is usually small, except at very high T (common ex-
ceptions are 2> NO, OH).
EXAMPLE C.l N 2 , 3a - 5 = 1 vibrational mode, with 6 S = 3392 K.
At T = 300 K, C F0 = 0.742 + 0.2968 + 0.0005 + ~0 ='1.0393 kJ/kg K.
At T = 1000 K, C n = 0.742 4- 0.2968 + 0.123 + «0 = 1-1618 kJ/kg K.
(an increase of 1 1.8% from 300 K).
EXAMPLE C.2 C0 2> 3a - 5 = 4 vibrational modes, with 6, = 960 K, 960 K, 1993 K, 3380 K
At T = 300 K f Cpo = 0.4723 + 0. 1 889 + 0. 1 826 + ~0 = 0.8438 kJ/kg K.
AtJ= 1000 K,C PQ = 0.4723 + 0.1889 + 0.5659 + «0 = 1,2271 kJ/kgK.
(an increase of 45.4% from 300 K).
NONLINEAR POLYATOMIC MOLECULES
(H 2 0, NH 3 , CH 4 , C 2 H 6 , ...)
Contributions to specific heat are similar to those for linear molecules, except that the ro-
tational contribution is larger and there are (3a - 6) independent vibrational modes, such
that
Cpo = C m + C^r + + Cn e = |tf + \R +f v (T) +f£T)
where the vibrational contribution is
MT) = R%\&'W ~ I) 2 ]. *f = %
i=i *■
and/ e (I) is usually small, except at very high temperatures.
EXAMPLE C.3 CH 4 , 3a - 6 = 9 vibrational modes, with 8 t = 4196 K, 2207 K (two modes), 1879 K
(three), 4343 K (three)
At T = 300 K, Cpo = 1.2958 + 0.7774 + 0.1527 + ~0 = 2.2259 kJ/kg K.
At T = 1000 K, C w = 1.2958 + 0.7774 + 2.4022 + =0 = 4.4754 kJ/kg K.
(an increase of 101.1% from 300 K).
Appendix D
Equations of State
Some of the most used pressure-explicit equations of state can be shown in a form with
two parameters. This form is known as a cubic equation of state and contains as a special
case the ideal-gas law:
p _ RT ... a
v - b v 2 + cbv + db %
where (a, b) are parameters and {c, d) define the model as shown in the following table
with the acentric factor (w) and
b = boRTJP, and a = a{fi 2 T%P c
The acentric factor is defined by the saturation pressure at a reduced temperature
T r = 0.7
In Pf at T r = 0.7
TABLE D.l
Equations of State
Model
c
d
«o
Ideal gas
van der Waals
1/8
27/64
Redlich-Kwong
t
0.08664
. 0.42748 T7 m
Soave
1
0.08664
0.42748 [l+f{\
- T\ a )f
Peng-Robinson
2
-1
0.0778
0.45724 [1
f=
0.48 + 1.574a>
- 0.176a) 2
for Soave
/ =
0.37464 + 1.54226w - 0.26992a) 2
for Peng-Robinson
12S
726 APPENDIX D EQUATIONS OF STATE
X
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ocNootNOOtnvoosn
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r~. \o o t~~
o> © -3-
>n oo on
to on
^ *o on o o "~t o
c-- <n o vo
CN t~- v> O O i >— <
in o c-- f-i
VN
1/1 on
CO
"Sj- VN —1
\o \o t--
00 -st <N
o o — <
o o
o © o
3
. . _ , _ » M h h
CN VO ■— iOOt~-VOO'*i - Osn 1 C
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N h (S N
i-t t- o <n «S <— O
<N O O O O t~~
ofN,coo<N^ot<— r>ooog
©OO^^cOOOin-^OOOO
o'ooo'ooo'^ojoooo
O On " co " "
in on oo
OO N Oi W d
f-< " O O CO
oo^H^HCNt-o^ooinoooo
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ooooooooooooo
2 3 3
00
on in >n
to i— '
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O O < O CO oo
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VI <t h- J CO f; _^ ^ „
TfcNror^oNcoO'-;co©o^
ooooo-— i^t'r^o
o o
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VN
CN
CN CO
O —
tn co
CN co
oo ol
On i—i
CO vo
00 CO
co
t— cooONtN^-^-OO
i~iM-inONi~icOt---00
VO CO
in co
oo co
©OOOOOC^CO'tOOOO
ft5
O to (N
3
o o o o
cn .-t
CO CO
r-
On CN
i — ' co *n ^ *o
in i o o cn ■=£
r-- co o oo tn no
« h Ol VI
in t-» C\ tJ- -^f in
r-l ^ H o o o
o co vo o
© in m OS r>
vi ft vi oi N ^
in co m ^ t-- oo
M (() H rt OO O
^h" CO NO NO O
s
ooooooooooooo
o o co o
to *o in co cn "^f
o oo on co
CO (N on On CO f»
rj- m — r-; vq
t--" ^ f-I co m
a
1
K K ffi K W K >C ^ T
B ^ o u u o or d 2 o
§ S k
£ £ £
S w w
p CQ
appendix D equations of state W 727
Table d.3
The Lee-Kesler Equation of State
The Lee-Kesler generalized equation of state is
Z^ =
l + C + D c A (
v' r v? v' r 5 TX 2 V v' r 2 t
T 2 T 3 r
^ c i j<
T 3
„ , d 2
D = d, + ^
in which
rj, T „ P V
1 = — r — ■ — I) = ■
T V r P e ' r RT C !P C
The set of constants is as follows:
Constant
Simple Fluids Constant
Simple Fluids
0.118 1193 a
0.0
b 2
0.265 728 c 4
0.042 724
0.154 790 rfj X 10 4
0.155 488
b A
0.030 323 d 2 X 10 4
0.623 689
0.023 674 4 /?
0.653 92
c 2
0.018 698 4 7
0.060 167
TABLE D.4
Saturated Liquid-Vapor Compressibilities, Lee-Kesler Simple Fluid
0.40
0.50
0.60
0.70
0.80
0.85
0.90
0.95
1
P r sat
2.7E-4
4.6E-3
0.028
0.099
0.252
0.373
0.532
0.737
1
6.5E-5
9.5E-4
0.0052
0.017
0.042
0.062
0.090
0.132
0.29
0.999
0.988
0.957
0.897
0.807
0.747
0.673
0.569
0.29
Table D,5
Acentric Factor for Some Substances
Substance
a)
Substance
Ammonia
NH 3
0.25
Water
H 2
0.344
Argon
At
0.001
w-Butane
0.199
Bromine
Br 2
0.108
Ethane
C 2 Hs
0.099
Helium
He
-0.365
Methane
CH 4
0.011
Neon
Ne
-0.029
R-32
0.277
Nitrogen
N 2
0.039
R-125
0.305
728 B appendix D Equations of State
FIGURE D.l Lee-Kesler Simple Fluid Compressibility Factor.
Appendix D Equations of State H 729
5.5 j- r -. r ..:.. rr ..;. f L ,__
0.01 0.1 1 10
Reduced pressure, P r
FIGURE D.2 Lee-Kesler Simple Fluid Enthalpy Departure.
730 ^ APPENDIX D EQUATIONS OF STATE
0.01 0.1 1 10
Reduced pressure, P r
FIGURE D.3 Lee-Kesler Simple Fluid Entropy Departure.
Appendix e
Figures
<
731
732 S Appendix E figures
FIGURE E.l Temperature-entropy diagram for water.
Appendix b Figures M 733
( e dH> 3HnSS3Hd
FIGURE E.2 Pressure-enthalpy diagram for ammonia.
734 M Appendix E Figures
FIGURE E.3 Pressure-enthalpy diagram for oxygen.
appendix E figures M 735
Appendix f
english unit tables
V
&
737
a.
738 U Appendix F English Unit tables
TABLE F.l
Critical Constants (English Units)
Villi QflfA
Formula
Molec.
Weight
Temp.
(R)
Pressure
(lbf/in. 2 )
Volume
(tf/lbm)
Ammonia
NH 3
17.031
729.9
1646
C\ A£07
Aigon
Ar
39.948
271.4
706
U.UJUU
Bromine
Br 2
159.808
1058.4
1 ACiA
n ni 7i
Carbon dioxide
co 2
44.010
547.4
10 iv
a niii')
U.U J4ji
Carbon monoxide
CO
28.010
239.2
DUB
ft ft^l^
Chlorine
Cl 2
70.906
750.4
1 1
Fluorine
F 2
31.991
259.7
7<*7
0,0279
Helium
He
4.003
9.34
ft 73ftft
Hydrogen (normal)
H 2
2.016
59.76
iSS.O
ft ^17(1
Krypton
Kr
83.800
376.9
ft 0174
Neon
Ne
20.183
79.92
4uU
ft ftllft
Nitric oxide
NO
30.006
324.0
O/tA
Nitrogen
N 2
28.013
227.2
ft
Nitrogen dioxide
N0 2
46.006
775.8
1465
U.UJ04
Nitrous oxide
N 2
44.013
557.3
1 fl C f\
IUjU
Oxygen
o 2
31.999
278.3
731
n CVXC.1
U.UjO /
Sulfur dioxide
so 2
64.063
775.4
J. 14J
ft ftlftfi
Water
H 2
18.015
1 1 £Z 1
U.UJUO
Xenon
Xe
131.300
521.5
QAH
84/
ft ftl^Ld
Acetylene
C 2 H 2
26.038
554.9
QOI
ft ft^O^
U.VU7J
Benzene
C 6 H 6
78.114
1012.0
709
A A^11
M-Butane
C 4 H 10
58.124
765.4
551
U.U/UJ
Chlorodifluoroethane (142b)
CHjCCIFj
100.495
/jS.j
O Iv
0.0368
ChtorodiSuoromethane (22)
CHC1F 2
C£ AAQ
721
0.0307
Dichlorodifluoroethane (141)
CH 3 CC1 2 F
116.950
866.7
658
0.0345
Dichlorotrifluoroethane (123)
CHC1 2 CF 3
152.930
822.4
532
0.0291
Difluoroethane (152a)
CHF 2 CH 3
66.050
695.5
656
0.0435
Difluoromethane (32)
CH 2 F 2
52.024
632.3
838
0.0378
Ethane
Oft
30.070
549.7
708
0.0790
Ethyl alcohol
QHjOH
46.069
925.0
891
0.0581
Ethylene
C 2 H 4
28.054
508.3
731
0.0744
n-Heptane
100.205
972.5
397
0.0691
n-Hexane
C 6 H 14
86.178
913.5
437
0.0688
Methane
CH 4
16.043
342.7
667
0.0990
Methyl alcohol
CH 3 OH
32.042
922.7
1173
0.0590
H-Octane
CgH 18
114,232
1023.8
361
0.0690
Pentafluoroethane (125)
CHF 2 CF 3
120.022
610.6
525
0.0282
M-Pentane
C S H 12
72.151
845.5
489
0.0675
Propane
C 3 H 3
44.094
665.6
616
0.0964
Propene
C 3 H 6
42.081
656.8
667
0.0689
Tetrafluoroethane (134a)
CF 3 CH 2 F
102.030
673.6
589
0.0311
Appendix F English Unit Tables H 739
TABLE F.2
TABLE F.3
Properties of Selected Solids at 77 F
Properties of Some Liquids at 77 F
Substance
P
Obm/ft 3 )
(Btu/lbm R)
Substance
P
(Ibm/ft 3 )
C p
(Btu/lbm R)
Asphalt
132.3
Ammonia
37.7
1.151
Brick, common
112.4
Benzene
54.9
0.41
Carbon, diamond
202.9
Butane
34.7
0.60
Carbon, graphite
125-156
U..l*tO
CC1 4
98.9
0.20
Coai
75-95
ft 3fls
co a
42.5
0.69
Concrete
137
0.21
Ethanol
48.9
0.59
Glass, plate
156
0.191
Gasoline
46.8
0.50
Glass, wool
12.5
0.158
Glycerine
78.7
0.58
Granite
172
0.212
Kerosene
50.9
0.48
Ice (32°F)
57.2
0.487
Methanol
49.1
0.61
Paper
43.7
0.287
H-octane
43.2
0.53
Plexiglas
73.7
0.344
Oil, engine
55.2
0.46
Polystyrene
57.4
0.549
Oil, light
57
0.43
Polyvinyl chloride
86.1
0.229
Propane
31.8
0.61
Rubber, soft
68.7
fl 1QQ
R-12
81.8
0.232
Sand, dry
93.6
0.191
R-22
74.3
0.30
Salt, rock
130-156
0.2196
R-32
60
0.463
Silicon
145.5
0.167
R-125
74.4
0.337
Snow, firm
35
0.501
R-134a
75.3
0.34
Wood, hard (oak)
44.9
0.301
Water
62.2
1.00
Wood, soft (pine)
Wool
31.8
6.24
0.33
0.411
Liquid Metals
Bismuth, Bi
627
0.033
Metals
Lead, Pb
665
0.038
Aluminum, duralumin
170
0.215
Mercury, Hg
848
0.033
Brass, 60-40
524
0.0898
NaK (56/44)
55.4
0.27
Copper, commercial
518
0.100
Potassium, K
51.7
0.193
Gold
1205
0.03082
Sodium, Na
58
0.33
Iron, cast
454
0.100
Tin, Sn
434
0.057
Iron, 304 St Steel
488
0.110
Zinc, Zn
410
0.12
Lead
708
0.031
Magnesium, 2% Mn
111
0.239
Nickel, 10% Cr
541
0.1066
Silver, 99.9% Ag
657
0.0564
Sodium
60.6
0.288
Tin
456
0.0525
Tungsten
1205
0.032
Zinc
446
0.0927
740 B Appendix F English unit Tables
TABLE F.4
Properties of Various Ideal Gases at 77 F, 1 aim* (English Units)
Chemical Molecular R p X 10 3 C rt
Gas
Formula
Mass
(ft-lbf/Ibm R)
(lbm/ft 3 )
(Btu/lbm R)
Steam
H 2
18.015
85.76
1 A AO
0.447
0.337
1.327
Acetylene
26.038
59.34
OJ. J 3
0.406
0.330
1.231
Air
2s. y /
11 OS
0.240
0.171
1.400
Ammonia
NH 3
17.031
Clf\ Tl
yu. 11.
0.509
0.392
1.297
Argon
: At
39.948
38.05
1 no 1
0.124
0.0745
1.667
Butane
C 4 H[
58,124
26.58
I jU.J
0.410
0.376
1.091
Carbon dioxide
C0 2
44.01
35.10
1 iU.O
0.201
0.156
1.289
Carbon monoxide
CO
28.01
55.16
/U.j
0.249
0.178
1.399
Ethane
C2H 6
30.07
51. jS
0.422
0.356
1.186
Ethanol
C2H5OH
4o.U6y
1 1 1.0
0.341
0.298
1.145
Ethylene
C2H4
J J.U l
71.04
0.370
0.299
1.237
Helium
He
4.003
386.0
10.08
1.240
0.744
1.667
Hydrogen
H 2
2.016
766.5
5.075
3.394
2.409
1.409
Methane
CH 4
16.043
96.35
40.52
0.538
0.415
1.299
Methanol
CHjOH
32.042
48.22
81.78
0.336
0.274
1.227
Neon
Ne
20.183
76.55
50.81
0.246
0.148
1.667
Nitric oxide
NO
30.006
51.50
75.54
0.237
0.171
1.387
Nitrogen
N 2
28.013
55.15
70.61
0.249
0.178
1.400
Nitrous oxide
N 2
44.013
35.10
110.8
0.210
0.165
1.274
jj-octane
114.23
13.53
5.74
0.409
0.391
1.044
Oxygen
2
31.999
48.28
80.66
0.220
0.158
1.393
Propane
C 3 H g
44.094
35.04
112.9
0.401
0,356
1.126
R-12
CCl 2 F 2
120.914
12.78
310.9
0.147
0.131
1.126
R-22
CHC1F 2
86.469
17.87
221.0
0.157
0.134
1.171
R-32'
CF 2 H 2
52.024
29.70
132.6
0.196
0.158
1.242
R-125
CHF 2 CF 3
120.022
12.87
307.0
0.189
0.172
1.097
R-134a
CF 3 CH 2 F
102,03
15.15
262.2
0.203
0.184
1.106
Sulfur dioxide
so 2
64.059
24.12
163.4
0.149
0.118
1.263
Sulfur trioxide
so 3
80.053
19.30
204.3
0.152
0.127
1.196
*Or saturation pressure if it is less than 1 arm.
Appendix ¥ English Unit Tables H 741
Table f,s
Ideal-Gas Properties of Air (English Units)Q, Standard Entropy at 1 aim = 101.325 kPa = 14.696 lbf/in. 2
T
a
It
4
T
u
h
4
/"O * * 1 /111 \
(litu/iDni)
LoUI/IDm H.)
(K)
(Utti/iomj
(mu/lom)
(mu/ibm K)
400
68.212
95.634
1.56788
1950
357.243
490.928
1.96404
440
75.047
105.212
1.59071
2000
367.642
504.755
1.97104
480
81.887
1 14.794
1.61155
2050
378.096
518.636
1.97790
520
88.733
124.383
1.63074
2100
388.602
532.570
1.98461
536.67
91.589
128.381
1.63831
2150
399.158
546.554
1.99119
540
92.160
129.180
1.63979
2200
409.764
560.588
1.99765
560
95.589
133.980
1.64852
2300
431.114
588.793
2.01018
600
102.457
143.590
1.66510
2400
452.640
617.175
2.02226
640
109.340
153.216
1.68063
2500
474.330
645.721
2.03391
680
116.242
162.860
1.69524
2600
496.175
674.421
2.04517
720
123.167
172.528
1.70906
2700
518.165
703.267
2.05606
760
130.118
182.221
1.72216
2800
540.286
732.244
2.06659
800
137.099
191.944
1.73463
2900
562.532
761.345
2.07681
840
144.114
201.701
1,74653
3000
584.895
790.564
2.08671
880
151.165
211.494
1.75791
3100
607.369
819.894
2.09633
920
158.255
221.327
1.76884
3200
629.948
849.328
2.10567
960
165.388
231.202
1.77935
3300
652.625
878.861
2.11476
1000
172.564
241.121
1.78947
3400
675.396
908.488
2.12361
1040
179.787
251.086
1.79924
3500
698.257
938.204
2.13222
1080
187.058
261.099
1.80868
3600
721.203
968.005
2.14062
1120
194.378
271.161
1.81783
3700
744.230
997.888
2.14880
1160
201.748
281.273
1.82670
3800
767.334
1027.848
2.15679
1200
209.168
291.436
1.83532
3900
790.513
1057.882
2.16459
1240
216.640
301.650
1.84369
4000
813.763
1087.988
2.17221
1280
224.163
311.915
1.85184
4100
837.081
1118.162
2.17967
1320
231.737
322.231
1.85977
4200
860.466
1148.402
2.18695
1360
239.362
332.598
1.86751
4300
883.913
1178.705
2.19408
1400
247.037
343.016
1.87506
4400
907.422
1209.069
2.20106
1440
254.762
353.483
1.88243
4500
930.989
1239.492
2.20790
1480
262.537
364.000
1.88964
4600
954.613
1269.972
2.21460
1520
270.359
374.565
1.89668
4700
978.292
1300.506
2.22117
1560
278.230
385.177
1.90357
4800
1002.023
1331.093
2.22761
1600
286.146
395.837
1.91032
4900
1025.806
1361.732
2.23392
1650
296.106
409.224
1.91856
5000
1049.638
1392.419
2.24012
1700
306.136
422.681
1.92659
5100
1073.518
1423.155
2,24621
1750
316.232
436.205
1.93444
5200
1097.444
1453.936
2.25219
1800
326.393
449.794
1.94209
5300
1121.414
1484.762
2.25806
1850
336.616
463.445
1.94957
5400
1145.428
1515.632
2.26383
1900
346.901
477.158
1.95689
742 M APPENDIX F ENGLISH UNIT TABLES
TABLE F.6
Ideal-Gas Properties of Various Substances (English Units), Entropies atl atm Pressure
NITROGEN, DIATOMIC (N 2 )
NITROGEN, MONATOMIC (N)
Btu/lb mol
203 216 Btu/lb mol
M
= 28.013
M = 14.007
T
R
Btu/lb mol
Btu/lbmol R
Btu/lb mol Btu/lbmol R
-3727
-2664
o
200
-2341
38.877
— 1671
31.689
400
-950
43.695
—679
35.130
537
45.739
o
36.589
600
441
46.515
314
37.143
800
1837
48.524
1307
38.571
1000
3251
50.100
2300
39.679
1200
4693
51.414
3293
40.584
1400
6169
52.552
4286
41.349
1600
7681
53.561
5279
42.012
1800
9227
54.472
6272
42.597
2000
10804 '
55.302
7265
43.120
2200
12407
56.066
8258
43.593
2400
14034
56.774
9251
44.025
2600
15681
57.433
10244
44.423
2800
17345
58.049
11237
44.791
3000
19025
58.629
12230
45.133
3200
20717
59.175
13223
45.454
3400
22421
59.691
14216
45,755
3600
6ft IS1
15209
46.038
3800
25857
60.647
16202
46.307
4000
27587
61.090
17195
46.562
4200
29324
61.514
18189
46.804
4400
31068
61.920
19183
47.035
4600
32817
62.308
20178
47.256
4800
34571
62.682
21174
47.468
5000
36330
63.041
22171
47.672
5500
40745
63.882
24670
48.148
6000
45182
64.654
27186
48.586
6500
49638 ,
65.368
29724
48.992
7000
54109
66.030
32294
49.373
7500
58595
66.649
34903
49.733
8000
63093
67.230
37559
50.076
8500
67603
67.777
40270
50.405
9000
72125
68.294
43040
50.721
9500
96658
68.784
45875
51.028
10000
81203
69.250
48777
51.325
Appendix f English unit Tables M
TABLE F.6 (continued)
Ideal-Gas Properties of Various Substances (English Units), Entropies at 1 aim Pressure
Oxygen, diatomic (0 2 )
Oxygen, Monatomic (0)
Btu/lb mol
107 124 Btu/lb mol
M
= 31.999
M= 16.00
T
S T
Btu/lb mol
Btu/lbmol R
Btu/lb mol Bto/lbniol R
-3733
-289I
200
-2345
42.100
-1829
33.041
400
-955
46.920
-724
36.884
537
48.973
o
38.442
600
446
49.758
330
39.023
800
1881
51.819
1358
1000
3366
53.475
2374
41.636
1200
4903
54.876
3383
42.556
1400
6487
56.096
4387
43.330
1600
8108
57.179
5389
43.999
1800
9761
58.152
6389
2000
11438
59.035
7387
45.114
2200
13136
59.844
8385
45.589
2400
14852
60.591
9381
46.023
2600
16584
61.284
10378
46.422
2800
18329
61.930
l Li Id
46.791
3000
20088
62.537
12369
47.134
3200
21860
63.109
13364
47.455
3400
23644
63.650
14359
47.757
3600
25441
64.163
15354
48.041
3800
27250
64.652
16349
48.310
4000
29071
65.119
17344
48.565
4200
30904
65.566 .
18339
48.808
4400
32748
65.995
19334
49.039
4600
34605
66.408
20330
49.261
4800
36472
66.805
21327
49.473
5000
38350
67.189
22325
49.677
5500
43091
• 68.092
24823
50.153
6000
47894
68.928
27329
50.589
6500
52751
69.705
29847
50.992
7000
57657
70.433
32378
51.367
7500
62608
71.116
34924
51.718
8000
67600
71.760
37485
52.049
8500
72633
72.370
40063
52.362
9000
77708
72.950
42658
52.658
9500
82828
73.504
45270
52.941
10000
87997
74.034
47897
53.210
744 U Appendix F English unit Tables
TABLE F.6 (continued)
Ideal-Gas Properties of Various Substances (English Units), Entropies at 1 aim Pressure
CARBON DIOXIDE (COj)
-169 184Btu/lbmol
M= 44.01
;,0 _
Carbon Monoxide (CO)
-47 518Btu/lb mol
M = 28.01
"JIJ37
T
,.0
T 1
s\
R
Btu/lb mol
uni/iDmoi k
Phi/lh mnl
JDlU/iU IOUI
Btu/lbmol R
-4026
-3728
200
-2636
43.466
-2343
40.319
400
-1153
48.565
-951
45.137
537
51.038
47.182
600
573
52.047
441
47.959
800
2525
54.848
1842
49.974
1000
4655
57.222
3266
51.562
1200
6927
59.291
4723
52.891
1400
9315
61.131
6220
54.044
1600
11798
62.788
7754
55.068
1800
14358
64.295
9323
55.992
2000
16982-
65.677
10923
56.835
2200
19659
66.952
12549
57.609
2400
22380
68.136
14197
58.326
2600
25138
69.239
15864
58.993
2800
27926
70.273
17547
59.616
3000
30741
71.244
19243
60.201
3200
33579
72.160
20951
60.752
3400
36437
73.026
22669
61.273
3600
39312
73.847
24395
61.767
3800
42202
74.629
26128
62.236
4000
45105
75.373
27869
62.683
4200
48021
76.084
29614
63.108
4400
50948
76.765
31366
63.515
4600
53885
77.418
33122
63.905
4800
56830
78.045
34883
64.280
5000
59784
78.648
36650
64.641
5500
67202
80.062
41089
65.487
6000
74660
81,360
45548
66.263
6500 ■
82155
82.560
50023
66.979
7000
89682
83.675
54514
67.645
7500
97239
84.718
59020
68.267
8000
104823
85.697
63539
68.850
8500
112434
86.620
68069
69.399
9000
120071
87.493
72610
69.918
9500
127734
88.321
77161
70.410
10000
135426
89.110
81721
70.878
APPENDIX P ENGLISH UNIT TABLES H 745
TABLE F.6 (continued)
Ideal-Gas Properties of Various Substances (English Units), Entropies at 1 atm Pressure
Water (H 2 0)
= —103 966Btu/lb mol
HYDROXYL (OH)
A/^7 = 16 761 Btu/ib mol
18.015
If
M —
1 1 fif\1
T
4
~ S T
XV
otu/io mot
mu/ibmol K
Btu/Jb mol
Btu/lbmol R
-4528
-3943
200
-2686
37.209
-2484
36.521
400
-1092
42.728
-986
41.729
537
45.076
43.852
600
509
45.973
452
44.649
800
2142
48.320
1870
46.689
1000
3824
50.197
3280
48.263
1200
5566
51.784
4692
49.549
1400
7371
53.174
6112
50.643
1600
9241
54.422
7547
51.601
1800
11178
55.563
9001
52.457
2000
13183
56.619
10477
53.235
2200
15254
57.605
11978
53.950
2400
17388
58.533
13504
54.614
2600
19582
59.411
15054
55.235
2800
21832
60.245
16627
55.817
3000
24132
61.038
18220
56.367
3200
26479
61.796
19834
56.887
3400
28867
62.520
21466
57.382
3600
31293
63.213
23114
57.853
3800
33756
63.878
24777
58.303
4000
36251
64,518
26455
58.733
4200
38774
65.134
28145
59.145
4400
41325
65.727
29849
59.542
4600
43899
66.299
31563
59.922
4800
46496
66.852
33287
60.289
5000
49114
67.386
35021
60.643
5500
55739
68.649
39393
61.477
6000
62463
69.819
43812
62.246
6500
69270
70.908
48272
62.959
7000
76146
71.927
52767
63.626
7500
83081
72.884
57294
64.250
8000
90069
73.786
61851
64.838
8500
97101
74.639
66434
65.394
9000 '
104176
75.448
71043
65.921
9500
111289
76.217
75677
66.422
10000
118440
76.950
80335
66.900
746 Appendix F English unit Tables
TABLE F.6 (continued)
Ideal-Gas Properties of Various Substances (English Units), Entropies at J aim Pressure
HYDROGEN (H 2 )
HYDROGEN, MONATOMIC (H)
1,0 _
u tstu/lb moi
»/537
= 93 723 Btu/lb mol
irl
— 9 fll £
— Z.U 1 □
M= 1.008
T
S T
S T
JBni/)b mol
Btu/lbmol R
Btu/lb mol Btu/lbmol R
-3640
-2664
200
-2224
24.703
-1672
22.473
400
-927
29.193
-679
25.914
537
31.186
27.373
600
438
31.957
314
27.927
800
1831
33.960
1307
29.355
1000
3225
35.519
2300
30.463
1200
4622
36.797
3293
31.368
1400
6029
37.883
4286
32.134
1600
7448
38.831
5279
32.797
1800
8884
39.676
6272
33.381
2000
10337 .
40.441
7265
33.905
2200
11812
41.143
8258
34.378
2400
13309
41.794
9251
34.810
2600
14829
42.401
10244
35.207
2800
16372
42.973
11237
35.575
3000
17938
43.512
12230
35.917
3200
19525
44.024
13223
36.238
3400
21133
44.512
14215
36.539
3600
22761
44.977
15208
36.823
3800
24407
45.422
16201
37.091
4000
26071
45.849
17194
37.346
4200
27752
46.260
18187
37.588
4400
29449
46.655
19180
37.819
4600
31161
47.035
20173
38.040
4800
32887
47.403
21166
38.251
5000
34627
47.758
22159
38.454
5500
39032
48.598
24641
38,927
6000
43513
49.378
27124
39.359
6500
48062
50.105
29606
39.756
7000
52678
50.789
32088
40.124
7500
57356
51.434
34571
40.467
8000
62094
52.045
37053
40.787
8500
66889
52.627
39535
41.088
9000
71738
53.182
42018
' 41.372
9500
76638
53.712
44500
41.640
10000
81581
54.220
46982
41.895
747
TABLE F.6 {continued)
Ideal-Gas Properties of Various Substances (English Units), Entropies at 1 attn Pressure
NITRIC OXIDE (NO)
NITROGEN DIOXIDE (NOJ
= 38 818 Btu/lb mol
M= 30.006
= 14 230 Btu/lb mol
M= 46.005
T
S T
R
Btu/lb mol Btu/lbmol R
Btu/lb mol Btu/lbmol R
— jyiz
ft
(J
— Hj ly
A
200
— 245 /
4j.Uoo
AQ 1Q1
400
— yfy
aq om
48. ZU/
_ 1 177
1 1 1 Z
jH. / oy
537
U
eft -1 1 1
ft
ins
600
A^t
4M
Jl.lU/
C£7
30/
JoJUt
800
1 Q8 1
1881
JJ.iOJ
OA fit}
£1 (lid
1000
C,i 7QO
04. /OS
£1 111
1200
A Q1A
30.1 jZ
D/ JJ
1400
03 12.
777
0(1/14
£>1 1 1 R
1600
/y4s
CO ion
£8 718
Uo./ 1 o
1800
C1CC7
jy.jJD
1 jyuj
7fl 1/18
/U. I OS
2000
111 Q1
i i iyj
1 £47 1
lOtZ 1
71 doi
2200
1 7c<i
£ft oso
1 HQ7fi
Lay fo
79 710
/Z. / 1 Z
2400
14jJZ
oi. / ly
71 *\£7
Zl 00/
71 838
2600
i £778
OZ. jy 1
7/11 87
ZH 1 5Z
74 RSS,
2800
17937
63.031
26819
75.861
3000
19657
63.624
29473
76.777
3200
21388
64.183
32142
77.638
3400
23128
64.710
34823
78.451
3600
24875
65.209
37515
79.220
3800
26629
65.684
40215
79.950
4000
28389
66.135
42923
80.645
4200
30154
66.565
45637
81.307
4400
31924
66.977
48358
81.940
4600
33698
67.371
51083
82.545
4800
35476
67.750
53813
83.126
5000
37258
68.113
56546
83.684
5500
41726
68.965
63395
84.990
600O
46212
69.746
.70260
86.184
6500
50714
70.467
77138
87.285
7000
55229
71.136
84026
88.306
7500
59756
71.760
90923
89.258
8000
64294
72.346
97826
90.149
8500
68842
72.898
104735
90.986
9000
73401
73.419
111648
91.777
9500
77968
73.913
118565
92.525
10000
82544
74.382
125485
93.235
748 H Appendix F English Unit Tables
table f.7
Thermodynamic Properties of Water
Table F.7.1
Saturated Water
Specific volume, iWlbm internal Energy, Btu/lbm
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat, Liquid
Evap.
Sat. Vai
(F)
(psia)
v f
v fs
v s
"/
"s
32
0.0887
0.01602
3301.6545
3301.6705
1021.21
1021.21
35
0.100
0.01602
2947.5021
2947.5181
2.99
1019.20
1022.19
40
0.122
0.01602
2445.0713
2445.0873
8.01
1015.84
1023.85
45
0.147
0.01602
2036.9527
2036.9687
13.03
1012.47
1025.50
50
0.178
0.01602
1703.9867
1704.0027
18.05
1009.10
1027.15
60
0.256
0.01603
1206.7283
1206.7443
28.08
1002.36
1030.44
70
0.363
0.01605
867.5791
867.5952
38.09
995.64
1033.72
80
0.507
0.01607
632.6739
632.6900
48.08
988.91
1036.99
90
0.699
0.01610
467.5865
467.6026
58.06
982.18
1040,24
100
0.950
0.01613
349.9602
349.9764
68.04
975.43
1043.47
110
1.276
0.01617
265.0548
265.0709
78.01
968.67
1046.68
120
1.695
0.01620
203.0105
203.0267
87.99
961.88
1049.87
130
2.225
0.01625
157.1419
157.1582
97.96
955.07
1053.03
140
2.892
0.01629
122.8567
122.8730
107.95
948.21
1056.16
150
3.722
0.01634
96.9611
96.9774
117.94
941.32
1059.26
160
4.745
0,01639
77.2079
77.2243
127.94
934.39
1062 32
170
5.997
0.01645
61.9983
62.0148
137.94
927.41
1065,35
180
7.515
0.01651
50.1826
50.1991
147.96
920.38
1068.34
190
9.344
0.01657
40.9255
40.9421
157.99
913.29
1071.29
200
11.530
0.01663
33.6146
33.6312
168.03
906.15
1074.18
210
14.126
0.01670
27.7964
27.8131
178.09
898.95
1077.04
212.0
14.696
0.01672
26.7864
26.8032
180.09
897.51
1077.60
220
17.189
0.01677
23.1325
23.1492
188.16
891.68
1079.84
230
20.781
0.01685
19.3677
19.3846
198.25
884.33
1082.58
240
24.968
0.01692
16.3088
16.3257
208.36
876.91
1085.27
250
29.823
0.01700
13.8077
13.8247
218.48
869.41
1087.90
260
35.422
0.01708
11.7503
11.7674
228.64
861.82
1090,46
270
41.848
0.01717
10.0483
10.0655
238.81
854.14
1092.95
280
49.189
0.01726
8.6325
8.6498
249.02
846.35
1095.37
290
57.535
0.01735
7.4486
' 7.4660
259.25
838.46
1097.71
300
66.985
0.01745
6.4537
6.4712
269.51
830.45
1099.96
310
77.641
0.01755
5.6136
5.6312
279.80
822.32
1102.13
320
89.609
0.01765
4.9010
4.9186
290.13
814.07
1104.20
330
103.00
0.01776
4.2938
4,3115
300.50
805.68
1106,17
340
117.94
0.01787
3.7742
3.7921
310.90
797.14
1108.04
350
134.54
0.01799
3.3279
3.3459
321.35
788.45
1109.80
Appendix F English Unit Tables 3 749
TABLE F.7.1 (continued)
Saturated Water
Enthalpy, Btu/lbm
ENTROPY, Btu/lbm R
Temp.
Press.
Sat. Liquid
Sat. Liquid
Lvap*
bat. Vspor
(F)
(psia)
h f
fs
K
g
s f
°fs
s„
s
32
f\ noon
1075.38
1075.39
2.1869
2.1869
35
A 1 AA
n AA
2,yy
1073.71
1076.70
0.0061
2.1703
2.1764
A A
4U
A 1
Q A1
B.U1
1070.89
1078.90
A Al £1
2.1430
2.1591
45
0.147
13.03
1068.06
1081.10
0.0262
2.1161
2.1423
50
A 1 TO
U.l la
1 Q AC
is. ID
1065.24
1083.29
A A1£1
U.U3 Ol
2.0898
2.1259
/TA
60
ft *x*c
U.256
10 AO
1059.59
1087.67
0.U555
2.0388
2.0943
art
/u
UJ63
1 O AA
38. oy
1053.95
1092.04
A AT A C
U.U /46
1.9896
2.0642
A £AT
U.jO/
.1 Q AO
48.08
1048.31
1096.39
A AA*3 1
1.9423
2.0356
90
A £AA
CO A£
58. 06
1042.65
1100.72
0.1 1 16
1.8966
2.0083
1 AA
10U
A ACA
£Q Ayl
68.04
1036.98
1105.02
A 1 **iftC
1.8526
1.9822
t i n
11U
1.2 /&
■"70 A1
/8.01
1031.28
1109.29
A 1 All
U.14/J
1.8101
1.9574
120
1.695
87.99
1025.55
1113.54
A 1 CA C
U.l 646
1.7690
1.9336
130
2,225
97.97
1019.78
1117.75
U.1K1 /
1.7292
1.9109
1 A A
140
2.892
T A"7 A£
lo/.yo
1013.96
1121.92
a 1 no<
u.iyos
1.6907
1.8892
150
3.722
1 17.95
1008.10
1 126.05
0.2150
1.6533
1.8683
1 £A
160
4.745
127.95
1002.18
1130.14
A ^1 1 1
0.2313
1.6171
1.8484
1 TA
170
5,yy/
13 /.yo
996.21
1134.17
0.24 1 5
1.5819
1.8292
180
7.515
147.98
990.17
1138.15
0.263 1
1.5478
1.8109
1 AA
iyo
9.344
ICO AO
155.
984.06
1142,08
a no/:
0,2/66
1.5146
1.7932
200
11.530
168.07
977.87
1145.94
0.2940
1.4822
1.7762'
210
14.126
178.13
971.61
1 149.74
0.3091
1.4507
1.7599
212.0
14.696
180.13
970.35
1150.49
0.3121
1.4446
1.7567
220
17.189
188.21
965.26
1 153.47
0.3240
1.4201
1.7441
230
20.781
198.31
958.81
1157.12
0.3388
1.3901
1.7289
240
24.968
208.43
952.27
1 160,70
0.3533
1.3609
1.7142
250
29.823
218.58
945.61
1164.19
0.3677
1.3324
1.7001
260
35.422
228.75
938.84
1167.59
0.3819
1.3044
1.6864
270
41.848
238.95
931.95
1170.90
0.3960
1.2771
1.6731
280
49.189
249.17
924.93
1174.10
0.4098
1.2504
1.6602
290
57.535
259.43
917.76 -
1177.19
0.4236
1.2241
1.6477
300
66.985
269.73
910.45
1180.18
0.4372
1.1984
1.6356
310
77.641
280.06
902.98
1183.03
0.4507
1.1731
1.6238
320
89.609
290.43
895.34
1185.76
0.4640
1.1483
1.6122
330
103.00
300.84
887.52
1188.36
0.4772
1.1238
1.6010
340
117.94
311.29
879.51
1190.80
0.4903
1.0997
1.5900
350
134.54
321.80
871.30
1193.10
0.5033
1.0760
1.5793
750 M appendix F English Unit Tables
TABLE F.7.1 {continued)
Saturated Water
Specific volume, ftVlbm internal energy, Btu/ibm
Temp.
Press.
Sat Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
(F)
(psia)
v f
v *
"f
«A
"s
360
152.93
0.01811
2.9430
2.9611
331.83
779.60
1111.43
370
173.24
0.01823
2.6098
2.6280
342.37
770.57
1112.94
380
195.61
0.01836
2.3203
2.3387
352.95
761.37
1114.31
390
220.17
0.01850
2.0680
2.0865
363.58
751.97
1115.55
400 .
347.08
0.01864
1.8474
1.8660
374.26
742.37
1116.63
410
276.48
0.01878
1.6537
1.6725
385.00
732.56
1117.56
420
308.52
0.01894
1.4833
1.5023
395.80
722.52
1118.32
430
343.37
0.01909
1.3329
1.3520
406.67
712.24
1118.91
440
381.18
0.01926
1.1998
1.2191
417.61
701.71
I i 19.32
450
422.13
0.01943
1.0816
1.1011
428.63
690.90
1119.53
460
466.38
0.01961
0.9764
0.9961
439.73
679.82
1119.55
470
514.11
0.01980
0.8826
0.9024
450.92
668.43
1119.35
480
565.50
0.02000
0.7986
0.8186
462.21
656.72
1118.93
490
620.74
0.02021
0.7233
0.7435
473.60
644.67
1118.28
500
680.02
0.02043
0.6556
0.6761
485.11
632.26
1117.37
510
743.53
0.02066
0.5946
0.6153
496.75
619.46
1116.21
520
811.48
0.02091
0.5395
0.5604
508.53
606.23
1114.76
530
884.07
0.021 17
0.4896
0.5108
520.46
592.56
1113.02
540
961.51
0.02145
0.4443
0.4658
532.56
578.39
1110.95
550
1044.02
0.02175
0.4031
0.4249
544.85
563.69
1108.54
560
1131.85
0.02207
0.3656
0.3876
557.35
548.42
1105.76
570
1225.21
0.02241
0.3312
0.3536
570.07
532.50
1102.56
580
1324.37
0.02278
0.2997
0.3225
583.05
515.87
1098.91
590.
1429.58
0.02318
0.2707
0.2939
596.31
498.44
1094.76
600
1541.13
0.02362
0.2440
0.2676
609.91
480.11
1090.02
610
1659.32
0.0241 1
0.2193
0.2434
623.87
460.76
1084.63
620
1784.48
0.02465
0.1963
0.2209
638.26
440.20
1078.46
630
1916.96
0.02525
0,1747
0.2000
653.17
418.22
1071,38
640
2057.17
0.02593
0.1545
0.1804
668.68
394.52
1063.20
650
2205,54
0.02673
0.1353
0.1620
684.96
368.66
1053.63
660
2362.59
0.02766
0.1169
0.1446
702.24
340.02
1042,26
670
2528.88
0.02882
0.0990
0.1*278
720.91
307.52
1028.43
680
2705.09
0.03031
0.0809
0.1112
741.70
269.26
1010.95
690
2891.99
0.03248
0.0618
0.0943
766.34
220.82
987.16
700
3090.47
0.03665
0.0377
0.0743
801.66
145.92
947.57
705.4
3203.79
0.05053
0.0505
872.56
872.56
appendix f English unit tables H 751
TABLE F.7.1 {continued)
Saturated Water
ENTHALPY) Btu/lbm ENTROPY, Btu/tbm R
Temp.
Press.
bat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Elvap.
Oat, V aUUI
(psia)
'Is
/.
"g
v -
V
Sr
"Is
t
360
152.93
332.35
862.88
1195.23
0.5162
1.0526
1.5688
370
173.24
342.95
854.24
1197.19
0.5289
1.0295
1.5584
380
195.61
353.61
845.36
1198.97
0.5416
1.0067
1.5483
390
220.17
364.33
836.23
1200.56
0.5542
0.9841
1.5383
400
247.08
375.11
826.84
1201.95
0.5667
0.9617
1.5284
410
276.48
385.96
817.17
1203.13
0.5791
0.9395
1.5187
420
308.52
396.89
807.20
1204.09
0.5915
0.9175
1.5090
430
343.37
407.89
796.93
1204.82
0.6038
0.8957
1.4995
440
381.18
418.97
786.34
1205.31
0.6160
0.8740
1.4900
450
422.13
430.15
775.40
1205.54
0.6282
0.8523
1.4805
460
466.38
441.42
764.09
1205.51
0.6404
0.8308
1.4711
470
514.11
452.80
752.40
1205.20
0.6525
0.8093
1.4618
480
565.50
464.30
740.30
1204.60
0.6646
0.7878
1.4524
490
620.74
475.92
727.76
1203.68
0.6767
0.7663
1.4430
500
680.02
487.68
714.76
1202.44
0.6888
0.7447
1.4335
510
743.53
499.59
701.27
1200.86
0.7009
0.7232
1.4240
520
811.48
511.67
687,25
1 198.92
0.7130
0.7015
1.4144
530
884.07
523.93
672.66
1196.58
0.7251
0.6796
1.4048
540
961.51
536.38
657.45
1193.83
0.7374
0.6576
1.3950
550
1044.02
549.05
641.58
1190.63
0.7496
0.6354
1.3850 "
560
1131.85
561.97
624.98
1186.95
0.7620
0.6129
1.3749
570
1225.21
575.15
607.59
1182.74
0.7745
0.5901
1.3646
580
1324.37
588.63
589.32
1 177.95
0.7871
0.5668
1,3539
590
1429.58
602.45
570.06
1172.51
0.7999
0.5431
1.3430
600
1541.13
616.64
549.71
1166.35
0.8129
0.5187
1.3317
610
1659.32
631.27
528.08
1159.36
0.8262
0.4937
1.3199
620
1784.48
646.40
505.00
1151.41
0.8397
0.4677
1.3075
630
1916.96
662.12
480.21
1142.33
0.8537
0.4407
1.2943
640
2057.17
678.55
453.33
1131.89
0.8681
0.4122
1.2803
650
2205.54
695.87
423.89
• 1119.76
0.8831
0.3820
1.2651
660
2362.59
714.34
391.13
1 105.47
0.8990
0,3493
1.2483
670
2528.88
734.39
353.83
1088.23
0.9160
0.3132
1.2292
680
2705.09
756.87
309.77
1066.64
0.9350
0.2718
1.2068
690
2891.99
783.72
253.88
1037.60
0.9575
0.2208
1.1783
700
3090.47
822.61
167.47
990.09
0.9901
0.1444
1.1345
705.4
3203.79
902.52
902.52
1.0580
1.0580
752 H appendix f English Unit tables
Table f.7.2
Superheated Vapor Water
Temp.
V
u
h
5
V
ii
h
s
(F)
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ftVlbm)
(Btu/ibm)
(Btu/lbm)
(Btu/lbm R)
lpsia (101.70)
5 psia (162.20)
Sat.
333,58
1044.02
1105.75
1.9779
73.531
1062.99
1131.03
1.8443
200
392.51
1077.49
1150.12
2.0507
78.147
1076.25
1148.55
1,0/ t J
240
416.42
1091.22
1168.28
2.0775
83.001
1090.25
U 67.05
1.8987
280
440.32
1105.02
1186.50
2.1028
87.831
1104.27
1185.53
1.9244
j2\)
404, iy
1118.92
1204.82
2.1269
92.645
1118.32
1204.04
1.9487
1£A
488.05
1132.92
1223.23
2.1499
97.447
1132.42
1222.59
1.9719
a nn
4UU
ci 1 n\
1 147.02
1241.75
2.1720
102.24
1146.61
1241,21
1.9941
a An
44 U
JJD. /O
1161.23
1260.37
2.1932
107.03
1160.89
1259.92
2.0154
^"71 ^1
3 / L.J J
1182.77
1288.53
2.2235
1 14.21
1182.50
1288.17
2.0458
^nn
£1 1 ii
1219.30
1336.09
2.2 /w
126.15
1219.10
1335.82
2.0930
■7fi n
/UU
oyu. iL
1256.65
1384.47
2.3 1^2
110 AO
138.08
1256.50
1384.26
2.1367
auu
h<{\ in
/DU.jU
1294.86
1433.70
O 1 </ffl
2.3o4y
150.01
1294.73
1433.53
2.1774
yuu
auy.os
1333.94
1483.81
2.5y32
161.94
1333.84
1483.68
2.2157
IUUU
ot>y.4>
1373.93
1534.82
' 2.4294
173.86
1373.85
1534.71
2.2520
1 1UU
rim A1
1414.83
1586.75
2.4638
185.78
1414.77
1586.66
2.2864
nnn
IzUU
yoa.oU
1456.67
1639.61
o Ana
197,70
1456.61
1639.53
2.3192
i inn
iU4o\l /
1499.43
1693.40
z.5281
209.62
1499.38
1693.33
2.3507
1 Ann
i t n*? t/t
I1U/./4
1543,13
1748.12
2.5584
221,53
1543.09
1748.06
2.3809
lOpsia (193.19)
14.696 psia (21 1.99)
Sat.
38.424
1072.21
1143.32
1.7877
26.803
1077.60
1150.49
1.7567
200
38.848
1074.67
1146,56
1.7927
—
--
240
41.320
1089.03
1165.50
1.8205
27.999
1087.87
1164.02
L7764
280
43.768
1103.31
1184.31
1.8467
29.687
1102.40
1183.14
1.8030
320
46.200
1117.56
1203.05
1.8713
31.359
1116.83
1202.11
1.8280
* 360
48.620
1131.81
1221.78
1.8948
33.018
1131.22
1221.01
1.8516
400
51.032
1146.10
1240.53
1.9171
34.668
1145.62
1239.90
1.8741
440
53.438
1160.46
1259.34
1.9385
36.313
1160.05
1258.80
1.8956
500
57.039
1182.16
1287.71
1.9690
38.772
1181.83
1287.27
1.9262
600
63.027
1218,85
1335.48
2.0164
42.857
1218.61
1335.16
1.9737
700
69.006
1256.30
1384.00
2.0601
46.932
1256.12
1383.75
2.0175
800
74.978
1294.58
1433.32
2.1009
51.001
1294.43
1433.13
2.0584
900
80.946
1333.72
1483,51
2.1392
55.066
1333.60
1483.35
2.0967
1000
86.912
1373.74
1534.57
2.1755
59.128
1373.65
1534.44
2.1330
1100
92.875
1414.68
1586.54
2.2099
63.188
1414.60
1586.44
2.1674
1200
98.837
1456.53
1639.43
2,2428
67.247
1456.47
1639.34
2.2003
1300
104.798
1499.32
1693.25
2.2743
71.304
1499.26
1693.17
2.2318
1400
110.759
1543.03
1747.99
2.3045
75,361
1542.98
1747.92
2.2620
1500
116.718
1587.67
1803.66
2.3337
79.417
1587.63
1803.60
2.2912
1600
122.678
1633.24
1860.25
2.3618
83.473
1633.20
1860.20
2.3194
Appendix F English Unit Tables
h 753
TABLE F.7.2 {continued)
Superheated Vapor Water
Temp,
V
u
h
s
V
u
h
s
(F)
(fP/lbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ffVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
20 psia (227.96)
40 psia (267.26)
Sat.
20.091
1082.02
1156.38
1.7320
10.501
1092.27
1170.00
1.6767
240
20.475
1086.54
1162.32
1.7405
280
21.734
1101.36
1181.80
1.7676
10.711
1097.31
1176.59
1.6857
320
22.976
1116.01
1201.04
1.7929
11.360
1112.81
1196.90
1.7124
360
24.206
1130.55
1220.14
1.8168
11.996
1127.98
1216.77
1.7373
AOI
1145.06
1239.17
1 SICK
17 £ni
1 142.95
1236.38
1.7606
AA(\
7£ f.A1
Z0.O4Z
1159.59
1258.19
1 ax;i i
l.sol I
1157.82
1255.84
1.7827
78
Zo/K>0
1181.46
1286.78
i.syiy
1 A 1 £A
14.154
1180.06
1284.91
1.8140
DUU
1218.35
1334.80
1 ^
1 J.DOJ
1217.33
1333.43
1 0£7 1
l.tSozl
J't.'tDO
1255.91
1383.47
17 1 Cl£
l/.lyo
1255.14
1382.42
1.9063
1294.27
1432.91
7 CVJA'X
io. /UI
1293.65
1432.08
1 (VI 7,1
i.y4/4
A(\ AKCl
1333.47
1483.17
Z.UoZo
on 7 A7
1332.96
1482.50
1 nnfi
zll All
1373.54
1534.30
7 1 TAft
Zl. /UU
1373.12
1533.74
7 fV777
2.0222
1 IUU
^ £ ,177
1414.51
1586.32
7 1 11A
77 1 Cl£
1414.16
15S5.86
2.0568
1ZUU
1456.39
1639.24
7 1 £.(L1
Z.IOOJ
ZH.WU
1456.09
1638.85
2.0S97
ICQ
1499.19
1693.08
z.iy/o
1 QA
ZO.I84
1498.94
1692.75
2.1212
^ 171
1542.92
1747.85
7 770A
Z.ZZaV
77 £7 7
Z/.Of /
1542.70
1747.56
2.1515
CO -JJT7
Do. 3 DZ
1587.58
1803.54
7 7£77
Z.ZO IZ
7ft 1 <cn
ZV.lby
1587.38
1803.29
2.1807
i firm
I DUU
fit m
1633.15
1860.14
Z.ZSJ4
jU.ooU
1632.97
1859.92
7 nnon
2.2Uoy
60 psia (292.73)
SO psia (312.06)
Sat.
7.177
1098.33
1178.02
1.6444
5.474
1102.56
1183.61
1.6214
320
7.485
1109.46
1 192.56
1.6633
5.544
1105.95
1188.02
1.6270
360
7.924
1125.31
1213.29
1.6893
5.886
1122.53
1209.67
1.6541
400
8.353
1 140.77
1233.52
1.7134
6.217
1138.53
1230.56
1.6790
440
8.775
1156.01
1253.44
1.7360
6.541
1154.15
1250.98
1.7022
500
9.399
1178.64
1283.00
1.7678
. 7.017
1177.19 •
1281.07
1.7346
600
10.425
1216.31
1332.06
1.8165
■ 7.794
1215.28
1330.66
1.7838
700
11.440
1254.35
1381.37
1.8609
8.561
1253.57
1380.31
1.8285
800
12.448
1293.03
1431.24
1.9022 '
9.322
1292.41
1430.40
1.8700
900
13.452
1332.46
1481.82
1.9408
10.078
1331.95
1481.14
1.9087
1000
14.454
1372.71
1533.19
1.9773
10.831
1372.29
1532.63
1.9453
1100
15.454
1413.81
1585.39
2.0119
11.583
1413.46
1584.93
1.9799
1200
16.452
1455.80
1638.46
2.0448
12.333
1455.51
1638.08
2.0129
1300
17.449
1498.69
1692.42
2.0764
13.082
1498.43
1692.09
2.0445
1400
18.445
1542.48
1747.28
2.1067
13.830
1542,26
1746.99
2.0749
1500
19.441
1587.18
1803.04
2.1359
14.577
1586.99
1802.79
2.1041
1600
20.436
1632.79
1859.70
2.1641
15.324
1632.62
1859.48
2.1323
1800
22.426
1726.69
1975.69
2.2178
16.818
1726.54
1975.50
2.1861
2000
24.415
1824.02
2095.10
2.2685
18.310
1823.88
2094.94
2.2367
754 M Appendix f English Unit Tables
TABLE F.7.2 (continued)
Superheated Vapor Water
Xcnip*
D
u
h
s
V
it
h
5
(F)
(ft^/lbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
100 psia (327.85)
150 psia (358.47)
Sat.
4.4340
1105.76
1187.81
1.6034
3.0163
1111.19
1194.91
1.5704
350
4.5917
1115.39
1200.36
1.6191
400
A {\1 A A
1136.21
1227.53
I. CO I l
1130.10
1219.51
450
5.2646
1156.20
1253.62
1.6812
3.4547
1151.47
1247.36
1.6312
500
5.5866
1175.72
1279.10
1,7085
3.6789
1171.93
1274.04
1.6598
550
5.9032
1195.02
1304.25
1.7340
3.8970
1191.88
1300.05
1.6862
600
6.2160
1214.23
1329.26
1.7582
4.1110
1211.58
1325.69
1.7110
700
6.8340
1252.78
1379.24
1.8033
4.5309
1250.78
1376.55
1.7568
800
7.4455
1291.78
1429.56
1.8449
4.9441
1290.21
1427.44
1 .7989
900
8.0528
1331.45
1480.47
1.8838
5.3529
1330.18
1478.76
1,8381
1000
8.6574
1371.87
1532.08
1,9204
5.7590
1370.83
1530.68
1.8750
1100
9.2599
1413.12
1584.47
1.9551
6.1630
1412.24
1583.31
1.9098
1200
9.8610
1455.21
1637.69
1.9882
6.5655
1454.47
1636.71
1.9430
1300
10.4610
1498.18
1691.76
2.0198
6.9670
1497.55
1690.93
1.9747
1400
11.0602
1542.04
1746.71
2.0502
7.3677
1541.49
1745.99
2.0052
1500
11.6588
1586.79
1802.54
2.0794
7.7677
1586.30
1801.91
2.0345
1600
12.2570
1632.44
1859.25
2.1076
8.1673
1632.00
1858.70
2.0627
1800
13.4525
1726.38
1975.32
2.1614
8.9657
1726.00
1974.86
2.1165
2000
14.6472
1823.74
2094.78
2.2120
9.7633
1823,38
2094.38
2.1672
200 psia (381.86)
300 psia (417.42)
Sat.
2.2892
1114.55
1199.28
1.5464
1.5441
1118.14
1203.86
1.5115
400
2.3609
1123.45
1210.83
1.5600
450
2.5477
1 146.44
1240.73
1.5938
1.6361
1135.37
1226.20
1.5365
500
2.7238
1167.96
1268.77
1.6238
1.7662
1159.47
1257.52
1.5701
K 550
2.8932
1188.65
1295.72
1.6512
1.8878
1181.85
1286.65
1.5997
600
3.0580
1208.87
1322.05
1.6767
2.0041
1203.24
1314.50
1.6266
700
3.3792
1248.76
1373.82
1.7234
2.2269
1244.63
1368.26
1.6751
800
3.6932
1288.62
1425.31
1.7659
2.4421
1285.41
1420.99
1.7187
900
4.0029
1328.90
1477.04
1.8055
2.6528
1326.31
1473.58
1.7589
1000
4.3097
1369.77
1529.28
1.8425
2,8604
1367.65
1526.45
1.7964
1100
4.6145
1411.36
1582.15
1.8776
3.0660
1409.60
1579.80
1.8317
1200
4.9178
1453.73
1635.74
1.9109
3.2700
1452.24
1633.77
1.8653
1300
5.2200
1496.91
1690.10
1.9427
3.4730
1495.63
1688.43
1.8972
1400
5.5214
1540.93
1745.28
1.9732
3.6751
1539.82
1743.84
1.9279
1500
5.8222
1585.81
1801.29
2.0025
3.8767
1584.82
1800.03
1.9573
1600
6.1225
1631.55
1858.15
2.0308
4.0777
1630.66
1857.04
1.9857
1800
6.7223
1725.62
1974.41
2.0847
4.4790
1724.85
1973.50
2.0396
2000
7.3214
1823.02
2093.99
2.1354
4.8794
1822.32
2093.20
2.0904
APPENDIX F ENGLISH UNIT TABLES
■ 755
TABLE F.7.2 (continued)
Superheated Vapor Water
Temp.
V
u
h
5
V
u
h
s
m
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ffVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
400 psia (444.69)
600 psia (486.33)
Sat.
1.1619
1 1 19.44
1205.45
1.4856
0.7702
1118.54
1204.06
1.4464
450
1.1745
1122.63
1209.57
1.4901
500
1.2843
1150.11
1245.17
1.5282
0.7947
1127.97
1216.21
1.4592
550
1.3S34
1174.56
1276.95
1.5605
0.8749
1158.23
1255.36
1.4990
6UU
1.476U
1197.33
1306.58
1.5892
0.9456
1184.50
1289.49
1.5320
7UU
1.6503
1240.38
1362.54
1.6396
1.0728
1231.51
1350.62
1.5871
SOU
1.M63
1282.14
1416.59
1.6844
1.1900
1275.42
1407.55
1.6343
yoo
1 .9776
1323.69
1470.07
1.7252
1.3021
1318.36
1462.92
1.6766
11KJU
2.1357
1365.51
1523.59
1.7632
1.4108
1361.15
1517.79
1.7155
1 lUU
1407.81
1577.44
1.7989
1.5173
1404.20
1572.66
1.7519
IzUU
1450.73
1631.79
1.8327
1.6222
1447.68
1627.80
1.7861
1 1 AA
lout)
2.5995
1494.34
1686.76
1.8648
1.7260
1491.74
1683.38
1.8186
140U
ji.75zU
1538.70
1742.40
1.8956
1.8289
1536.44
1739.51
1.8497
1500
2.9039
1583.83
1798.78
1.9251
1.9312
1581.84
1796.26
1.8794
1600
3.0553
1629.77
1855.93
1.9535
2.0330
1627.98
1853.71
1,9080
1 TAA
1 /UO
J.2U64
1676.52
1913.86
1.9810
2,1345
1674.88
1911.87
1.9355
1 ot\r\
liSUU
3.3573
1724.08
1972.59
2.0076
2.2357
1722.55
1970.78
1.9622
oaaa
2UUU
1821.61
2092.41
2.0584
2.4375
1820.20
2090.84
2.0131
800 psia (518.36)
1000 psia (544.74)
Sat.
0.5691
1115.02
1 199.26
1.4160
0.4459
1 109.86
1192.37
1.3903
550
0.6154
1138.83
1229.93
1.4469
0.4534
1114.77
1198.67
1.3965
600
0.6776
1170.10
1270.41
1.4861
0.5140
1153.66
1248.76
1.4450
650
0.7324
1197.22
1305.64
1.5186
0.5637
1184.74
1289.06
1.4822
700
0.7829
1222.08
1337.98
1.5471
0.6080
1212.03
1324.54
1.5135
800
0.8764
1268.45
1398.19
1.5969
0.6878
1261.21
1388.49
1.5664
900
0.9640
1312.88
1455.60
1.6408
0.7610
1307.26
1448.08
1.6120
1000
1.0482
1356.71
1511.88
1.6807
0.8305
1352.17
1505.86
1.6530
1100
1.1300
1400.52
1567.81
1.7178.
0.8976
1396.77
1562.88
1.6908
1200
1.2102
1444.60
1623.76
1.7525
0.9630
1441.46
1619.67
1.7260
1300
1.2892
1489.11
1679.97
1.7854
1.0272
1486.45
1676.53
1.7593
1400
1.3674
1534.17
1736.59
1.8167
1.0905
1531.88
1733.67
1.7909
1500
1.4448
1579.85
1793.74
1,8467
1.1531
1577.84
1791.21
1.8210
1600
1.5218
1626.19
1851.49
1.8754
1.2152
1624.40
1849.27
1.8499
1700
1.5985
1673.25
1909.89
1.9031
1.2769
1671.61
1907.91
1.8777
1800
1.6749
1721.03
1968.98
1.9298
1.3384
1719.51
1967.18
1.9046
2000
1.8271
1818.80
2089.28
1.9808
1.4608
1817.41
2087.74
1.9557
756 H Appendix F English Unit Tables
TABLE F.7.2 (continued)
Superheated Vapor Water
Temp.
(F)
(ftrVlbm)
u
(Btu/lbm)
h
(Btu/lbm)
(Btu/lbm R)
V
(f^/lbm)
H
(Btu/lbm)
h
(Btu/lbm)
5
(Btu/lbm R)
1500 psia (596.38)
2000 psia (635.99)
Sat.
0.2769
1091.81
1168.67
1.3358
0.1881
1066.63
1136.25
1.2861
650
0.3329
1146.95
1239.34
1.4012
0.2057
1091.06
1167.18
1.3141
700
0.3716
1183.44
1286.60
1.4429
0.2487
1 147.74
1239.79
1.3782
750
0.4049
1214.13
1326.52
1.4766
" 0.2803
1187.32
1291.07
1.4216
800
0.4350
1241.79
1362.53
1.5058
0.3071
1220.13
1333.80
1.4562
850
0.4631
1267.69
1396.23
1.5321
0.3312
1249.46
1372.03
1.4860
900
0.4897
1292.53
1428.46
1.5562
0.3534
1276.78
1407.58
1.5126
1000
0.5400
1340.43
1490.32
1.6001
0.3945
1328.10
1474.09
1.5598
1100
0.5876
1387.16
1550.26
1.6398
0.4325
1377.17
1537.23
1.6017
1200
0.6334
1433.45
1609.25
1.6765
0.4685
1425.19
1598.58
1.6398
1300
0.6778
1479.68
1667.82
1.7108
0.5031
1472.74
1658.95
1.6751
1400
0.7213
1526.06
1726.28
1.7431
0.5368
1520.15
1718.81
1.7082
1500
0.7641
1572.77
1784.86
1.7738
0.5697
1567.64
1778.48
1.7395
1600
0.8064
1619.90
1843.72
1.8301
0.6020
1615.37
1838.18
1.7692
1700
0.8482
1667.53
1902.98
1.8312
0.6340
1663.45
1898.08
1.7976
1800
0.8899
1715.73
1962.73
1.8582
0.6656
1711.97
1958.32
1.8248
1900
0.9313
1764,53
2023.03
1.8843
0.6971
1760.99
2018.99
1.8511
2000
0.9725
1813.97
2083.91
1.9096
0.7284
1810.56
2080.15
1.8765
4000
psia
8000 psia
650
0.02447
657.71
675.82
0.8574
0.02239
627.01
660.16
0.8278
700
0.02867
742.13
763.35
0.9345
0.02418
688.59
724.39
0.8844
750
0.06332
960.69
1007.56
1.1395
0.02671
755.67
795.21
0.9441
800
0.10523
1095.04
1172.93
1.2740
0.03061
830.67
875.99
1.0095
850
0.12833
1156.47
1251.46
1.3352
0.03706
915.81
970.67
1.0832
900
0.14623
1201.47
1309.71
1.3789
0.04657
1003.68
1072.63
1.1596
950
0.16152
1239.20
1358.75
1.4143
0.05721
1079.59
1164.28
1.2259
1000
0.17520
1272.94
1402.62
1.4449
0.06722
1141.04
1240.55
1.2791
1100
0.19954
1333,90
1481.60
1.4973
0.08445
1236.84
1361.85
1.3595
1200
0.22129
1390.11
1553.91
1.5423
0.09892
1314.18
1460.62
1.4210
1300
0.24137
1443.72
1622.38
1.5823
0.11161
1382.27
1547.50
1.4718
1400
0.26029
1495.73
1688.39
1.6188
0.12309
1444.85
1627.08
1.5158
1500
0.27837
1546.73
1752.78
1.6525
0.13372
1503.78
1701.74
1.5549
1600
0.29586
1597.12
1816.11
1,6841
0.14373
1560.12
1772.89
L5904
1700
0.31291
1647.17
1878.79
1.7138
0.15328
1614.58
1841.49
1.6229
1800
0.32964
1697.11
1941.11
1.7420
0.16251
1667.69
1908.27
1.6531
1900
0.34616
1747.10
2003.32
1.7689
0.17151
1719.85
1973.75
1.6815
2000
0.36251
1797.27
2065.60
1.7948
0.18034
1771.38
2038.36
1.7083
Appendix F English Unit Tables El 757
TABLE F.7.3
Compressed Liquid Water
Temp.
V
u
h
s
It
(F)
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(tf/lbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lb
500 psia (467.12)
1000 psia (544.74)
Sat.
0.01975
447.69
449.51
0.6490
0.02159
538.37
542.36
0.74318
32
0.01599
0.00
1.48
0.0000
0.01597
0.02
2.98
0.0000
50
0.01599
18.02
19.50
0.0360
0.01599
17.98
20.94
0.0359
75
0.0160
42.98
44.46
0.0838
0.0160
42.87
45.83
0.0836
100
0.0161
67.87
69.36
0.1293
0.0161
67.70
70.67
0.1290
125
0.0162
92.75
94.24
0.1728
0.0162
92.52
95.51
0.1724
150
0.0163
117.66
119.17
0.2146
0.0163
117.37
120.39
0.2141
175
0.0165
142.62
144.14
0.2547
0.0164
142.28
145.32
0.2542
200
0.0166
167.64
168.18
0.2934
0.0166
167.25
170.32
0.2928
225
0.0168
192.76
194.31
0.3308
0.0168
192.30
195.40
0.3301
250
0.0170
217.99
219.56
0.3670
0.0169
217.46
220.60
0.3663
275
0.0172
243.36
244.95
0.4022
0.0171
242.77
245.94
0.4014
300
0.0174
268.91
270.52
0.4364
0.0174
268.24
271.45
0.4355
325
0.0177
294.68
296.32
0.4698
0.0176
293.91
297.17
0.4688
350
0.0180
320.70
322.36
0.5025
0.0179
319.83
323.14
0.5014
375
0.0183
347.01
348.70
0.5345
0.0182
346.02
349.39
0.5333
400
0.0186
373.68
375.40
0.5660
0.0185
372.55
375.98
0.5647
425
0.0190
400.77
402.52
0.5971
0.0189
399.47
402.97
0.5957
450
0.0194
428.39
430.19
0.6280
0.0193
426.89
430.47
0.6263
200O psia (635.99) ____ 8000 psia
Sat.
0.02565
662.38
671.87
0.8622
50
0.01592
17.91
23.80
0.0357
0.01563
17.38
40.52
0.0342
75
0.0160
42.66
48.57
0.0832
0.0157
41.42
64.65
0.0804
100
0.0160
67.36
73.30
0.1284
0.01577
65.49
88.83
0.1246
125
0.0161
92.07
98.04
0.1716
0.01586
89.62
113.10
0.1670
150
0.0162
116.82
122.84
0.2132
0.01597
113.81
137.45
0.2078
175
0.0164
141.62
147.68
0.2531
0.01610
138.04
161.87
0.2471
200
0.0165
166.48
172.60
0.2916 -
0.01623
162.31
186.34
0.2849
225
0.0167
191.42
197.59
0.3288
0.01639
186.61
210.87
0.3214
250
0.0169
216.45
222.69
0.3648
0.01655
210.97
235.47
0.3567
275
0.0171
241.61
247.93
0.3998
0.01675
235.39
260.16
0.3909
300
0.0173
266.92
273.33
0.4337
0.01693
259.91
284.97
0.4241
325
0.0176
292.42
298.92
0.4669
0.01714
284.53
309.91
0.4564
350
0.0178
318.14
324.74
0.4993
0.01737
309.29
335.01
0.4878
400
0.0184
370.38
377.20
0.5621
0.01788
359.26
385.73
0.5486
450
0.0192
424.03
431.13
0.6231
0.01848
409.94
437.30
0.6069
500
0.0201
479.84
487.29
0.6832
0.01918
461.56
489.95
0.6633
600
0.0233
605.37
613.99
0.8086
0.02106
569.36
600.53
0.7728
758 M APPENDIX F ENGLISH UNIT TABLES
Table F.7.4
Saturated Solid-Saturated Vapor, Water (English Units)
Specific volume, ftVlbm internal energy, Btu/lbm
Temp.
Press.
Sat. Solid
Sat. Vapor
Sat. Solid
Evap.
Sat. Vapor
(F)
(ibfiW)
v t
v g X 10" 3
u k
32.02
0.08866
0.017473
3.302
-143.34
1164.5
1021.2
32
0.08859
0.01747
3.305
-143.35
1164.5
1021.2
30
0.08083
0.01747
3.607
-144.35
1164.9
1020.5
25
0.06406
0.01746
4.505
-146.84
1165.7
1018.9
20
0.05051
0.01745
5.655
-149.31
1166.5
1017.2
15
0.03963
0.01745
7.133
-151.75
1167.3
1015.6
10
0.03093
0.01744
9.043
-154.16
1168.1
1013.9
5
0.02402
0.01743
11.522
-156.56
1168.8
1012.2
0.01855
0.01742
14.761
-158.93
1169.5
1010.6
-5
0.01424
0.01742
19.019
-161.27
1 170.2
1008.9
-10
0.01086
0.01741
24.657
-163.59
1170.8
1007.3
-15
0.00823
0.01740
32.169
- 165.89
1171.5
1005.6
-20
0.00620
0.01740
42.238
-168.16
1172.1
1003.9
-25
0.00464
0.01739
55.782
-170.40
1172.7
1002.3
-30
0.00346
0.01738
74.046
- 172.63
1173.2
1000.6
-35
0.00256
0.01737
98.890
-174.82
1173.8
998.9
-40
0.00187
0.01737
134.017
-177.00
1174.3
997.3
APPENDIX F ENGLISH UNIT TABLES H 759
TABLE F.7.4 {continued)
Saturated Solid-Saturated Vapor, Water (English Units)
enthalpy, Btu/lbm entropy, Btu/lbm R
Temp.
Press.
Sat. Solid
Evap.
Sat. Vapor
Sat Solid
Evap.
Sat. Vapor
(F)
(lbf/in. 2 )
hi
h
J /
s g
32.02
0.08866
-143.34
1218.7
1075.4
-0.2916
2.4786
2.1869
32
0.08859
-143.35
1218.7
1075.4
-0.2917
2.4787
2.1870
30
0.08083
-144.35
1218.8
1074,5
-0.2938
2.4891
2.1953
25
0.06406
-146.84
1219.1
1072.3
-0.2990
2.5154
2.2164
20
0.05051
-149.31
1219.4
1070.1
-0.3042
2.5422
2.2380
15
0.03963
-151.75
1219.6
1067.9
-0.3093
2.5695
2.2601
10
0.03093
-154.16
1219.8
1065.7
-0.3145
2.5973
2.2827
5
0.02402
-156.56
1220.0
1063.5
-0.3197
2.6256
2.3059
0.01855
-158.93
1220.2
1061.2
-0.3248
2.6544
2.3296
-5
0.01424
-161.27
1220.3
1059.0
-0.3300
2.6839
2.3539
-10
0.01086
-163.59
1220.4
1056.8
-0.3351
2.7140
2.3788
-15
0.00823
-165.89
1220.5
1054.6
-0.3403
2.7447
2.4044
-20
0.00620
-168.16
1220.5
1052.4
-0.3455
2.7761
2.4307
-25
0.00464
-170.40
1220.6
1050,2
-0.3506
2.8081
2.4575
-30
0.00346
-172.63
1220.6
1048.0
-0.3557
2.8406
2.4849
-35
0.00256
-174.82
1220.6
1045.7
-0.3608
2.8737
2.5129
-40
0.00187
-177.00
1220.5
1043.5
-0.3659
2.9084
2.5425
760 H AppendixF English Unit Tables
Table F.8
Thermodynamic Properties of Ammonia
TABLE F.8.1
Saturated Ammonia
Temp.
Press.
Specific Volume. fWlbm
INTERNAL ENERGY. Btu/lbm
F
psia
Sat. Liquid
Evap.
Sat, Vapor
Sat. Liquid
Evap.
Sat. Vapor
T
P
v f
"A
"/
"ft
u t
-60
5.547
0.02277
44.7397
44.7625
-20.92
564.27
543.36
-50
7.663
0.02299
33.0702
33.0932
-10.51
556.84
546.33
-40
10.404
0.02322
24.8464
24.8696
-0.04
549.25
549.20
-30
13.898
0.02345
18.9490
18.9724
10.48
541.50
551.98
-28.0
14.696
0.02350
17.9833
18.0068
12.59
539.93
552.52
—20
18.289
0.02369
14.6510
14.6747
21.07
533.57
554.64
— 10
23.737
0.02394
11.4714
11.4953
31.73
252.47
557.20
30.415
0.02420
9.0861
9.1103
42.46
517.18
559.64
10
38.508
0.02446
7.2734
7.2979
53.26
508.71
561.96
20
48.218
0.02474
5.8792
5.9039
64.12
500.04
564.16
30
59.756
0.02502
4.7945
4.8195
75.06
491.17
566.23
40
73,346
0.02532
3.9418
3.9671
86.07
482.09
568.15
50
89.226
0.02564
3.2647
3.2903
97.16
472.78
569.94
60
107.641
0.02597
2.7221
2.7481
108.33
463.24
571.56
70
128.849
0.0263 1
2.2835
2.3098
119.58
453.44
573.02
80
153.116
0.02668
1.9260
1.9526
130.92
443.37
574.30
90
180.721
0,02706
1.6323
1.6594
142.36
433.01
573.37
100
211.949
0.02747
1.3894
1.4168
153.89
422.34
576.23
110
247.098
0.02790
1.1870
1.2149
165.53
411.32
576.85
120
286.473
0.02836
1.0172
1.0456
177.28
399.92
577.20
130
330.392
0.02885
0.8740
0.9028
189.17
388.10
577.27
140
379.181
0.02938
0.7524
0.7818
201.20
375.82
577.02
150
433.181
0.02995
0.6485
0.6785
213.40
363.01
576.41
160
492.742
0.03057
0.5593
0.5899
225.80
349.61
575.41
170
558.231
0.03124
0.4822
0.5135
238.42
335.53
573.95
180
630.029
0.03199
0.4153
0.4472
251.33
320.66
571.99
190
708.538
0.03281
0.3567
0.3895
264.58
304.87
569.45
200
794.183
0.03375
0.3051
0.3388
278.24
287.96
566.20
210
887.424
0.03482
0.2592
-0.2941
292.43
269.70
562.13
220
988.761
0.03608
0.2181
0.2542
307.28
249.72
557.00
230
1098.766
0.03759
0.1807
0.2183
323.03
227.47
550.50
240
1218.113
0.03950
0.1460
0.1855
340.05
202.02
542.06
250
1347.668
0.04206
0.1126
0.1547
359.03
171.57
530.60
260
1488.694
0.04599
0.0781
0.1241
381.74
131.74
513.48
270.1
1643.742
0.06816
0.0682
446.09
446.09
Appendix F English Unit Tables ffl 761
TABLE F.8.1 (continued)
Saturated Ammonia
Temp.
F
T
Press.
psia
P
Enthalpy, Btu/lbm
ENTROPY, Btu/lbm R
Sat. Liquid
></
Sat VsinfiT*
£
Oof T innir)
Sf
Eyap.
fs
Sat. Vapof
-60
5.547
-20.89
610,19
589.30
-0.0510
1.5267
1.4758
-50
7.663
-10.48
603.73
593.26
-0.0252
1.4737
1.4485
~40
10.404
597.08
597.08
1.4227
1.4227
-30
13.898
10.54
590.23
600.77
0.0248
1.3737
1.3985
-28.0
14.696
12.65
: 588.84
601.49
0.0297
1,3641
1.3938
-20
18.289
21.15
583.15
604.31
0.0492
1.3263
1.3755
-10
23.737
31.84
575.85
607.69
0.0731
1.2806
1.3538
30.415
42.60
568.32
610.92
0.0967
1.2364
1.3331
10
38.508
53.43
560.54
613.97
0,1200
1.1935
1.3134
20
48.218
64.34
552.50
616.84
0.1429
1.1518
1.2947
30
59.756
75.33
544.18
619.52
0.1654
1.1113
1.2768
40
73.346
86.41
535.59
622.00
0.1877
1.0719
1.2596
50
89.226
97.58
526.68
624.26
0.2097
1.0334
1.2431
60
107.641
108.84
517.46
626.30
0.2314
0.9957
1.2271
70
128.849
120,21
507.89
628.09
0.2529
0.9589
1.2117
80
153.116
131.68
497.94
629.62
0.2741
0.9227
1.1968
90
180.721
143.26
487.60
630.86
0.2951
0.8871
1.1822
100
211.949
154.97
476.83
631.80
0.3159
0.8520
1.1679
110
247.098
166.80
465.59
632.40
0.3366
0.8173
1.1539 -
120
286.473
178.79
453.84
632.63
0,3571
0.7829
1.1400
130
330.392
190.93
441,54
632.47
0.3774
0.7488
1.1262
140
379.181
203.26
428.61
631.87
0.3977
0.7147
1.1125
150
433.181
215.80
415.00
630.80
0.4180
0.6807
1.0987
160
492.742
228.58
400.61
629.19
0.4382
0.6465
1.0847
170
558.231
241.65
385.35
627.00
0.4586
0.6120
1.0705
180
630.029
255.06
369.08
624.14
0.4790
0.5770
1.0560
190
708.538
268.88
351.63
620.51
0.4997
0.5412
1.0410
200
794.183
283.20
332.80
616.00
0.5208
0.5045
1.0253
210
887.424
298.14
312.27
610.42
0.5424
0.4663
1.0087
220
988.761
313.88
289.63
' 603.51
0.5647
0.4261
0.9909
230
1098.766
330.67
264.21
594.89
0.5882
0.3831
0.9713
240
1218.113
348.95
234,93
583.87
0.6132
0.3358
0.9490
250
1347.668
369.52
199.65
569.17
0.6410
0.2813
0.9224
260
1488.694
394.41
153.25
547.66
0.6743
0.2129
0.8872
270.1
1643.742
466.83
466.83
0.7718
0.7718
762 H Appendix F English Unit Tables
Table f.8.2
Superheated Ammonia
Temp.
V
h
s
V
h
s
V
h
s
r
ir/lorn
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/lbm R
irviDin
Btu/ibm Btu/lbm R
5 psia (-63.09)
lOpsia (-41.33)
15 psia (—
27.27)
Sat
49.32002
588.05
1.4846
25 80648
596.58
1.4261
17.66533
601.75
1.3921
-40
52.3487
599.56
1.5128
25.8962
597.27
1.4277
—20
54.9506
609.53
1.5360
27.2401
607.60
1.4518
17.9999
605.63
1.4010
57.5366
619.51
1.5582
28,5674
617.88
1.4746
18.9086
616.22
1.4245
20
60.1099
629.50
1.5795
29.8814
628.12
1.4964
19.8036
626.72
1.4469
40
62.6732
639.52
1.5999
31.1852
638.34
1.5173
20.6880
637.15
1.4682
60
65.2288
649.57
1.6197
32.4809
648.56
1.5374
21.5641
647.54
1.4886
80
67.7782
659.67
1.6387
33.7703
658.80
1.5567
22.4338
657.91
1.5082
100
70.3228
669.84
1.6572
35.0549
669.07
1.5754
23.2985
668.29
1.5271
120
72.8637
680.06
1.6752
36 3356
679.38
1.5935
24.1593
678.70
1.5453
140
75.4015
690.36
1.6926
37.6133
689.75
1.6111
25.0170
689.14
1.5630
160
77.9370
700.74
1.7097
38.8886
700.19
1.6282
25.8723
699.64
1.5803
180
80.4706
711.20
1.7263
40 1620
710.70
1.6449
26.7256
710.21
1.5970
200
83.0026
721.75
1.7425
41.4338
721.30
1.6612
27.5774
720.84
1.6134
220
85.5334
732.39
1.7584
42.7043
731.98
1.6771
28.4278
731.56
1.6294
240
88.0631
743.13
1.7740
43.9737
742.74
1.6928
29.2772
742.36
1.6451
260
90.5918
753.96
1.7892
45,2422
753.61
1.7081
30.1256
753.24
1.6604
280
93.1199
764.90
1.8042
46.5100
764.56
1.7231
30.9733
764.23
1.6755
20 psia(- 16.63)
25psia (-7.95)
30 psia (-
0.57)
Sat.
13.49628
605.47
1.3680
10.95013
608.37
1.3494
9.22850
610,74
1.3342
U
14. U/ /H
614.54
1.3881
11.1771
612.82
1.3592
611.06
1.3349
.20
14.7635
625.30
1.4111
11.7383
623.86
1.3827
9.7206
622.39
1.3591
40
15.4385
635.94
1.4328
12.2881
634.72 '
1.4049
10.1872
633.49
1,3817
60
16.1051
646.51
1.4535
12.8291
645.46
1.4260
10.6447
644.41
1.4032
80
16.7651
657.02
1.4734
13.3634
656.12
1.4461
1 1.0954
655,21
1.4236
100
17.4200
667.51
1.4925
13.8926
666.73
1.4654
11.5407
665.93
1.4431
120
18.0709
678.01
1.5109
14.4176
677.32
1.4840
11.9820
676.62
1.4618
140
18.7187
688.53
1.5287
14.9395
687.91
1.5020
12.4200
687.29
1.4799
160
19.3640
699.09
1.5461
15.4589
698.54
1.5194
12.8554
697.98
1.4975
180
20.0073
709.71
1.5629
15.9763
' 709.20
1.5363
13.2888
708.70
1.5145
200
20.6491
720.39
1.5794
16.4920
719.93
1.5528
13.7206
719.47
1.5311
220
21.2895
731.14
1,5954
17.0065
730.72
1.5689
14.1511
730.29
1.5472
240
21.9288
741.97
1.6111
17.5198
741.58
1.5847
14.5804
741.19
1.5630
260
22.5673
752.88
1.6265
18.0322
752.52
1.6001
15.0088
752.16
1.5785
280
23.2049
763.89
1.6416
18.5439
763.55
1.6152
15.4365
' 763.21
'1.5936
300
23.8419
774.99
1.6564
19.0548
774.67
1.6301
15.8634
774.36
1.6085
320
24.4783
786.18
1.6709
19.5652
785.89
1.6446
16.2898
785.59
1.6231
Appendix F English unit Tables
m 763
TABLE F.8.2 (continued)
Superheated Ammonia
Temp.
F
Sat.
40
60
80
100
120
140
160
ISO
200
220
240
260
280
300
320
340
360
v
ftVlbm
h
Btu/lbm
Btu/Jbm R
v
ftVlbm
h
Btu/lbm
Btu/lbm R
v
f^/lbm
/(
Btu/lbm
35 psia (5.89)
Btu/lbm R
40 psia (11.66)
50 psia (21.66)
Sat.
7.98414
612.73
1.3214
7.04135
614.45
1.3103
5.70491
617.30
20
8.2786
620.90
1.3387
7.1964
619.39
1.3206
40
8.6860
632.23
1.3618
7.5596
630.96
1.3443
5.9814
628.37
60
9.0841
643.34
1.3836
7.9132
642.26
1.3665
6.2731
640.07
80
9.4751
654.29
1.4043
8.2596
653.37
1.3874
6.5573
651.49
100
9.8606
665.14
1.4240
8.6004
664.33
1.4074
6.8356
662.70
120
10.2420
675.92
1.4430
8.9370
675.21
1.4265
7.1096
673.79
140
10.6202
686.67
1.4612
9.2702
686.04
1.4449
7.3800
684.78
160
10.9957
697.42
1.4788
9.6008
696.86
1.4626
7.6478
695.73
180
11.3692
708.19
1.4959
9.9294
707.69
1.4798
7.9135
706.67
200
11.7410
719.01
1.5126
10.2562
718.54
1.4965
8.1775
717.61
220
12.1115
729.87
1.5288
10.5817
729.44
1.5128
8.4400
728.59
240
12.4808
740.80
1.5447
10.9061
741.40
1.5287
8.7014
739.62
260
12.8493
751.80
1.5602
11.2296
751.43
1.5442
8.9619
750.70
280
13.2169
762.88
1.5753
11.5522
762.54
1.5594
9.2216
761.86
300
13.5838
774.04
1.5902
11.8741
773.72
1.5744
9.4805
773.09
320
13.9502
785.29
1.6049
12.1955
785.00
1.5890
9.7389
784.40
340
14.3160
796.64
1.6192
12.5163
796.36
1.6034
9.9967
795.80
60 psia (30.19)
70 psia (37.68)
1.2917
1.3142
1.3372
1.3588
1.3792
1.3986
1.4173
1.4352
1.4526
1.4695
1.4859
1.5018
1.5175
1.5327
1.5477
1.5624
1.5769
80 psia (44.38)
4.80091
619.57
1.2764
4.14732
621.44
1.2635
3.65200
623.02
4.9277
625.69
1.2888
4.1738
622.94
' 1.2665
5.1787
637.82
1.3126
4.3961
635.52
1.2912
3.8083
633.16
5.4217
649.57
1.3348
4.6099
647.62
1.3140
4.0005
645.63
5.6586
661.05
1.3557
4.8174
659.37
1.3354
4.1861
657.66
5.8909
672.34
- 1.3755
5.0201
670.88
1.3556
4.3667
669.39
6.1197
683.50
1.3944
5.2191
682.21
1.3749
4.5435
680.90
6.3456
694.59
1.4126
5.4153
693.44
1.3933
4.7174
692.27
6.5694
705.64
1.4302
5.6093-
704.60
1.4110
4.8890
703.55
6.7915
716.68
1.4472
5.8014
715.73
1.4281
5.0588
714.79
7.0121
727.73
1.4637
5.9921
726,87
1.4448
5.2270
726.00
7.2316
738.83
1.4798
6.1816
738.03
1.4610
5.3941
737.23
7.4501
749.97
1.4955
6.3702
749.23
1.4767
5.5602
748.50
7.6678
761.17
1.5108
6.5579
760.49
1.4922
5.7254
759.80
7.8848
772.45
1.5259
6.7449
771.81
1.5073
5.8900
771.17
8.1011
783.80
1.5406
6.9313
783.21
1.5221
6.0538
782.61
8.3169
795.24
1.5551
7.1171
794.68
1.5366
6.2172
794.12
8.5323
806.77
1.5693
7.3025
806.24
1,5509
6.3801
805.71
1.2523
1.2721
1.2956
1.3175
1.3381
1.3577
1.3763
L3942
1.4115
1.4283
1.4446
1.4604
1.4759
1.4911
1.5059
1.5205
1.5348
764 n appendix F English Unit Tables
TABLE F.8.2 (continued)
Superheated Ammonia
Temp.
V
It
s
V
h
s
V
h
s
F
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/lbm R
tr/lbm
Btu/lbm
Btu/lbm R
90 psia (50.45)
100 psia (56.02)
125 psia (68.28)
Sat.
3.26324
624.36
1.2423
2.94969
foe fO
OZJ.JZ
t in/I
2.37866
627.80
1.2143
603.3503630.74
1.2547
2.9831
628.25
1.2387
SO
3.5260
643.59
1.2790
3.1459
641.51
1.2637
2.4597
636.11
1.2299
100
3.6947
655.92
1.3014
3.3013
654.16
1.2867
2.5917
649.59
1.2544
120
3.8583
667.88
1.3224
3.4513
666.36
1.3082
2.7177
662.44
1.2770
140
4.0179
679.58
1.3423
3.5972
678.24
1,3283
2.8392
674.83
1.2980
160
4.1745
691.10
1.3612
3.7400
689.91
1.3475
2.9574
686.90
1.3178
180
4.3287
702.50
1.3793
3.8804
701.44
1.3658
3.0730
698.74
1.3366
200
4.4811
713.83
1.3967
4.0188
712.87
1.3834
3.1865
710.44
1.3546
220
4.6319
725.13
1.4136
4.1558
724.25
1.4004
3.2985
722.04
1.3720
240
4.7816
736.43
1.4300
4.2915
735.63
1.4169
3.4091
733.59
1.3887
260
4.9302
747.75
1.4459
4.4261
747.01
1.4329
3.5187
745.13
1.4050
280
5.0779
759.11
1.4615
4.5599
758.42
1.4485
3.6274
756.68
1.4208
300
5.2250
770.53
1.4767
4.6930
769.88
1.4638
3.7353
768.27
1.4362
320
5.3714
782.01
1.4916
4.8254
781.40
1.4788
3.8426
779.89
1.4514
340
5.5173
793.56
1.5063
4.9573
792.99
1.4935
3.9493
791.58
1.4662
360
5.6626
805.18
1.5206
5.0887
804.66
1.5079
4.0555
803.33
1.4807
380
5.8076
816.90
1.5348
5.2196
816.40
1.5220
4.1613
815.15
1.4949
150 psia (78.79)
175 psia (88
.03)
200 psia (96.31)
Sat.
1.99226
629.45
1.1986
1.71282
630.64
1.1850
1.50102
631.49
1.1731
80
1.9997
630.36
1.2003
100 -
2.1170
644.81
1.2265
1.7762
639.77
1.2015
1.5190
634.45
1.1785
120
2.2275
658.37
1.2504
1.8762
654.13
1.2267
1.6117
649.71
1.2052
.140
2.3331
671.31
1.2723
1.9708
667.67
1.2497
1.6984
663.90
1.2293
160
2.4351
683.80
1.2928
2.0614
680.62
1.2710
1. 7807
677.36
1.2514
180
2.5343
695.99
1.3122
2.1491
693.17
1.2909
1.8598
690.30
1.2719
200
2.6313
707.96
1.3306
2.2345
705.44
1.3098
1.9365
702.87
1.2913
220
2.7267
719.79
1.3483
2.3181
717.51
1.3278
2.01 14
715.20
1.3097
240
2.8207
731.54
1.3653
2.4002
729.46
1.3451
2.0847
727.35
1.3273
260
2.9136
743.24
1.3818
2.4813
741.33
1.3619
2.1569
739.39
1.3443
280
3.0056
754.93
1.3978
2.5613
753.16
1.3781
2.2280
751.38
1.3607
300
3.0968
766.63
1.4134
2.6406
764.99
1.3939
2.2984
763,33
1.3767
320
3.1873
778.37
1.4287
2.7192
776.84
1.4092
2.3680
775.30
1.3922
340
3.2772
790.15
1.4436
2.7972
788.72
1.4243
2.4370
787.28
1.4074
360
3.3667
801.99
1.4582
2.8746
800.65
1.4390
2.5056
799.30
1.4223
380
3.4557
813.90
1.4726
2.9516
812.64
1.4535
2.5736
811.38
1.4368
400
3.5442
825.88
1.4867
3.0282
824.70
1.4677
2.6412
823.51
1.4511
APPENDIX F ENGLISH UNIT TABLES
m 765
TABLE F.8,2 {continued)
Superheated Ammonia
Temp.
V
n
s
V
h
s
V
ft
$
f
ffVlhm
Btu/lbm
Btu/lbmR
Btu/lbm
Btu/lbm R
tf/lbm
Btu/lbm
Btu/lbm R
250 psia (110.78)
300 psia (123.20)
350 psia (134.14)
Sat.
1.20063
632.43
1.1528
0.99733 632.63
1.1356
0.85027 632.28
1.1205
120
1.2384
640.21
1.1663
140
1.3150
655.95
1.1930
1.0568
647.32
1.1605
0.8696
637.87
1.1299
160
1.3863
670.53
1.2170
1.1217
663.27
1.1866
0.9309
655.48
1.1588
180
1.4539
684.34
1.2389
1.1821
678.07
1.2101
0.9868
671.46
1.1842
200
1.5188
697.59
1.2593
1.2394
692.08
1.2317
1.0391
686.34
1,2071
220
1.5815
710.45
1.2785
1.2943
705.55
1.2518
1.0886
700.47
1.2282
240
1.6426
723.05
1.2968
1.3474
718.63
1.2708
1.1362
714.08
1.2479
260
1.7024
735.46
1.3142
1.3991
731.44
1.2888
1.1822
727.32
1.2666
280
1.7612
747.76
1.3311
1.4497
744.07
1.3062
1.2270
740.31
1.2844
300
1.8191
759.98
1.3474
1.4994
756.58
1.3228
1.2708
753.12
1.3015
320
1.8762
772.18
1.3633
1.5482
769.02
1,3390
1.3138
765.82
1.3180
340
1.9328
784.37
L3787
1.5965
781.43
1.3547
1.3561
778.46
1.3340
360
1.9887
796.59
1.393S
1.6441
793.84
1.3701
1.3979
791.07
1.3496
380
2.0442
808.83
1.4085
1.6913
806.27
1.3850
1.4391
803.67
1.364S
400
2.0993
821.13
1.4230
-1.7380
818.72
1.3997
1.4798
816.30
1.3796
420
2.1540
833.48
1.4372
1.7843
831.23
1.4141
1.5202
828.95
1.3942
440
2.2083
845.90
1.4512
1.8302
843.78
1.4282
1.5602
841.65
1.4085
400psia (143.97)
600 psia (175.93)
800 psia (200.65)
Sat.
0.73876
631.50
1.1070
0.47311
625.39
1.0620
0.33575
615.67
1.0242
160
0.7860
647.06
1.1324
ISO
0.8392
664.44
1.1601
0.4834
630.48
1.0700
200
0.8880
680.32
1.1845
0.5287
652.67 ■
1.1041
220
0.9338
695.21
1.2067
0.5680
671.78
1.1327
0.3769
642.62
1,0645
240
0.9773
709.40
1.2273
0.6035
689.03
1.1577
0.4115
665.08
1.0971
260
1.0192
723.10
1.2466
0.6366
705.06
1.1803
0.4419
684.62
1.1246
280
1.0597
736.47
1.2650
0.6678
720.26
1.2011
0.4694
702,36
1.1489
300
1.0992
749.60
1.2825
0.6976
- 734.8'8
1,2206
0.4951
718.93
1.1710
320
1.1379
762.58
1.2993
0.7264
749.09
1.2391
0.5193
734.69
1.1915
340
1.1758
775.45
1.3156
0.7542
763.02
1.2567
0.5425
749.89
1.2108
360
1.2131
788.27
1.3315
0.7814
776.75
1.2737
0.5648
764.68
1.2290
380
1.2499
801.06
1.3469
0.8079
790.34
1.2901
0.5864
779.19
1.2465
400
1.2862
813.85
1.3619
0.8340
803.86
1.3060
0.6074
793.50
1.2634
420
1.3221
826.66
1.3767
0.8595
817.32
1.3215
0.6279
807.68
1.2797
440
1.3576
839.51
1.3911
0.8847
830.76
1.3366
0.648O
821.76
1.2955
460
1.392S
852.39
1.4053
0.9095
844.21
1.3514
0.6677
835.80
1.3109
480
1.4277
865.34
1.4192
0.9340
857.67
1.3658
0.6871
849.80
1,3260
766 B APPENDIX F ENGLISH UNIT TABLES
Table F.9
Thermodynamic Properties ofR-22
Table F.9,1
Saturated R-22
SPECIFIC VOLUMB, flVlbm INTERNAL ENERGY, Btu/lbm
Temp.
F
Press,
psia
Sat. Liquid
Evil p.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
T
i
p
Vr
is
5
-100
2.398
0.01066
18.4219
18.4326
— 14.57
yy. /b
1
o j. i y
-90
3.423
0.01077
13.2243
13.2351
— 12.22
OS 12
OO. IO
-80
4.782
0.01088
9.6840
9.6949
—9.85
0£L OS
yo.ys
-70
6.552
0.01099
7.2208
7.2318
—7.44
yo.54
88 1fi
-60
8.818
0.01111
5.4733
5.4844
— 5.01
OA (Yl
yH.Vi
8Q ftfi
-50
11.674
0.01124
4.2111
4.2224
—2.54
yt.jo
90.02
-41.4
14.696
0.01135
3.3944
3.4058
—0,37
yi.zz
Qfl 8^
-40
15.222
0.01136
3.2844
3.2957
—0.03
oi m
y i.ui
QO, Q7
-30
19.573
0.01150
2.5934
2.6049
2.51
QCl A 1
Q1 Q1
y l.y i
-20
24.845
0.01163
2.0709
2.0826
5.08
on n£
ol.lt>
QO 84
-10
31.162
0.01178
1.6707
1.6825
7,68
50, U I
y j* 1 J
38.657
0.01193
L3603
1.3723
10.32
QA 11
54.
QA
10
47.464
0.01209
1.1170
i i inn
82.53
95.53
20
57.727
0.01226
0.9241
0.9363
15.71
80.67
96.38
30
69.591
0.01243
0.7697
0.7821
18.45
78.76
97.21
40
83.206
0.01262
0.6449
0.6575
21.23
76.79
98.02
50
98.727
0.01282
0.5432
0.5561
24.04
74.75
98.79
60
116.312
0.01303
0.4597
0.4727
26.89
72.65
99.54
70
136.123
0.01325
0.3905
0.4037
29.78
70.46
100.24
80
158.326
0.01349
0.3327
0.3462
32.71
68.19
100.91
90
183.094
0.01375
0.2841
0.2979
35.69
65.83
101,53
100
210.604
0.01404
0.2430
0.2570
38.72
63.37
102.09
110
241.042
0.01435
0.2079
0.2222
41.81
60.78
102.59
120
274.604
0.01469
0.1777
0.1924
44.96
58.05
103.01
130
311.496
0.01508
0.1515
0.1666
48.19
55.14
103.33
140
351.944
0.01552
0.1287
0.1442
51.52
52.02
103.54
150
396.194
0.01603
0.1085
0.1245
54.97
48.63
103.60
160
444.525
0.01663
0.0904
0.1070
58.58
44.88
103.46
170
497.259
0.01737
0.0739
' 0.0913
62.42
40.62
103.04
180
554.783
0.01833
0.0585
0.0768
66.62
35.57
102.18
190
617.590
0.01973
0.0431
0.0628
71.46
29.10
100.55
200
686,356
0.02244
0.0250
0.0474
78.01
18.81
96.83
204.8
720.698
0.03053
0.0305
87.30
87.30
L_
Appendix f English Unit Tables B 767
TABLE F.9.1 (continued)
Saturated R-22
Temp.
r
T
Press.
psia
p
Enthalpy, Btu/ibm
Entropy, Btu/lbm R
Sat. Liquid
fly
Evap.
Sat, Vapor
"*
Sat. Liquid
s f
Evap.
Sat. Vapor
-100
2.398
-14.56
107.94
93.37
-0.0373
0.300!
0.2627
-90
3.423
-12.22
106.76
94.54
-0.0309
0.2888
0.2579
-80
4.782
-9.84
105.55
95.71
-0.0246
0.2780
0.2534
-70
6.552
-7.43
104.30
96.87
-0.0183
0.2676
0.2493
-60
8.818
-4.99
103.00
98.01
-0.0121
0.2577
0.2456
-50
11.674
-2.51
101.66
99.14
-0.0060
0.2481
0.2421
-41.4
14.696
-0.34
100.45
100.11
-0.0008
0.2401
0.2393
-40
15.222
100.26
100.26
0.2389
0.2389
-30
19.573
2.55
98.80
101.35
0.0060
0.2299
0.2359
-20
24.845
5.13
97.28
102.42
0.0119
0.2213
0.2332
-10
31.162
7.75
95.70
103.46
0.0178
0.2128
0.2306
38.657
10.41
94.06
104.47
0.0236
0.2046
0.2282
10
47.464
13.10
92.34
105.44
0.0293
0.1966
0.2259
20
57.727
15.84
90.55
106.38
0.0350
0.1888
0.2238
30
69.591
18.61
88.67
107.28
0.0407
0.1811
0.2218
40
83.206
21.42
86.72
108.14
ft ftAfQ
ft tile
u.ziyy
50
98.727
24.27
84.68
108.95
0.0519
0.1661
0.2180
60
116.312
27.17
82.54
109.71
0.0574
0.1588
0.2163
70
136.123
30.12
80.30
110.41
0.0630
0.1516
0.2146
80
158.326
33.11
77.94
111.05
0.0685
0.1444
0.2129
90
183.094
36.16
. 75.46
111.62
0.0739
0.1373
0.2112
100
210.604
39.27
72.84
112.11
0.0794
0.1301
0.2096
110
241.042
42.45
70.05
112.50
0.0849
0.1230
0.2079
120
274.604
45.71
67.08
112.78
0.0904
0.1157
0.2061
130
311.496
49.06
63.88
112.94
0.0960
0.1083
0.2043
140
351.944
52.53
60.40
112.93
0.1016
0.1007
0.2023
150
396.194
56.14
56.58
1 12.73
0.1074
0.0928
0.2002
160
444.525
59.95
52.32
112.26
0.1133
' 0.0844
0.1978
170
497.259
64.02
47.42
111,44
0.1196
0.0753
0.1949
180
554.783
68.50
41.57
110.07
0,1263
0.0650
0.1913
190
617.590
73.71
34.02
107.73
0.1341
0.0524
0.1865
200
686.356
80.86
21.99
102.85
0.1446
0.0333
0.1779
204.8
720.698
91.38
91.38
0.1602
0.1602
768 M Appendix f English unit tables
Table f.9.2
Superheated R-22
Temp.
V
5
V
It
s
V
h
s
r
ftVlbm
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/lbm R
fP/lbm
Btu/lbm Btu/lbm R
5 psia (-78.62)
10 psia (-55.59)
15 psia (—
4U.3/J
Sat.
9.30117
95.87
0.2528
4.87779
98.52
0.2440
3.34121
1 fifl 1 Q
-40
10.2935
101.09
0.2659
5.0838
100.69
0.2493
3.3463
100.28
0.2393
-20
10.8034
103.89
0.2724
5.3460
103.53
0.2559
3.5261
103.16
0.2460
11.3114
106.73
0.2787
5.6060
106.41
0.2623
3.7037
106.09
0.2525
20
11.8177
109.64
0.2849
5.8643
109.36
0.2686
3.8794
109.06
0.2588
40
12.3227
112.61
0.2910
6.1212
112.35
0.2747
4.0537
112,09
0.2650
60
12.8265
115.64
0.2969
6.3769
115.40
0.2807
4.2268
115.17
0.27 10
80
13.3293
118.72
0.3027
6.6316
118.51
0.2865
4.3989
118.30
0.2769
100
13.8313
121.87
0.3085
6.8855
121.67
0.2923
4.5701
121.48
0.2827
120
14.3327
125.07
0.3141
7.1387
124.89
0.2979
4.7406
124.72
0.2884
140
14.8335
128.33
0.3196
7.3913
128.17
0.3035
4.9105
128.00
0.2940
160
15.3337
131.64
0.3250
7.6434
131.49
0.3089
5.0799
131.34
0.2995
180
15.8336
135.01
0.3304
7.8951
134.87
0.3143
5.2489
134.74
0.3049
200
16.3331
138.44
0.3357
8.1464
138.31
0.3196
5.4174
138.18
0.3102
220
16.8323
141.92
0.3409
8.3974
141.80
0.3248
5.5857
141.68
0.3154
240
17.3312
145.45
0.3460
8.6481
145.34
0.3300
5.7537
145,23
0.3205
260
17.8298
149.03
0.3510
8.8986
148.93
0.3350
5.9215
148.83
0.3256
280
18.3283
152.67
0.3560
9.1489
152.57
0.3400
6.0890
152.48
0,3306
20 psia (-29.12)
25 psia (-19.73)
30 psia (—
11.71)
Sat.
2.55270
101.44
0.2357
2.07040
102.44
0.2331
1.74388
103.28
0.2310
u
105.76
0.2454
Z. loUo
105.42
0.2397
1.7997
105.08
0.2350
20
2.8867
108.77
0.2518
2.2908
108.47
0.2462
1.8933
108.16
0.2415
40
3.0198
111.83
0.2580
2.3992
111.56
0.2525
1.9853
111.29
0,2479
60
3.1516
114.93
0.2641
2.5063
1 14.69
0.2586
2.0760
114.45
0.2541
80
3.2823
118.08
0.2701
2.6123
117.86
0.2646
2.1655
117.64
0.2602
100
3.4122
121.28
0.2759
2.7175
121.09
0.2705
2.2542
120.89
0.2661
120
3.5414
124.54
0.2816
2.8219
124.36
0.2762
2.3421
124.18
0.2718
140
3.6700
127.84
0.2872
2.9257
127.68
0.2819
2.4294
127.51
0.2775
160
3.7981
131.19
0.2927
3.0289
131.04
0.2874
2.5162
130.89
0.2830
180
3.9257
134.60
0.2981
3.1318
■134.46
0.2928
2.6025
134.32
0.2885
200
4.0529
138.05
0.3034
3.2342
137.93
0.2982
2.6884
137.80
0.2938
220
4.1799
141.56
0,3087
3.3363
141.44
0.3034
2.7739
141.32
0.2991
240
4.3065
145.12
0.3138
3.4382
145.01
0.3086
2.8592
144.90
0.3043
260
4.4329
148.73
0.3189
3.5397
148.62
0.3137
2.9443
148.52
0.3094
280
4.5591
152.38
0.3239
3.6411
152.28
0.3187
3.0291
' 152.19
0.3144
300
4.6851
156.09
0.3288
3.7423
155.99
0.3236
3.1138
155.90
0.3194
320
4.8109
159.84
0.3337
3.8434
159.75
0.3285
3.1983
159.67
0.3243
APPENDIX F ENGLISH UNIT TABLES
m 769
TABLE F.9.2 (continued)
Superheated R-22
Temp.
v
5
//
s
V
It
5
F
fP/lbm
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm Btu/lbm R
11 /lUfll
Btu/lbm
Btu/lbm R
40psia(1.63)
50 psia (12.61)
60 psia (22.03)
Sat.
1.32853
104.63
0.2278
1.07436
105.69
0.2253
0.90223
106.57
0.2234
20
107.54
0.2340
1.0968
106.90
0.2279
40
1.4676
110.73
0.2405
1.1564
110.16
0.2346
0.9486
109.58
0.2295
60
1.5378
113.95
0.2468
1.2145
113.44
0.2410
0.9987
112.92
0.2361
80
1.6068
117.20
0.2530
1.2714
116.74
0.2472
1.0475
116.28
0.2424
100
1.6749
120.48
0.2589
1.3272
120.08
0.2533
1.0952
119.66
0.2486
120
1 .7423
123.81
0.2648
1.3822
123.44
0.2592
1.1420
123.06
0.2545
140
1.8090
127.18
0.2705
1.4366
126.84
0.2649
1.1882
126.50
0.2604
160
1.8751
130.59
0.2761
1.4903
130.28
0.2706
1.2338
129.96
0.2660
180
1.9407
134.04
0.2816
1.5436
133.75
0.2761
1.2788
133.47
0.2716
200
2.0060
137.54
0.2869
1.5965
137.28
0.2815
1.3235
137.01
0.2771
220
2.0709
141.08
0.2922
1.6491
140.84
0.2869
1.3678
140.60
0.2824
240
2.1356
144.67
0.2974
1.7013
144.45
0.2921
1.4118
144.22
0.2877
260
2.2000
148.31
0.3026
1.7533
148.10
0.2972
1.4556
147.89
0.2928
280
2.2641
151.99
0.3076
1.8051
151.80
0.3023
1.4991
151.60
0.2979
300
2.3281
155.72
0.3126
1.8567
155.54
0.3073
1.5424
155.35
0.3029
320
2.3919
159.49
0.3175
1.9081
159.32
0.3122
1.5856
159.15
0.3079
340
2,4556
163.31
0.3223
1.9594
163.15
0.3171
1.6286
162.99
0.3127
70 psia (30.32)
80 psia (37.76)
100 psia (50.77)
Sat
0.77766
107.31
0.2217
0.68319
107.95
0.2203
0.54908
109.01
0.2179
40
0.7998
108.97
0.2251
0.6878
108.35
0.2211
60
0.8443
112.39
0.2318
0.7282
111.84
0.2279
0.5650
110.70
0.2212
80
0.8874
115.81
0.2382
0.7671
115.32
0.2345
0.5982
114.32
0.2280
100
0.9293
119.23
0.2445
0.8048
118.80
0.2408
0.6300
117.91
0.2345
120
0.9704
122.68
0.2505
0.8415
122.29
0.2470
0.6608
121.49
0.2408
140
1.0107
126.15
0.2564
0.8775
125.80
0.2529
0.6908
125.08
0.2469
160
1.0504
129.65
0.2621
0.9129
129.33
0.2587
0.7201
128.67
0.2528
180
1.0896
133.18
0.2677
0.9477 •
132.89
0.2643
0.7488
132.29
0.2586
200
1.1284
136.75
0.2732
0.9821
136.48
0.2699
0.7771
135.93
0.2642
220
1.1669
140.35
0.2786
1.0161
140.10
0.2753
0.8050
139.60
0.2696
240
1.2050
143.99
0.2839
1.0498
143.76
0.2806
0.8326
143.30
0.2750
260
1.2428
147.68
0.2891
1.0833
147.46
0.2858
0.8599
147.03
0.2803
280
1.2805
151.40
0.2942
1.1165
151.20
0.2909
0.8869
150.80
0.2854
300
1.3179
155.17
0.2992
1.1495
154.98
0.2960
0.9137
154.61
0.2905
320
1.3552
158.97
0.3042
1.1823
158.80
0.3009
0.9404
158.45
0.2955
340
1.3923
162.82
0.3090
1.2150
162.66
0.3058
0.9669
162.33
0.3004
360
1.4292
166.72
0.3138
1.2476
166,56
0.3106
0.9932
166.25
0.3052
770 H Appendix f English Unit tables
TABLE F.9.2 (continued)
Superheated R-22
Temp.
V
h
s
V
h
V
h
s
F
ftVlbm
Btu/lbm
Btu/lbm R
ftVlbni
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/lbm R
125 psia (64.53)
150 psia (76.38)
175 psia (86.85)
Sat.
0.43988
110.04
0.2155
0.36587
110.83
0.2135
0.31224
111.45
0.2117
80
0.4622
112.99
0.2210
0.3705
111.56
0.2148
1 nn
0.4896
116.74
0.2279
0.3953
115.50
0.2220
0.3273
114.18
0.2167
120.
0.5158
120.46
0.2344
0.4187
119.37
0.2288
0.3488
118.22
0.2238
140
0,5411
124.15
0.2406
0.4410
123.18
0.2353
0.3691
122.17
0.2305
160
0.5656
127.84
0.2467
0.4624
126.97
0.2415
0.3884
126.07
0.2369
180
0.5896
131.53
0.2526
0.4832
130.74
0.2475
0.4070
129.94
0.2430
200
0.6130
135.23
0.2583
0,5034
134.52
0.2533
0.4250
133.79
0.2489
220
0.6360
138.96
0.2638
0.5232
138.31
0.2589
0.4426
137.64
0.2547
240
0.6587
142.71
0.2693
0.5426
142.11
0.2645
0.4597
141.49
0.2603
260
0.6810
146.48
0.2746
0.5618
145.93
0.2698
0.4765
145.36
0.2657
280
0.7032
150.29
0.2798
0.5806
149.78
0.2751
0.4930
149.25
0.2711
300
0.7251
154.13
0.2849
- 0.5993
153.65
0.2803
0.5094
153.16
0.2763
320
0.7468
158.00
0.2900
0.6177
157.55
0.2854
0.5255
157.10
0.2814
340
0.7683
161.91
0.2949
0.6360
161.49
0.2903
0.5414
161.06
0.2864
360
0.7898
165.85
0.2998
0.6541
165.46
0.2952
0.5572
165.06
0.2913
380
0.8111
169.83
0.3046
0.6721
169.46
0.3001
0.5728
169.08
0.2962
400
0.8322
173.85
0.3093
0.6900
173.50
0.3048
0.5884
173.14
0.3010
200 psia (96.27)
250 psia (112.76)
300 psia (126.98)
Sat.
0.27150
111.93
0.2102
0.21352
112.59
0.2074
0.17400
112.90
0.2049
100
0.2755
112.75
0.2116
120
0.2959
117.00
0.2191
0.2204
114.30
0.2104
140
0.3149
121.11
0.2261
0.2379
118.82
0.2180
0.1852
1 16.20
0.2104
160
0.3327
125.13
0.2327
0.2540
123.14
0.2251
0.2006
120.94
0.2182
180
0.3497
129.10
0.2390
0.2690
127.34
0.2318
0.2146
125.44
0.2253
200
0.3661
133.03
0.2450
0.2833
131.46
0.2381
0.2276
129.78
0.2320
220
0.3820
136.95
0.2509
0.2969
135.53
0.2442
0.2398
134.04
0.2384
240
0.3974
140.87
0.2566
0.3100
139.58
0.2501
0.2515
138.22
0.2445
260
0.4125
144.79
0.2621
0.3228
143.60
0.2558
0.2628
142.38
0.2503
280
0.4273
148.72
0.2675
0.3352
147.63
0.2613
0.2737
146.50
0.2560
300
0.4419
152.67
0.2727
0.3474
151.66
0.2667
0.2843
150.62
0.2615
320
0.4563
156.64
0.2779
0.3593
155.70
0.2719
0.2947
154.74
0.2668
340
0.4705
160.63
0.2830
0.3711
159.76
0.2770
0.3048
158.86
0.2720
360
0.4845
164.65
0.2879
0.3827
163.83
0.2821
0.3148
163.00
0.2771
380
0.4984
168.70
0.2928
0.3942
167.93
0.2870
0.3247
167.15
6.2821
400
0.5122
172.78
0.2976
0.4055
172.05
0.2919
0.3344
171.31
0.2870
420
0.5259
176.89
0.3023
0.4167
176.20
0.2966
0.3440
175.51
0.2919
Appendix F English Unit Tables m 771
TABLE F.9.2 (continued)
Superheated R-22
Temp.
V
A
5
D
s
V
h
s
F
ftVlbm
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/lbm R
ftVlbm
Btu/lbm
Btu/ibm R
400 psia (150.82)
500 psia (170.50)
600 psia (187.29)
Sat.
0.12297 112.70
0.2000
0.09053
1 1 1.38
0.1947
0.06663
1 AS3 <: 1
A f OOA
U. I SHU
160
ft 1 1 ft 1 ;
1 1 J.J/.
0.2046
180
171 ft 5
1Z1.UI
0.2133
0.0987
115.06
0.2005
200
ft 1 <;^7
U.J JO/
17£ ft7
izo. UZ
0.2210
0.1122
121.43
0.2103
0.0791
115.12
ft 1 QS 1
220
ft 1 fni
1 3ft 7£
0.2281
0.1232
126.95
0.2186
0.0919
122.32
240
ft 177R
0.2347
0.1328
132.05
0.2260
0.1020
128.29
0.2175
260
no 7£
0.2410
0.1416
136.89
0.2328
0.1106
133.70
ft 97S9
280
ft 1 qa^
\ AA I 3
144. t.J
0.2470
0.1498
141.57
0.2392
0.1184
138.79
0.2321
300
ft 7rK7
HO JC
14K.40
0.2527
0.1576
146.14
0.2453
0.1256
143.67
a 91 n
1 ^7 7/1
0.2583
0.1650
150.63
0.2512
0.1323
148.41
ft 9448
340
ft 77 1 Q
\ ^7 ft T
13 /.Ul
0.2637
0.1721
155.08
0.2568
0.1387
153.05
0.2507
360
ft T)OQ
u.zzyy
loi.Z/
0.2690
0,1789
159.49
0.2622
0.1449
157.63
ft 9561
380
ft 777Q
1 <;/t
0.2741
0.1856
163.88
0.2675
0.1508
162,16
0.26 1 8
400
ft 7^ J \ < ;
loy.e 1
0.2791
0.1921
168.26
0.2727
0.1566
166.66
ft 9671
420
ft 9S1A
1 7/1 fid
1 /4.uy
0.2841
0.1985
172.63
0.2777
0.1622
171.15
ft 7797
v.i / zz
440
0.2605
178.38
0.2889
0.2048
177.01
0.2826
0.1677
175.62
9771
460
0.2679
182.69
0.2936
0.2109
181.40
0.2875
0.1730
180.09
0.2822
480
0.2752
187.02
0.2983
0.2170
185.80
A 9Q97
0.1783
154.5/
0.2870
700 psia (201
88)
800 psia
900 psia
Sat.
0.04365
101.02
0.1750
220
0.0671
116.05
0.1975
0.0422
104.39
0.I78S
0.0305
95.43
0.1648
240
0.0788
123.79
0.2087
0.060O
118.02
0.1986
0.0434
109.91
0.1857
260
0.0879
130.09
0.2176
0.0702
125.88
0.2097
0.0557
120.85
0.2011
280
0.0956
135.73
0.2253
0.0782
132.34
0.2186
0.0643
128.53
0.2117
300
0.1025
141.01
0.2324
0.0851
138.13
0.2263
0.0713
135.01
0.2203
320
0.1089
146.05
0.2389
0.0913
143.54
0.2333
0.0774
140.87
0.2279
340
0.1149
150.93
0.2451
0.0970
148.70
0.2399
0.0830
146.36
0.2349
360
0.1206
155.69
0.2510
0.1023 •
153.69
0.2460
0.0881
151.60
0.2413
380
0.1260
160.39
0.2566
0.1074
158,56
0.2519
0.0929
156.67
0.2475
400
0.1312
165.02
0.2621
0.1122
163.34
0.2575
0.0975
161.62
0.2533
420
0.1363
169.62
0.2674
0.1169
168.07
0.2630
0.1018
166.49
0.2589
440
0.1412
174.20
0.2725
0.1214
172.76
0.2682
0.1060
171.29
0.2643
460
0.1460
178.76
0.2775
0.1258
177.41
0.2734
0.1101
176.05
0.2695
480
0.1507
183.31
0.2824
0.1301
182.05
0.2783
0.1141
180.77
0.2746
500
0.1553
187.87
0.2872
0.1343
186.67
0.2832
0.1179
185.47
0.2795
520
0.1599
192.42
0.2919
0.1384
191.30
0.2880
0.1217
190.17
0.2844
772 H Appendix F English Unit Tables
Table F.iO
Thermodynamic Properties ofR-134a
Table F.10.1
Saturated R-134a
SPECIFIC VOLUME, ff/lbm INTERNAL ENERGY, Btu/lbm
Temp.
Press.
Sat. Liquid
Evap.
Sat. Vapor
Sat. Liquid
Evap.
Sat. Vapor
F
(psia)
v f
»/
"fs
"s
-100
0.951
0.01077
39.5032
39.5139
50.47
94.15
144.62
-90
1.410
0.01083
27.3236
27.3345
52.03
93.89
145,92
-80
2.047
0.01091
19.2731
19.2840
53.96
93.27
147.24
-70
2.913
0.01101
13.8538
13.8648
■.: 56.19 .
92.38
148.57
-60
4.067
0.01111
10.1389
10.1501
.' . 58.64
91.26
149.91
-50
5.575
0.01122
: 7.5468
7.5580
61.27
89.99
151.26
-40
7.511
0.01134
5.7066
5.7179
64.04
88.58
152.62
-30
9.959
0.01 146
4.3785
4.3900
66.90
87.09
153.99
-20
13.009
0.01159
3.4049
3.4165
69.83
85.53
155.36
-15.3
14.696
0.01166
3.0350
3.0466
71.25
84.76
156.02
-10
16.760
0.01173
2,6805
2.6922
72.83
83.91
156.74
21.315
0.01187
2.1340
2.1458
75.88
82.24
158.12
10
26.787
0.01202
1.7162
1.7282
78.96
80.53
159.50
20
33.294
0.01218
1.3928
1.4050
82.09
78.78
160.87
30
40.962
0.01235
1.1398
1.1521
85.25
76.99
162.24
40
49.922
0.01253
0.9395
0.9520
88.45
75.16
163.60
50
60.311
0.01271
0.7794
0.7921
91.68
73.27
164.95
60
72.271
0.01291
0.6503
0.6632
94.95
71.32
166.28
70
85.954
0.01313
0.5451
0.5582
98.27
69.31
167.58
SO
101.515
0.01335
0.4588
0.4721
101.63
67.22
168.85
.90
119.115
0.01360
0.3873
0.4009
105.04
65.04
170.09
100
138.926
0.01387
0.3278
0.3416
108.51
62.77
171.28
110
161.122
0.01416
0.2777
0.2919
112.03
60.38
172.41
120
185.890
0.01448
0.2354
0.2499
115.62
57.85
173.48
130
213.425
0.01483
0.1993
0.2142
119.29
55.17
174.46
140
243.932
0.01523
0.1684
0.1836
123.04
52.30
175.34
150
277.630
0.01568
0.1415
0.1572
126.89
49.21
176.11
160
314.758
0.01620
0.1181
0.1343
130.86
45.85
176.71
170
355.578
0.01683
0.0974
•0.1142
134.99
42.12
177.11
180
400.392
0.01760
0.0787
0.0963
139.32
37.91
177.23
190
449.572
0.01862
0.0614
0.0801
143.97
32.94
176.90
200
503.624
0.02013
0.0444
0.0645
149.19
26.59
175.79
210
563.438
0.02334
0.0238
0.0471
156.18
16.17
172.34
214.1
589.953
0.03153
0.0315
164.65
o
164.65
Appendix F English Unit tables H 773
TABLE F.10.1 [continued)
Saturated R-l 3 4a
Enthalpy, Btu/lbm Entropy, Btu/lbm R
Temp. Press. Sat. Liquid Evap. Sat. Vapor Sat. Liquid Evap. Sat. Vapor
(F) (P sia ) h f "fg K s f s /g Sg
-100
0.951
50.47
101.10
151.57
0.1563
0.2811
0.4373
-90
1.410
52.04
101.02
153.05
0.1605
0.2733
0.4338
-80
2.047
53.97
100.58
154.54
0.1657
0.2649
0.4306
-70
2.913
56.19
99.85
156.04
0.1715
256?
-60
4.067
58.65
98.90
157.55
0.1777
0.2474
0.4251
-50
5.575
61,29
97.77
159.06
0.1842
0.2387
ft 4229
-40
7.511
64.05
96.52
160.57
0.1909
0.2300
0.4208
-30
9.959
66,92
95.16
162.08
0.1976
0.22 1 5
0.4191
-20
13.009
69.86
93.72
163.59
0.2044
0.2132
0.4175
-153
14.696
71.28
93.02
164.30
0.2076
0.2093
0.4169
-10
16.760
72.87
92.22
165.09
0.2111
0.2051
0.4162
21.315
75.92
90.66
166.58
0.2178
0.1972
ft 4t SO
10
26.787
79.02
89.04
168.06
0.2244
0. 1 896
0.4140
20
33.294
82.16
87.36
169.53
0.2310
0.1821
0.4132
30
40.962
85.34
85.63
170.98
0.2375
0.1749
0.4124
40
: 49.922
88.56
83.83
172.40
: 0.2440
0.1678
0.4118
50
60.311
; 91.82
81.97
173.79
0.2504
0.1608
0.4112
60
72.271
95.13
80.02
175.14
0.2568
0.1540
0.4108
70
85.954
98.48
77.98
176.46
0.2631
0.1472
0.4103
80
101.515
101.88
75.84
177.72
0.2694
0.1405
0.4099
90
119.115
105.34
73.58
178.92
0.2757
0.1339
0.4095
100
138.926
108.86
71.19
180.06
0.2819
0.1272
0.4091
110
161.122
112.46
68.66
181.11
0.2882
0.1205
0.4087
120
185.890
116.12
65.95
182.07
0.2945
0.1138
0.4082
130
213.425
119.88
63.04
182.92
0.3008
0.1069
0.4077
140
243.932
123.73
59.90
183.63
0.3071
0.0999
0.4070
150
277.630
127.70
56.49
184.18
0.3135
0.0926
0.4061
160
314.758
131,81
; 52.73
184.53
0.3200
0.0851
0.4051
170
355.578
136.09
48.53
184.63
0.3267
0.0771
0.4037
180
400.392
140.62
43.74 '
184.36
0.3336
0.0684
0.4020
190
449.572
145.52
38.05
183.56
0.3409
0.0586
0.3995
200
503.624
151.07
30.73
181.80
0.3491
0.0466
0.3957
210
563.438
158.61
18.65
177.26
0.3601
0.0278
0.3879
214.1
589.953
168.09
168.09
0.3740
0.3740
774 H Appendix F English Unit Tables
TABLE F.10.2
Superheated R-I34a
Temp.
v
tt
ft
s
V
u
It
<F)
(tf/lbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ftMbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
5 psia ( _
jj.ji )
15 psia (-14.44)
Sat.
8.3676
150.78
158.53
0.4236
2.9885
156.13
164,42
0.4168
-20
9.1149
156.03
164.47
0.4377
ci
u
1 Do. 1 1
j. iimj
158.58
167.19
ft AOIQ
20
9.9881
162.58
171.83
0.4537
3.2586
162.01
171.06
0.4311
40
10.4202
165.99
175.63
0.4615
3.4109
165.51
174.97
0.4391
60
10.8502
169.48
179.52
0.4691
3.5610
169.07
178.95
0.4469
80
11.2786
173.06
183.50
0.4766
3.7093
172.70
183.00
0.4545
100
11.7059
176.73
187.56
0.4840
3.8563
176.41
187.12
0.4620
120
12.1322
180.49
191.71
0.4913
4.0024
180.20
191.31
0.4694
140
12,5578
184.33
195.95
0.4985
4.1476
184.08
195.59
0.4767
160
12.9828
188.27
200.28
0.5056
4.2922
188.03
199.95
0.4838
180
13.4073
192.29
204.69
0.5126
4.4364
192.07
204.39
0.4909
200
13.8314
196.39
209.19
0.5195
4.5801
196.19
208.91
0.4978
220
14.2551
200.58
213.77
0.5263
4.7234
200,40
213.51
0.5047
240
14.6786
204.86
218.44
0.5331
4.8665
204.68
218.19
0.5115
260
15.1019
209.21
223.19
0.5398
5.0093
209.05
222.96
0.5182
280
15.5250
213.65
228.02
0.5464
5.1519
213.50
227.80
0.5248
300
15.9478
218.17
232.93
0.5530
5.2943
218,03
232.72 ■
0.5314
320
16.3706
222.78
237.92
0.5595
5.4365
222.64
237.73
0.5379
30psia(15.15)
40 psia (28.83)
Sat.
1.5517
160.21
168.82
0.4136
1.1787
162.08
170.81
0.4125
20
1.5725
161.09
169.82
0.4157
40
1.6559
164.73
173.93
0.4240
1.2157
164.18
173.18
0.4173
60
1.7367
168.41
178.05
0.4321
1.2796
167.95
177.42
0.4256
80
1.8155
172.14
182.21
0.4400
1.3413
171.74
181.67
0.4336
100
1.8929
175.92
186.43
0.4477
1.4015
175.57
185.95
0.4414
120
1.9691
179.77
190.70
0.4552
1.4604
179.46
190.27
0.4490
140
2.0445
183.68
195.03
0.4625
1.5184
183.41
194.65
0.4565
160
2.1192
187.68
199.44
0.4697
1.5757
187.43
199.09
0.4637
180
2.1933
191.74
203.92
0.4769 -
1.6324
191.52
203.60
0.4709
200
2.2670
195.89
208.48
0.4839
1.6886
195.69
208.18
0.4780
220
2.3403
200.12
213.11
0.4908
1.7444
199.93
212.84
0.4849
240
2.4133
204.42
217.82
0.4976
1.7999
204.24
217.57
0.4918
260
2.4860
208.80
222.61
0.5044
1.8552
208.64
222.37
0.4985
280
2.5585
213.27
227.47
0.5110
1.9102
213.11
227.25
0.5052
300
2.6309
217.81
232.41
0.5176
1.9650
217.66
232.20
0.5118
320
2.7030
222.42
237.43
0.5241
2.0196
222.28
237.23
0.5184
340
2.7750
227.12
242.53
0.5306
2.0741
226.99
242.34
0.5248
360
2.8469
231.89
247.70
0.5370
2.1285
231.76
247.52
0.5312
APPENDIX F ENGLISH Unit Tables m 775
TABLE F.10.2 {continued)
Superheated R-lS4a
Temp.
V
u
h
s
V
u
It
s
(T)
(ftVlbm)
(Btu/lbni)
(Btu/lbm)
(Btu/lbm R)
(tf/ibm)
(Btu/lbm)
(Btu/lbm)
(Btu/lb
60 psia (49.72)
80 psia (65.81)
Sat.
0.7961
164.91
173.75
0.4113
0.5996
167.04
175.91
0.4105
60
0.8204
166.95
176.06
0.4157
80
0.8657
170.89
180.51
0.4241
0.6262
169.97
179.24
U+T 1 uo
100
0.9091
174.85
184.94
0.4322
0.6617
174.06
183.86
0.4252
120
0.9510
178.82
189.38
0.4400
0.6954
178.15
188.44
0.4332
140
182.85
193.86
u./z/y
182.25
193.03
0,4410
-LOU
1 fti i s
186.92
198.38
U.455U
0.7595
186.39
197.64
0.4485
180
1 0719
191.06
202.95
a Afr>-\
U.4DiJ
u, /yui
190.58
202.28
0.4559
200
1.1 100
195.26
207.59
ft fl9A<
194.83
206.98
0.4632
220
1.1484
199.54
212.29
ft Alfid
U.6JUJ
199.14
211.72
0.4702
240
1.1865
203.88
217.05
ft dfi^
fl B9Q<
u.a /yo
203.51
216.53
0.4772
260
1.2243
208.30
221.89
9 400,9
ft oaqt
207.95
221.41
0.4841
280
1.2618
212.79
226.80
A 01*7<
u.yj / j
212.47
226.34
0.4909
300
1.2991
217.36
231.78
ft ^ni^
U. J U J J
a a*;*: 1
u.y&oi
217.05
231.35
0.4975
320
1.3362
222.00
236.83
ft t sni
A flA/K
u.yy45
221.71
236.43
0.5041
340
1.3732
226.71
241.96
U. J loo
1.0227
226.44
241.58
0.5107
360
1.4100
231.51
247.16
n <;9ia
U.JZJl/
i.U5U5
231.24
246.80
0.5171
380
1.4468
236.37
252.43
n ^90/i
1.U/8S
236.12
252.09
0.5235
400
1.4834
241.31
257.78
ft ^T?9
U.J J J /
1 1 A£T£
241.07
257.46
0.5298
100 psia (79.08)
125 psia (93.09)
Sat.
0.4794
168.74
177.61
0.4100
0.3814
170.46
179.28
0.4094
80
0.4809
168.93
177.83
0.4104
100
0.5122
173.20
182.68
0.4192
0.3910
172.01
181.06
0.4126
120
0.5414
177.42
187.44
0.4276
0.4171
176.43
186.08
0.4214
140
0.5691
181.62
192.15
0.4356
0.4413
180.77
190.98
0.4297
160
0.5957
185.84
196.86
0.4433
0.4642
185.10
195.84
0.4377
180
0.6215
190.08
201.58
0.4508
0.4861
189.43
200.68
0.4454
200
0.6466
194.38
206.34
0.4581
0.5073
193.79
205.52
0.4529
220
0.6712
198.72
211.15
0.4653
0.5278
198.19
210.40
0.4601
240
0.6954
203.13
216.00
0.4723
0.5480
202.64
215.32
0.4673
260
0.7193
207.60
220.91
0.4792
0.5677
207.15
220.28
0.4743
280
0.7429
212.14
225.88
0.4861
0.5872
211.72
225.30
0.4811
300
0.7663
216.74
230.92
0.4928
0.6064
216.35
230.38
0.4879
320
0.7895
221.42
236.03
0.4994
0.6254
221.05
235.51
0.4946
340
0.8125
226.16
241.20
0.5060
0.6442
225.81
240.71
0.5012
360
0.8353
230.98
246.44
0.5124
0.6629
230.65
245.98
0.5077
380
0.8580
235.87
251.75
0.5188
0.6814
235.56
251.32
0.5141
400
0.8806
240.83
257.13
0.5252
0.6998
240.53
256.72
0.5205
776 B appendix F English Unit Tables
TABLE F.10,2 {continued)
Superheated R-134a
Temp.
V
h
s
V
u
It
(F)
(ftVlbm)
(Btu/Jbm)
(Btu/Ibm)
(Btu/lbm R)
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/ibm R)
150 psia (105.13)
200 psia (125.25)
Sat.
0.3150
171.87
180.61
0.4089
0.2304
174.00
182.53
0.4080
120
0.3332
175.33
184,57
0.4159
140
0.3554
179.85
189.72
A A^l AiZ
0.4Z46
(\ f)A cn
0.2459
177.72
186.82
A A 1
0.4152
160
0.3761
184.31
194.75
0.4328
0,2645
182.54
192.33
0.4242
ISO
0,3955
188.74
199.72
0.44U/
A 1 A
0,26 14
187.23
197.64
0.4327
200
0.4141
193.18
204.67
A A A QA
0.44 S4
A 1AT1
0.2971
191.86
202.85
A A A cm
0,440/
220
0.4321
197.64
209.63
0.455 o
0.3120
196.46
208.01
f\ A AO A
0.4484
240
0,4496
202.14
214.62
0.4 oiO
0.3262
201.08
213.15
A A C CA
0,4559
260
0.4666
206.69
219.64
a Aim
0.4 /Ul
0.:S400
205.72
218.31
A AiZ'y 1
0.4oj 1
280
0.4833
211.29
224.70
0.4770
0.3534
210.40
223.48
0.4702
300
0.4998
215.95
229.82
A A 01 O
U.45JC
A ICC A
215.13
228.69
0.4772
320
0.5160
220.67
235.00
0.4906
0,3792
219.91
233.94
A A OA A
0.4 84 U
340
0.5320
225.46
240.23
0.4972
0.3918
224.74
239.24
0.4907
360
0.5479
230.32
245.52
0.5037
0.4042
229.64
244.60
0,4973
380
0.5636
235.24
250.88
0.5102
0.4 1 65
234.60
250.01
0,5038
400
0.5792
240.23
256.31
0.5166
0.4286
239.62
255.48
0.5103
250 psia (141.87)
300 psia (156.14)
Sat.
0.1783
175,50
183.75
0.4068
0.1428
176.50
184.43
0.4055
160
0.1955
180.42
189.46
0.4162
0.1467
177.70
185.84
0.4078
180
0.2117
185.49
195.28
0.4255
0.1637
183.44
192.53
0.4184
200
0.2261
190.38
200.84
0.4340
0.1779
188,71
198.59
0.4278
220
0.2394
195.18
206.26
0.4421
0.1905
193.77
204.35
0.4364
240
0.2519
199.94
211.60
0.4498
0.2020
198.72
209.93
0.4445
260
0.2638
204.70
216.90
0.4573
0.2128
203.62
215.43
0.4522
280
0.2752
209.47
222.21
0.4646
0.2230
208.50
220.88
0.4597
300
0.2863
214.27
227.52
0,4717
0.2328
213.39
226.31
0.4669
320
0.2971
219.12
232.86
0.4786
0.2423
218.30
231.75
0.4740
340
0.3076
224.01
238.24
0.4854
0.2515
223.25
237.21
0.4809
360
0.3180
228.95
243.66
0.4921
0.2605
228.24
242.70
0.4877
380
0.3282
233.95
249,13
0.4987
0.2693
233.29
248.24
0.4944
400
0.3382
239.01
254.65
0.5052
0.2779
238.38
253.81
0.5009
B 111
TABLE F.10.2 (continued)
Superheated R-134a
JTcinp*
tt
//
s
v
n
h
s
<F>
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
(ftVlbm)
(Btu/lbm)
(Btu/lbm)
(Btu/lbm R)
400 psia (179.92)
500 psia (199.36)
Sat.
0.0965
177.23
184.37
0.4020
0.0655
175.90
181.96
0.3960
180
0.0966
177.26
184.41
0.4020
—
—
200
0.1146
184.44
192.92
0.4152
0.0666
176.38
182.54
0.3969
220
0.1277
190.41
199.86
0.4255
0.0867
185.78
193.80
0.4137
240
0.1386
195.92
206.19
0.4347
0.0990
192.46
201.62
0.4251
260
0.1484
201.21
212.20
0.4432
0.1089
198.40
208.47
0.4347
280
0.1573
206.38
218,03
0.4512
0.1174
204.00
214.86
0.4435
300
0.1657
211.49
223.76
0.4588
0.1252
209.41
220.99
0.4517
320
0.1737
216.58
229.44
0.4662
0.1323
214.74
226.98
0.4594
340
0.1813
221.68
235.09
0.4733
0.1390
220.01
232.87
0.4669
360
0.1886
226.79
240.75
0.4803
0.1454
225.27
238.73
0.4741
380
0.1957
231.93
246.42
0.4S72
0.1516
230.53
244.56
0.4812
400
0.2027
237.12
252.12
0.4939
0.1575
235.82
250.39
0.4880
750 psia
1000 psia
ISO
0.01640
136.22
138.49
0.3285
0.01593
134.77
137.71
0.3262
200
0.01786
144.85
147.32
0.3421
0.01700
142.70
145.84
0.3387
220
0.02069
155.27
158.14
0.3583
0.01851
151.26
154.69
0.3519
240
0.03426
173.83
178.58
0.3879
0.02102
160.95
164.84
0.3666 .
260
0.05166
' 187.78
194.95
0.4110
0.02603
172.59
177.40
0.3843
280
0.06206
196.16
204,77
0.4244
0.03411
184.70
191.01
0.4029
300
0.06997
203.08
212.79
0.4351
0.04208
194.58
202.37
0.4181
320
0.07662
209.37
220,00
0.4445
0.04875
202.67
211.69
0.4302
340
0.08250
215.33
226.78
0.4531
0.05441
209.79
219.86
0.4406
360
0.08786
221.11
233.30
0.4611
0.05938
216.36
227.35
0.449S
380
0.09284
226.78
239.66
0.4688
0.06385
222.61
234.43
0.4583
400
0.09753
232.39
245.92 '
0.4762
0.06797
228.67
241.25
0.4664
778 S APPENDIX F ENGLISH UNIT TABLES
Table F.ll
Enthalpy of Formation and Absolute Entropy of Various Substances at 77 F, 1 atm Pressure
Substance
Formula
M
State
Btu/lbmol
s f
Btu/lbmol R
Water
H 2
18.015
gas
-103 966
45.076
water
H 2
18.015
liq
-122 885
16.707
Hydrogen peroxide
H 2 02
34.015
gas
-58 515
55.623
Ozone
Oj
47.998
gas
+ 61 339
57.042
Carbon (graphite)
/1
O
12.011
solid
1.371
Carbon monoxide
CO
28.011
gas
-47 518
47.182
Carbon dioxide
C0 2
44.010
gas
-169 184
51.038
Methane
CH 4
16.043
gas
-32 190
44.459
Acetylene
C,H 2
26.038
gas
+97 477
47.972
Ethene
C2H4
28.054
gas
+22.557
52.360
Ethane
Q>H 6
30.070
gas
-36 432
54.812
Propene
42.081
gas
+ 8 783
63.761
Propane
C 3 H S
44.094
gas
-44 669
64.442
Butane
C 4 H [0
58.124
gas
-54 256
73.215
Pentane
C5H12
72.151
gas
-62 984
83.318
Benzene
C6H 6
78.114
gas
+35 675
64.358
Hexane
86.178
gas
-71 926
92.641
Heptane
100.205
gas
-80 782
102.153
K-Octane
C S H 13
114.232
gas
-89 682
111.399
H-Octane
C 3 H 18
114.232
liq
-107 526
86.122
Methanol
CH 3 OH
32.042
gas
-86 543
57.227
Ethanol
C 2 HjOH
46.069
gas
-101 032
67.434
Ammonia
NH 3
17.031
gas
-19 656
45.969
T-T-Diesel
C[4.^jH 2 4 9
198.06
liq
-74 807
125.609
Sulfur
s
32.06
solid
7.656
Sulfur dioxide
S0 2
64.059
gas
-127 619
59.258
Sulfur trioxide
S0 3
80.058
gas
-170148
61.302
Nitrogen oxide
N 2
44.013
gas
+35 275
52.510
Nitromethane
CH3NO2
61.04
liq
-48 624
41.034
Answers to Selected
Problems
2.27 19.613 N; 0.102 kg
2.30 0.37%
2.33 4800 N; 3.82 s
2.36 2300N
2.39 1.28 kg/m 3
2.42 0.5 mVkg; 16 m 3 /kmol
2.45 198 kPa ■
2.48 24.38 m/s 1
2.51 a. 149.9 kPa
b. 162.8 kPa
c. 145.6 kPa
2.54 250 kPa; 99.6 kPa
2.60 199 kPa; 125.5 kPa
2.63 17.66 m
2.66 3.476 m
2.69 8.72 m
2.72 25.84 kPa
2.75 106.4 kPa
2.78 0.18%
2.81 23.94 kPa
2.84 370 kPa
2.93E 1 165 lbf; 3.42 s
2.96E lSftMbm^SOfWibmol
2.99E 16.86 Ibflin. 2
2.102E 141.3 lbfiW
2.105E 0.28 in.
3.21 7.38 MPa; 472 kg/m 3 ; 6.14 MPa; 275 kg/m 3
3.24 a. 1234 K
b. 1356 K
3.36 a. 0.2319mVkg
b. 0.000 733 mVkg
c. 0.090 58 mVkg
d. 0.4922 mVkg
e. 1.4153 m 3 /kg
3.39 a. 0.012 28 mVkg
b. 0.044 93 m 3 /kg
c. 0.005 563 mVkg
d. 0.013 58 mVkg
3.42 23.8 MPa
3.48 120.2°C; 0.05 m
3.51 18.9 kPa
3.54 771 kPa; 1318 kPa, 0.4678
3.57 0.948
3.60 1.318 MPa, 93.295 kg
3.63 900 kPa, 1.6263 m 3
3.69 2152 kPa
3.72 0.603 kg
3.75 2.75%; 9.4%; 25.3%
3.78 204.4 kPa
3.81 4.5%; 1.4%
3.84 356 kg/m 3
3.87 925 kPa, 0.007 m 3 /kg
3.90 a. 120°C; 0.516
b. 1 17 K; 0.959
3.93 a. 71.4°C
b. 858 kPa, 0.666
3.96 a. 99.9 kg
b. 149 kPa
3.99 1450 kPa
3.102 1554 kPa, 0.1 18
3.105 641°C
3.111 a. 372.5 K
b. 53 mm
3.114 60%; 1%
779
780 M Answers To Selected Problems
3.117 a. 6.TC, 0.1729 m 3 /kg
b. 1.9878 m 3 /kg
c. 336.8 kPa
3.120 0.9578 mVkg
3.13SE a. 1.752 fWlbm
b. 0.011 78 ftVlbm
c. 1.0669 ftVlbm
d. 21.564 fP/lbm
3.141E 70.36 lbm
3.144E 0.11 libra
3.147E 1966.88 lbm
3.150E 0.021 54 lbm
3.153E 2171bf/in. 2
3.156E ~8.1 F, 1.761 ftVlbm
4.21 9807 J
4.24 0,000 833 m 3 ; 0.0833 m, 0.0278 m
4.27 303.5 m
4.30 500 N; 0.05 m, 0.083 33 m; 25 J
4,33 40 kJ
4.36 -128.7 kJ
4.39 0.2375 kJ
4.42 a. 404°C
d. 163.35 kJ
4.45 0.0427 kJ
4.48 -80.4 kJ
4.51 1.969; -5I.SkJ/kg
4.54 1 17.5 kJ
4.57 43.2kJ/kg
4.60 -49.4 kJ
4.63 OkJ; -20.22 kJ
4.66 -38.64 kJ/kg
4.69 583.2 kJ
4.72 0.318m, 17.47J
4.75 5.89 X 10" 5 J
4. 84 186 W
4.87 98.8 km/b
4.90 1500 W
4.93 lkW
4.96 15.8°C
4.99 480 m 2
4.102 45°C
4.105 725°C
4.108 396 kPa; -26.7 kJ
4.111 -13.4 kJ
4.1 14 143.6°C; 0.4625 m 3 ; 145 kJ
4.123E 154.2 Btu; 12.85 Btu
4.126E 0.030 873 ft 3 ; 0.3087 ft, 0.1029 ft
4.129E -1 0.49 Btu
4.132E 3.33 Btu
4.135E 44.42 Btu
4.138E 2.94ftlbf
4.141E 17 325Btu/h
5.21 31 kJ
5.24 463 kJ, 1 17.7 kJ
5.27 9.9 m/s; -5.49 m
5.30 a. 1435 kJ/kg, 0.2645 mVkg
b. 1374.5 kJ/kg, 1.4153 mVkg
c. 1200 kPa, 1383 kJ/kg
5.33 a. 6S.l°C t 0.043 87 m 3 /kg, 208.1 kJ/kg
b. 680.7 kPa, 0.0289 m 3 /kg, 219.7 kJ/kg, 0.8287
c. 1017 kPa, 0.0177 m 3 /kg, 382.1 kJ/kg, 0.8788
5.36 133.6°C,0kJ, -1148 kJ
5.39 -263.3 kJ
5.42 -23.5 kJ
5.45 877.8 kJ, OkJ
5.48 287.7 m 3
5.51 179.9 a C, 1684 kJ
5.54 0.0327 m 3 ; 0.23 m 3 ; 605 kJ
5.57 803°C,587kJ
5.60 25°C
5.63 25 510kJ
5.66 -8395 kJ
5.69 a. 0.93 14 kg, 0.5798 m 3
b. 85.24 kJ
c. 588 kJ
5.72 191.3 kJ
'5.75 41.82 MJ
5.78 65.9°C
5.81 80°C
5.84 a. 397.2 kJ/kg
b. 490.1 kJ/kg
c. 485.3 kJ/kg
5.87 1. 1237.7 kJ/kg
b. 1839kJ/kg
c. 1767,8 kJ/kg
d. 1764.2 kJ/kg
5,90 188.3 kPa, 498,4 K
5.93 220.7 kJ
5.96 2900 K; 224 kJ
5.99 0.03 m 3 ; 1913 kPa; -83.5 kJ; -319.9 kJ
5.102 b. 1000 kPa, 1200 K
d. 32.3 kJ, 212 kj
5.105 -0.192 kJ; -0.072 kJ
5.108 668 K; 121.4 kJ/kg, 60.5 kJ/kg
5.111 1491 kPa; 41.5 kJ; 1024.9 kJ
5.114 27.3 kJ
5.117 4.2 min.
5.120 15.08 h
5.123 8°C
5.126 -3115 kJ, -32276kJ
5.129 51.2°C, 33.2°C, 0.1 m\ 359.8 kJ/kg
5.132 361 kPa; 2080 kJ; 60 kJ
5.135 a. 0.0132 m 3
b. 8.48 kJ
c. 129.8 kJ
d. 43.3 min.
5.138 OkJ; -15.7 kJ
5.I47E 22.28 Btu
5J50E a. 225 Ibffin. 2 , 0.2104 ftVlbm
b. 628.4 Btu/lbm, 6.457 ftVlbm
c. 55 lbf/in. 2 , 0.999 ftVlbm, 185.2 Btu/tbm
5.153E -223 Btu
5.156E 12 218 ft 3
5J59E a. 0.249
b. 662 Btu
5.1 62E 163 lbf/in. 2 , 633.5 Btu/lbm, Btu/lbm, 312.9
Btu/lbm
5.165E 337.4 Btu
5.168E -575.5 Btu
5.I71E 74.1 Btu/lbm, 288.3 Btu/lbm;
Btu/lbm, -214.2 Btu/lbm
5.I74E 122.1 Ibtfin. 2 , 825.9 R, -0.0139 Btu
5.177E 16.3 Btu, 34.6 Btu
5.180E 32.9 s
5.183E 4452 Btu
6.24 60 min.
6.27 6.36 m/s
answers To Selected Problems H 781
6.30 381.9 m/s
6.33 581.8 m/s
6.36 123.9kPa;319.7K
6.39 22.9°C; 215.7 kPa
6.42 -20°C;3679%
6.45 0.0579; 240 m/s
6.48 482.3 kJ/kg; 964.6 kW
6.51 664 m 3 /s
6.54 -9.9 kW
6.57 3.333 kg/s
6.60 a. 1.538
b. - 140 kJ/kg; 73.8 kJ/kg
6.63 1374.5 kJ/kg
6.66 0.866 kW; 0.99 kW
6.69 0.106 kg/s
6.72 0.98 kW
6.75 29.4 m/s; 2.31 kg/s
6.78 91.565 MW
6.81 131.2 m/s; 1056 kW
6.84 69 kg/s
6.87 1.815 kg/s
6.90 426.3 K
6.93 I19.6°C;3.0m 3 /s
6.96 0.964 kg/s
6.99 49 m/s; 24 800 kW
6.102 a. 2673.9 kJ/kg, 0.9755
b. 22.489 MW
c. 18.394 MW
d. 0.26
6.105 123 075 kg/h
6.108 16.96 kg; -468.9 kJ
6.111 520°C; 0.342 m 3
6.114 -379636kJ
6.117 8.90 kg; 25.459 MJ
6.120 0.903 m
6.123 270 390 kJ
6.126 8.743 kg/s
6.129 a. - 1 1 8.77 MW
b. -127.8 kW
6.132 773.7 K
6.135 0.52°C
6.138 1.79 m 1 ; 4.39 m 1
782 m answers To Selected problems
6.144E 3,0ft/s
6.147E 1.453 Btu/s; 4.255 Btu/s; 20.64 ftVs 2
6.150E 17.72 lbf/in. 2 ; 569.7 R
6.153E 7.571bm/h
6.156E 0.432 hp
6.I59E 0.264 Ibm/s
6.I62E 61 ibm/s
6.165E 1.904 Ibm/s
6.168E 4.75 X 10 7 Btu/h; 2.291 X 10 8 Btu/h
6.171E 2.667 ft 3 ; 15.8 Btu
6.174E 47 479 Btu
7.18 0.225
7.21 2.164
7.24 1.53 g/s; 42.9 kW
7.27 0.1875
7.30 26.8 kJ; 35.8 s
7.42 7.07
7.45 0.051
7.48 2.56 k\V
7.54 3 J; 0.000 33
7.57 5886.7 kJ; 0.0142
7.60 0,731
7.63 9.76 kW
7.75 38.4'C
7.78 378.4 kJ/kg; 126.1 kJ/kg; 0.667
7.81 27.2°C; I0.4°C
7.84 0.687
7.87. 153.1 kW
7.99 23.06 Btu; 32.4 s
*7.102E 85 100 Btu
7.108E 33.8 Btu/s
7.11 IE 0.58 Btu
7.1 14E 505 680 Btu; 0.28;
7.117E 1591 Btu
8.27 a. 2573.8 kJ/kg; 65.0°C; 0.981
b. 682°C; 0.7139 kJ/kg K
c. 0.7325 kJ/kgK
d. 0.045 06 ra 3 /kg; 1.3022 kJ/kgK
e. 1356.7 kJ/kg
8.33 1940 kJ/kg; 0.46 kJ
8.39 363.75 kJ; 396.97 kJ
8.42 7.11 kJ; 59.65 kJ
8.45 2000 kPa; 471.2 kJ
8.48 30°C; -31.6kJ/kg
8.51 -108.4U
8.54 172°C; -132 kJ
8.57 0.385 m 3
8.60 4910 kJ; 1290.3 kJ
8.63 312.2°C;0.225kJ/K
8.66 a. 3.662 kJ/K; 3.950 kJ/K; 12.929 kJ/K
8.69 -66.2 kJ; -824.1 kJ; 0.716 kJ/K
8.72 0.395 kJ/kgK
8.75 334.6 kJ/kg; 1.0091 kJ/kg K; 334.4 kJ/kg;
1.0086 kJ/kgK
8.78 2.57 kJ/K
8.81 8I 946kJ
8.84 0.202 kJ/K
8.87 660.8 kJ; 0.661 kJ/K
8.90 1.92 cm 3 ; 0.145 kJ/K
8.93 b. 11.48 cm 3 ; 405,3 K; 8.44 cm 3 ; 298 K
8.96 48.6 cm; 0.2935 kJ
8.99 0.315 kJ/K
8.102 6.52 kg
8.105 300 kPa; 400 K; 0.52 kJ/K
8.111 1.81 kJ; -0.96 kJ
8.1 14 1.538; -182.1 kJ; 147.3 U; 0.0936 kJ/K
8.117 -191.3 kJ/kg; -47.9 kJ/kg;
0.0374 kJ/kg
8.120 0.1 kW/K;0.1kW/K
8.123 0.05 W/K; 0.11 W/K; 0.168 W/K
8.129 3.33 kJ; 30.43 kJ; 9.04 kJ
8.132 12.18 kJ/K
8.135 a. 2.54 kJ
8.138 0.768 kJ/K
1.1 50E
b. 0.214; 0.9326
c. 7.995
i.I53E
100 Btu;
:.156E
2802.7 Btu; 543.3 Btu
I.159E
0.0645 Btu/R.
I.162E
0.0956 Btu/lbm R
M65E
0.1277 Btu/R
:.168E
12.392 atm; 0.1614 ft 3
;.171E
10.53 lbm
I.174E
69.4 Ibfiin. 2 ; -235.6 Btu/lbm;
-377 Btu/lbm
Answers to Selected Problems H 783
8.177E 1.305; 0.854 ft 5 ; -23.35 Btu;
-5.56 Btu; 0.002 26 Btu/R
8.180E 0.000 878 Btu/R s; 0.002 59 Btu/R s
9.21 349.7°C; 953.9 kJ/kg
9.24 -4.1 kW
9.27 259.1 K; 93.6 kPa
9.30 149.6 kPa; 322.9 K
9.33 69.29 kW; 69.29 kW
9.36 167.2 kW; 661.2 kW
9.39 a. 989.3 kJ/kg
b. 510 kPa
9.42 a. 1569 kJ/kg; 0.8433
b. -20.1 kJ/kg; 187.6 kJ/kg
c. 0.428
9.45 744.4 K
9.48 0; 1 87.2 kJ/kg; 0.1626 kJ/kg K
9.51 0.017 kW/K
9.54 47.29 kg/min; 8.94 kJ/K
9.60 0.952 kg/s; 4.05 kg/s; 0.852 kW/K
9.63 0.506 kJ/kgK
9.66 13 186 kJ; 12.37 kJ/K
9.69 1.076 X 10 s ; 680 K; -2.322 X 10 8 kJ
9.72 0.361 kg; 332.9°C; 0.186 kJ/K
9.75 0.466 kJ/K
9.78 132.2 k?a; 500 kW
9.81 14.08 m/s; 10.1m
9.84 -4.0 kW
9.87 17.6°C; 100.67 kPa
9.90 365.8 kW
9.93 -51.3kJ/kg; 135.8 kJ/kg; 0.117 kJ/kg K
9.96 0.854; 0.1492 kJ/kg K
9.99 49.7 kW
9.102 a. 1333.7 kJ/kg; 0.943
b. 23.7kJ/kg; 191.2 kJ/kg
c. 0.362
9.105 951 K; 0.1268 kJ/kgK
9.108 0.92; 0.028 kJ/kg K
9.111 -180.3 kJ/kg;55.9°C
9.114 -2.77 kW
9.117 28 1°C; 0.724 kW/K
9.120 1.867; 0.9639
9.123 0.157 m 3 ; 0.3469 kJ/K
9.126 a. 0.0952
. b. -0.6 kJ/kg; 328.1 kJ/kg
c. 0.3088 kJ/kgK
9.129 12.022 kg; 362.3 kJ/kg; 4140 kPa; -539.2 kJ; 4.423
kJ/K
9.132 a. 353.3 K
b. 35.9 kW; 19.2 kW
c. 0.0163 kW/K
9.144E 21.6 lbf/in. 2 ; 579.3 R
9.147E 59.59 Btu/s; 59.59 Btu/s
9.150E 12.2 Btu/R s
9.153E 917.8 Btu/s; 2.033 Btu/R s
9.156E 2.132 X 10 6 lbm; 1221 R;
182.8 lbf/in. 2 ; -2.158 X 10 s Btu
9.159E -3.53 hp
9. 1 62E 4S4.6 Btu/s; 94.8 Btu/s
9.165E 63.13 lbf/in 2 ; 1349.2 R
9.168E 55 lbf/in. 2 ; 137.3 F
9.1 7 IE 1.003 X 10 6 Btu; 77.2 Btu/R s
10.21 40 kW
10.24 -48.2kJ/kg
10.27 -0.504 kW
10.30 1483.9 kJ/kg; 1636.8 kJ/kg
10.33 120.3 kJ
10.36 419.9 kW
10.39 1269 kW
10.42 7444 kW
10.45 46.3°C; 19.8 kJ/kg
10.48 10.0 kg/s; 0.057
10.51 6.237 kJ/kg
10.54 1787.5 kJ/kg; 218.5 kJ/kg; 1.51 kJ/kg; 21.61 kJ/kg
10.57 271.96 kJ/kg
10.60 -36.1 kJ/kg
10.63 a. 500 W
b. 244 W
c.
10.66 300.7 K; -44.0 kJ
10.69 0.92
10.72 1.007 kg/s; 0.77
10.75 0.918; 0.871
10.78 55.3 kJ/kg; 0.91
10.81 0.403
10.84 377.5 kJ/kg
784 M ANSWERS TO SELECTED PROBLEMS
10.87 10 k\V f 5 k\V, 5 k\V; 6.25 kW, 5 kW, 0.194 kW
10.90 0.94
10.93 96.33 kJ
10.96 6.6 kJ; 271.4 kJ
10.99 702.4 kJ/kg; 764.8 kJ/kg; 0.319; 0.479
10.102 1.4 MPa; 1085.8 kJ; 1147.6 kJ
10.105 44.9°C; 125.4 kPa
10.114E -5.43 Btu/lbm; -19,3 Btu/lbm
I0.117E 541.9 R; 16 895 Btu
10.120E 31.68 Btu/s; 5467 Btu/s
10.123E 580.3 R; 8.7 Btu/lbra
10.126E 1.136 Btu/lbm
10.129E a. 500 W
b. 225 W
c.
10.132E 171.3 Btu
I0.135E 21.3 Btu/lbm; 0.946
10.138E 0.335; 0.70
10.I41E 157.2 Btu; 213.2 Btu
I0.144E 2102 ft/s; 0.95
11.21 0.323
11.27 0.103
11,30 15.43 kW
11.33 0.308
11.36 6487 kW; 16 475kW
11.39 0.3574; 0.059; 0.3587, 0.0913
11.42 0.1661; 1.0 kJ/kg; 4.5 kJ/kg
11.45 .0.357; 896.2 kJ/kg
11.48 7.257kg/s
U.51 0.191; 49.3 kW
11.54 0.31; 0.803
11.57 0.274
11.60 0.289; 0.117kJ/kgK
11.63 40.3°C; 29.185 MW; 11.606 MW
11.66 7424kW
11.69 3036 kW; 7320 kW; 0.484
11.72 1596.5 K; 26.66 kg/s
11.75 375 kPa; 442.2 kJ/kg; 0.339 kg/s;
958.8 K; 0.687
11.78 253 kJ/kg;27IkJ/kg
11.81 -163.3 kJ/kg; -133.2 kJ/kg;
-133.2 kJ/kg
1 1 .84 360.4 kPa; 425.7 kJ/kg; 0.352 kg/s; 975.2 K; 0.678
11.87 544.8 kPa; 1011 m/s
1 1 .90 530 kPa; 1229.8 K; 830 K; 957 m/s
11.93 2502 K; 6338 kPa
11.96 2677 K; 1458 kJ/kg; 1165K
11.99 7.67; 262 kJ/kg; 4883 kPa
11.102 7946 kPa; 1303.6 kJ/kg; 1055 kPa
11.105 7351 kPa; 2660 K; 0.578
11.108 6298 kPa; 550,5 kJ/kg; 0.653
11.111 20.92; 894.8 kPa
11.114 0.458
11.117 900 K; 429.9 kJ/kg; 15.6
11.120 3.198; 3.172
11.123 0.0397 kg/s
11.126 0.0403 kg/s; 5.21 kW; 0.26 kg/s
11.129 15.55 kW
11.132 0.0917 kW/K; 0.04 kW/K
11.135 1.015 kJ
11.138 0.369
11.141 26.584 kg/s; 133.6°C
11.144 1.843; 0.44
11.147 0.5657
11.150 0.765
11.153 a. 6600 kW; 125 kW
b. 22 579 kg/s; 23 432 kg/s
c. 0.033
11.156 0.368; 1033.4 kJ/kg
11.159 304.9 kPa; 323.2 kJ/kg; 0.464 kg/s; 886.8 K; 0.451
11.162 276 kPa; 529.6 kJ/kg; 0.536
11.165 a. 1.433
b. 1.032
11.168E 0.277
11.171E 0.104
11.174E 0.45; 0.7515
11.177E 12.656 Ibm/s
11.180E 0.284; 0.0153 Btu/lbm R
I1.I83E 2813 R; 63.985 Ibm/s
1 1.186E 61.9 Btu/lbm; 275 Btu/lbm; 0.775
11.189E 1033 lbflin. 2 ; 5789 R; 0.541;
188 lb£W
11.192E 1033 Ibfyin. 2 ; 5789 R; 0.488;
169.5 lbtfin. 2
1I.I95E 12.24; 0.5843; 139.8 lbtfm. 2
11.198E 0.458
Answers To Selected problems
U.201E 1620 R; 198.0 Btu/lbm; 18.1
1L204E 3.206
1 1 .207E 206.3 Btu/lbm; 528.9 Btu/lbm; 0.61
I1.210E 1.94 Btu/lbm; 67.76 Btu/lbm;
50.58 Btu/lbm
11.213E 0.357; 421.2 Btu/lbm
1 1.2I6E 242.7 Btu/lbm; 449.5 Btu/lbm; 0.427
12.21 0.5427, 0.2094, 0.2479; 0.3221 kJ/kg K; 5.065 m 3
12.24 0.381,0.180, 0.439; 0.096 45 kJ/kgK;
0.8298 kJ/kg K; 0.7334 kJ/kg K
12.27 0.2513 kJ/kg K; 1.005 m 3
12.30 72.586; 0.114 54kJ/kgK; 1.1655
12.33 1.675 m 3 ; 372.8 kj
12.36 334.5 K; 305.7 kPa
12.39 1096 kW
12.42 1247 kW
12.45 34.575 kJ/kmol; 679 K
12.48 2.863 kj; 20.16 kj
12.51 279 kPa; 418.6 K; —231 kJ
12.54 320.4 K; -46.16 kJ/kg
12.57 13 236 kPa; 0.933; -1595.7 kJ;
-I988kJ; -6.7656 kJ/K
12.60 304.7 K; 0. 1789 kJ/kg K
12.63 301.2 K
12.66 2831 kW
12.69 616 K; -0.339 kW/K
12.72 697.7 kPa; 3647.2 kJ; 5.4 kJ/K
12.75 0.56
12.78 0.387; 15.15 kW
12.81 0.007 73; 0.0155 kg/s;29.I°C
12.84 -3.055 kW
12.87 -5.15 kJ; 0.0123 kg; 155.9 kPa
12.90 0.0189; 0.0108; 48.6 kJ/kg
12.93 0.019; 0.0099 kg/kg air; 35 kJ/kg air
12.96 a. 25°C
b. 0.71; 0.0143
12.99 33.33 kg/s; 0,323 kg/s
12.102 18.4 kg/h; -4.21 kW
12.105 46°C, 12%; 1.165 kW
12.108 a. 25°C
b. 0.008,21.4 o C
c, 71%; 0.016
12.111 1 7%, 1 6 kJ/kg; 1 00%, - 1 5 kJ/kg
12.114 a. 20.6°C, 23.5°C
b. 0.0086, 20.2°C
c. 37%,27.2°C
d. 80%, 8.2°C
12.117 54.936 kW; 38.087 kW
12.120 361.3 K; -2.4 kJ
12.123 0.3857 kJ/K
12.126 SI 1.7 kg/s; 0.0163
12.129 0.001 05 kg/s; 4.537 kW; 25.0 kJ/kg; 42°C; 0.30
12.132 -3.48 kJ; 0.201 kJ/K
12.144E 72.586;21.285 ft ibf/lbmR; 1.1656
12.147E 1123.7 Btu/s
12.150E 38 lbf/in. 2 ; 565.1 R
12.153E 1184Btu/s
12.156E 0.836 Btu/R
12.159E 7.8 F; -1.49 Btu
12.162E -4.18 Btu; 0.0227 lbm; 20.76 lbf/in. 2
12.165E 1.235 Btu/s; -0.78 Btu/s
12.168E 0.123 Ibm/min; 0.04 lbm/min; 94 F; 9%
13.21 0.0098 kPa
13.27 4.247 kPa
13.30 - 150.6 kW
13.33 3.83 X I0 _2 Pa
13.45 2.44 kJ
13.48 5.23 X I0 _s l/kPa
13.51 HOOm/s; -66.7 J/kg
13.66 0.9743 kJ/kmol K
13.69 0.354; 13.51 kg
13.72 1.063 MPa; 0.002 38 kg; 0.753 kJ
13.75 -62.1 kJ/kg; -379 kJ/kg
13.78 -0.4703 R
13.81 5.0°C;4.7°C
13.84 0.471 kg; 36.0 kJ; 101.2 kJ
13.87 55.4 kJ
13.90 -1644 kJ
13.93 5.04
13.96 a. 0.044 m 3 ; 0.040 33 m 3
13.99 0.88; 0.638
13.102 -8309 kW
13.105 a. -7.71 kj; -7.71 kJ
b. -10.85 kj; -7.81 kJ
13.108 -981.4 kJ
786 u Answers To Selected Problems
13,111 62.57 kW
13.114 575 kPa
13.117 a. 254 K
b. 469 806 kJ
c. 259.1 K; -451 523 kJ
13.120E 0.2912 Ibf/in. 2 ; 132 ftVlbm
13.I23E 27.4 F
13.126E 384Sft/s
13.129E -123.9 Btu/lbmol
13.132E SI7.4R; 98.8 Btu
13.135E -78.4 Btu/lbm; -202.1 Btu/lbm
13.138E -42,1 Btu/lbm; -149.1 Btu/lbm
13.141E -704 Btu
13.144E -892 Btu
14.21 8.943 kg/kg; 14.28 kmot/kmol
14.24 0.80; 125%
14.27 824.1 kg; 23.765 kmol; 32.778 kmol
14.30 145%
14.33 9.444 kg/kg
14.36 0.718 kmoi/kmol gas
14.39 43.2°C; 0.0639 kg/kg fuel
14.42 2854 K
14.45 -I 234 557 kJ/kmol
14.48 -255 816kJ/kmol
14.51 -372 738 kJ/kmol
14.54 1.1123; -1 182 480 kJ
14.57 232 009 kJ/kmol
14.60 20 9S6kJ/kg
14.63 -627 058 kJ/kmol
14.66 1668kJ/m 3
14.69 13 101 kJ/kg fuel; 13 101 kj/kg fuel; 1216 K
14.72 72.6°C; 2524.5 K
14.75 1646 K
14.78 2048 K
14.81 a. 2909 K
b. 7400 K
14.84 5712 K
14.87 148.6%; 287 MJ/kmol fuel
14.90 2134 kJ/Kkmol fuel
14.93 a. 5.757
b. 1414.3 kJ/K
14.96 238.3 kPa; -1.613 X 10 s kJ; 4070 K
14.99 2039 K
14.102 4IO%;30.5°C
14.105 1.109 V
14.108 0.013 26; -149931 kJ
14.111 0.328; 0.414
14.114 238%
14.120 a. 2011 kPa; 666.4 K
b. 2907 K; 8772 kPa; 512.62 kJ/K
c. 152 860 kJ
14.126E 125.8 lbf/in. 2 ; -194 945 Btu
14.129E -369 746 Btu/lbmol;
-337 570 Btu/lbmol
14.132E 6.91; 9.689
I4.135E 79.85 Btu/ft 3
14.138E 3628 R
14.141E 3510 R
14.144E a. 5.07
b. 307.99 Btu/R
14.147E 34.9Btu/s; -67.45 Btu/s
14.150E 0.353; 0.419
15.21 34.36 MPa
15.24 29.682 MPa
15.27 exp(-12.8407)
15.30 2980 K
15.33 50.3% H
15.36 1444 K
15.39 1108 kPa; 93.7% 2 ; 97 681 kJ
15.42 0.0237
15.45 exp(-2,1665);exp(-2,4716)
15,48 exp(- 8.293)
15.51 69.1% NH 3 , 21.8% N 2 , 9.1% H 2
15.54 176 811 kJ
15.57 30.87% CH 4 ; 30.87% H 2 0, 28.7% H 2) 9.56% CO
15.60 66.21% H 2 0, 32.42% (^H,,
1.37%C2H 5 OH;41 330 kJ
15.63 a. 6.202
b. 85.44% NH 3} 10.92% H 2 , 3.64% N 2
15.69 75.904% N 2] 10.065% 2 , 7.764% H 2 0, 6.21 1% C0 2 ,
0.055% NO, 0.001% N0 2
1 5.72 66.1 1% H 2 0, 12.9 1 % H 2 , 9.9% OH, 5.68% H, 5.4%
2
Answers To Selected Problems S 787
15.75 0.009 67
15.81 16.92 N 2 , 3.3538 H 2 0, 2.6462 C0 2t 0.6462 H 2)
0.3538 CO
15.84 1.96
15.87 214 306k/kmolair
15.90E exp(-185.85);exp(+5.127)
15.93E 0.1015
15.96E 75 360BM
15-.99E a. 6.826
b. 85.91% NH 3t 10.57% H 2 ,
3.52% N 2
15.102E 66.24% H 2 0, 12.88% H 2l 9.87% OH, 5.63% H,
5.38% 2
15.105E 73.7%; 0.128%
INDEX
Note: Chapter 16 is available on the website:
www.wiley.com/college/sonntag and has been indexed
with "WI6" before the page number.
Absolute entropy, 5S7
Absolute temperature scale, 33, 231
Absorption refrigeration cycle, 441
Acentric factor, 725
Adiabatic compressibility, 525
Adiabatic flame temperature, 585
Adiabatic process, definition, 100
Adiabatic saturation process, 488
Aftercooler, 6, 194
Air, ideal gas properties, 135, 662, 723, 740
Air-conditioner, 6, 238
Air fuel ratio, 565
Air preheater, 609
Air-standard power cycles, 410
Air-standard refrigeration cycle, 442
Air-water mixtures, 480
Allotropic transformation, 49
Ammonia, properties, 692, 760
Ammonia-absorption cycle, 441
Appendix contents, 65 1
Atmosphere, standard, definition, 25
Availability, 355
Available energy, 343
Avogadros's number, endpapers
Back pressure, Wl 6- 1 7
Back work, 413
Bar, definition, 25
Benedict-Webb-Rubin equation of state, 68, 531, 726
Bernoulli equation, 314, W16-9
Binary cycle, 446
Binary mixtures, 541
Black body, 102
Boiler, steam, 2, 204, 205
Bottoming cycle, 446
Boyle temperature, 527
Brayton cycle, 411
British thermal unit, definition, 100, 117,653
Bulk modulus, 525
Calorie, 100, 653
Carbon dioxide, properties, 663, 666, 744
Carbon monoxide, properties, 666, 744
Carnot cycle, 227, 232
Cascade refrigeration, 447
Celsius Scale, 32
Centigrade Scale, 32
Chemical equilibrium, 624
Chemical potential, 541
Cheng cycle, 509,616
Choked flow, W 16- 17
Clapeyron equation, 511
Clausius, inequality of, 251
Clausius statement, 221
Coal, 571, 605
Coal gasifier, 564, 604
Coefficient of performance, 219, 436
Cogeneration, 409
Combined cycle, 446
Combustion, 564
Combustion efficiency, 599
Comfort zone, 492
Compressed liquid, 45, 52
Compressibility chart, 65, 528, 728
Compressibility factor, 65, 531, 728
Compressible flow, W16-1
Compression ratio, 426
Compressor, 178, 308
Concentration, 473
Condenser, 170, 194, 204, 330
Conduction, 101
Conservation of mass, 143, 162
Constant-pressure specific heat, 134, 659, 723, 740
Constant-volume specific heat, 134, 659, 740
Continuity equation, 164
Continuum, 16
Control mass, definition, 14
Conversion factors, 653
Control volume, definition, 14
Convection, 101
Cooling tower, 1,501,505
Crank angle, 426
Critical constants, 656, 738
789
790 u Index
Critical point, 46
Cycle, definition, 18
Dalton's model, 476
Dehumidified 482, 486, 502, 506
Density:
critical, 656, 738.W16-15
definition, 23
of solids and liquids, 24, 657, 739
Rackett equation, S3
Desuperheater, 310
Dew point, 480, 567
Diatomic molecule, 22, 723
Diesel cycle, 431
Diffuser, 174, 194
Diffiiser efficiency, W 16-27
Discharge coefficient, W16-27
Displacement, 426
Dissociation, 638
Drip pump, 402
Dry-bulb temperature, 490
Drying, 486, 505, 506
Dual cycle, 463
Economizer, 194
Efficiency:
combustion, 599
compressor, 321
cycle, 217
diffuser, W16-27
nozzle, 323, W16-26
pump, 325
regenerator, 420
Second-law, 355
steam generator, 599
v thermal, 217
turbine, 318
Electrical work, 98
Electromotive force, 596
Emissivity, 102
Energy:
available, 343
chemical, 585
electronic, 22
internal, 119,
kinetic, 119, 120
potential, 119, 120
total, 119
English engineering system of units, 20
Enthalpy:
of combustion, 581, 582
definition, 130
of evaporation, 132
of formation, 573
of ideal gas, 135, 660, 662, 664
stagnation, 167, W16-1
total, 167
Enthalpy chart, generalized, 533
Entropy:
absolute, 587
definition, 256
general comment, 323
generation, 266
of ideal gas, 273, 660, 662, 664
of mixing, 478
net change of, 268
principle of increase, 268, 316
of solids and liquids, 272
Entropy chart, generalized, 535
Equation of state:
Benedict-Webb-Rubin, 68, 531
ideal gas, 62
Lee-Kesler, 531,727
Peng-Robinson, 725
real gas, 527
Redlich-Kwong, 531,725
Soave, 725
van derWaals, 530, 725
virial, 528
Equilibrium:
chemical, 624
definition, 17, 617
mechanical, 17, 618
metastable, 623
phase, 619
requirements for, 619
thermal, 17
thermodynamic, 18
Equilibrium constant:
definition, 627
table of, 671
Equivalence ratio, 566
Ericsson cycle, 421
Evaporator, 194, 206, 436
Excess air, 566
Exergy, 356, 364
Expansion engine, 9, 443
Extensive property, 17
Extraction, 398
Fahrenheit temperature scale, 33
Fanno Iine,W16-21
Feedwater heater, 397 '
closed, 401
open, 398
First law of thermodynamics:
for a control volume, 162, 165
for a cycle, 117
for a control mass, 118
Flame temperature, 585
Flash evaporator, 194, 206, 337
Fourier's law, 101
Freon, —12, —22. See Refrigerants
Friction, 224
Fuel air ratio, 565
Fuel-celt, 2, 596, 598
Fuels, 564, 582
Fusion line, 48
Gas, ideal, 61
Gas constant, definition, 62
Gas constants, tables of, 658, 740
Gasoline, 563, 582
Gasoline engine, 427
Gas thermometer, 233
Gas turbine cycle, 41 1, 421
Gauge pressure, 27
Generalized charts:
compressibility, 65, 527, 728
enthalpy, 729
entropy, 730
low-pressure 528
Geothermat energy, 206
Gibbs function:
definition, 517
partial molal, 626
Gibbs relations, 264
Heat:
capacity. See Specific heat
definition, 100
of reaction, 581
Heat engine, 214, 236
Heat exchanger, 9, 164, 194
Heating value, 581
Heat pump, 214, 236
Heat transfer:
conduction, 101
convection, 102
radiation, 102
Heat transfer coefficient:
convection, 102
Heat transfer rate, 101
Helmholtz function, 517
Horsepower, definition, 86
Humidifier, 488, 501
Humidity:
- ratio, 481
relative, 481
specific, 481
Hydraulic line, 30
Hydrides, 646
Hydrocarbons, 562
Hydrogen fuel cell, 598
Hypothetical ideal gas, 543
Ice point, 32
Ideal gas:
definition, 61
enthalpy, 135
entropy, 273
internal energy, 135
mixtures of, 473
properties, 135, 660,741
temperature scale, 233
Incompressible liquid, 135
Increase of entropy, 268,
Inequality of Clausius, 25 1
Intensive property, 17
Inter-cooling, 329, 423
Internal combustion engine, 426
Internal energy, 119, 124
of combustion, 581
International temperature scale, 33
Irreversibility, 346
Isentropic efficiency, 318, 325
Isentropic process, definition, 260
Isobaric process, definition, 18
Isochoric process, definition, 18
Isolated system, 15, 269
Isothermal compressibility, 525
Isothermal process, definition, 18
Jet ejector, 337
Jet engine, 10, 197,424
Jet propulsion cycle, 424
Joule, definition, 86
Kalina cycle, 447
Kays rule, 545
Kelvin-Planck statement, 220
Kelvin temperature scale, 33, 231
Kinetic energy, 119, 120
Latent heat. See Enthalpy of evaporation
Lee-Kesler equation, 53 1, 727
Liquids, properties, 657, 739
Lost work, 267
792 m index
Mach number, W16-12
Macroscopic point of view, 15
Mass, 19
Mass conservation, 143, 163
Mass flow rate, 164
Mass fraction, 473
Maxwell relations, 516
Mean effective pressure, 41 1, 427
Mercury density, 39
Metastable equilibrium, 623
Methane properties, 718
Metric system, 20
Microscopic point of view, 16
Mixtures, 473,543
Moisture separator, 205
Mole, 19
Molecular mass, table of, 656, 738
Mole fraction, 473
Mollier diagram, 257
Momentum equation, W16-2
Monatomic gas, 22, 723
Multistage compression, 422, 456
Natural gas, 564
Newton, definition, 20
Newton's law of cooling, 102
Newton's second law, 20
Nitrogen, properties, 714
Nonideat mixtures, 540
Normal shock, W 16-20
table of functions, W16-36
Nozzle efficiency, 323, W 16-26
Nozzle flow, 172, Wl 6-16
table of functions, W16-35
Nuclear reactor, 2, 4, 205
Open feedwater heater, 398
Orifice 38, W16-28
Otto cycle, 427
Oxygen, P-h diagram, 734
Partial molal properties, 541
Partial pressure, 476
Pascal, definition, 25
Perpetual motion machine, 222
Phase, definition, 16
Physical constants, endpapers
Pinch point, 451
Pitot tube, W16-31
Polytropic exponent, 89, 279
Potytropic process, 89, 279, 280, 315
Potential energy, 119,120
Power plant, 2, 180, 204, 205, 384
Prefixes, endpapers
Pressure:
cooker, 75
critical, 46, 656, 738, W16-15
definition, 25
gauge, 27
mean effective, 411, 427
partial, 476, 541
reduced, 66
relative, 661
saturation, 44
Wagners correlation, 82
Process:
definition, 17
polytropic, 89, 279, 280
quasi-equilibrium, 18
reversible, 223
Properties, computerized, 69
Properties, independent, 5 1
Property relation, 263
Property, thermodynamic, definition, 17
Pseudocritical properties, 543
Pseudopure substance, 543
Psychrometric,chart, 491, 735
Pump:
efficiency of, 321, 325, 403
operation of, 178
reversible, 314
work, 314
Pure substance, definition, 44
Quality, definition, 45, 47
Quasi-equilibrium process, 1 8
Rackett equation, 83
Radiation, 102
Rankine cycle, 384
Rankine temperature scale, 33, 232
Ratio of specific heats, 278
Rayleigh line, W 16-21
Reactions, see chemical equilibrium
Redlich-Kwong equation of state, 531, 725
mixture, 545
Reduced properties, 66
Refrigerants:
R-12 tables, 698
R-22 tables, 702, 766
R-134a tables, 708, 772
Refrigeration cycles, 183, 435, 441
Regenerative cycle, 396
Regenerator, 418
Reheat cycle, 393
Relative humidity, 481
INDEX H 793
Relative pressure, 661
Relative volume, 661
Residual volume, 528
Reversible process, definition, 223
Reversible work, 314, 346, 351
Rocket engine, 1 1
Rotational energy, 22, 723
Saturated liquid, 45
Saturated vapor, 45
Saturation pressure, 44, 511, 538
Saturation temperature, 44, 5 1 1
Second law efficiency, 355, 358, 360
Second law of thermodynamics:
for control mass, 267
for control volume, 302
for cycle, 220, 254
Simple compressible substance, 44
Simultaneous reactions, 634
SI system of units, 20
Solids, properties, 657, 739
Sonic velocity, 525.W16-I4
Specific heat:
constant-pressure, 134
constant-volume, 134
equations, 659
of ideal gases, 136, 658, 659, 740
of solids and liquids, 657, 739
temperature dependency, 137, 658
thermodynamic relations, 263, 51 1
Specific humidity, 481
Specific volume, 23
Speed of sound, 525
Spring force, 92
Stagnation enthalpy, 167, W16-1
Stagnation pressure, 333, W16-2
Stagnation properties, W16-2
State of substance, 17
Steady-state process, 167, 313
Steam drum, 2
Steam generator, 2, 4, 194, 410
efficiency of, 599
Steam power plant, 2, 180, 204, 384
Steam tables, 51, 674,748
Steam turbine, 2, 304, 384
Stirling cycle, 433
Stoichiometeric coefficiencts, 565
Stoichiometeric mixture, 566
Stretched wire, 96
Subcooled liquid, 45
Sublimation, 48, 514
Supercharger, 322, 337
Supercritical Rankine cycle, 452
Superheated vapor, 45
Superheater, 194,200
Supersaturation, 623
Surface tension, 97
System definition, 14
Tank charging, 186
Temperature:
critical, 46, 545, 656, 738
equality, 3 1
reduced, 66
saturation, 44
thermodynamic scale, 32, 230
various scales, 32
Theoretical air, 565
Thermal efficiency, 217
Thermistor, 42
Thermodynamics, definition, 14
Thermodynamic property relation, 263, 511
Thermodynamic surface, 59
Thermodynamic tables, 51
development of, 538
Thermodynamic temperature scale, 230
Thermoelectric devices, 7
Third law of thermodynamics, 587
Throttling process, 174, 194
Thrust, W16-7
Topping cycle, 446
Torque, 86
Transient process, 183
Translation energy, 22
Trap (liquid), 401,402
Triple point, 48
Turbine:
adiabatic, 304,317
efficiency of, 317, 358
gas, 10,418
liquid, 332
operation of, 175
steam, 2, 358
Turbocharger, 328, 342
Units, 19, 653
Universal gas constant, 62
Unrestrained expansion, 225
Valve flow, 174, 194
Van der Waals equation of state, 530, 725
mixture, 545
Van't Hoff equation, 645
Vapor-compression refrigeration, 435
Vapor-pressure curve, 45, 49, 511, 538
Velocity of light, 143, endpapers
794 m Index
Velocity of sound, 525, W16-11
Velocity coefficient, W16-27
Vibrational energy, 22, 723
Virial coefficient, 528
Virial equation of state, 528
Volume:
critical, 46, 656
reduced, 53 1
relative, 661
residual, 528
saturated liquid correlation, 83
specific, 23
Volume expansivity, 524
Wagner's correlation, 82
Water, properties, 674, 748
Watt, definition, 86
Wet-bulb temperature, 490
Work:
definition, 84
flow, 166
nonequilibrium process, 99
reversible, 313, 343
various forces, 98
Zeroth iaw of thermodynamics, 3 1
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